9702Physics

AS Level Physics, one topic at a time

Notes for all eleven AS topics of the Cambridge 9702 syllabus, with the key equations, worked examples, simulations you can play with, and exam-style questions. Multiple-choice questions mark themselves; structured questions come with a mark scheme written the way examiners write them.

Paper 1: Multiple Choice

1 h 15 min40 marks31% of AS

40 four-option questions on the AS topics.

Paper 2: AS Structured Questions

1 h 15 min60 marks46% of AS

Short-answer and structured questions on the AS topics.

Paper 3: Advanced Practical Skills

2 h40 marks23% of AS

Practical work in the lab: measuring, recording, graphing and evaluating.

Topics

How the marks work

9702 mark schemes label every mark with a code. The mark schemes on this site use the same codes, so you learn what examiners are looking for.

CodeMeaning
B1An independent mark for a correct fact, statement or value, given regardless of anything else.
M1A method mark for a correct method, such as substituting into the right equation.
A1An accuracy mark for the correct answer. It depends on the method mark being earned.
C1A compensation mark: given for a correct intermediate step if your final answer is wrong. A correct final answer earns all the C1 marks automatically.
Always show your working

If your final answer is wrong, the C1 and M1 marks for your working can still earn you most of the credit. A bare wrong number earns nothing.

Data you are given in the exam

These constants are printed at the start of Papers 1, 2 and 4, so you don't need to memorise their values.

QuantityValue
acceleration of free fallg = 9.81 m s−2
speed of light in free spacec = 3.00 × 108 m s−1
elementary chargee = 1.60 × 10−19 C
unified atomic mass unit1 u = 1.66 × 10−27 kg
rest mass of protonmp = 1.67 × 10−27 kg
rest mass of electronme = 9.11 × 10−31 kg
Avogadro constantNA = 6.02 × 1023 mol−1
molar gas constantR = 8.31 J K−1 mol−1
Boltzmann constantk = 1.38 × 10−23 J K−1
gravitational constantG = 6.67 × 10−11 N m2 kg−2
permittivity of free spaceε0 = 8.85 × 10−12 F m−1, and 14πε0 = 8.99 × 109 m F−1
Planck constanth = 6.63 × 10−34 J s
Stefan–Boltzmann constantσ = 5.67 × 10−8 W m−2 K−4

Each topic lists its key equations. Those marked On formulae page are printed on the paper; those marked Learn it are not, so you must memorise them.

About the questions

The questions here are written in the style of 9702 papers, testing the same skills. They are not copies of official papers. For genuine past papers and mark schemes, use the Cambridge School Support Hub or ask your teacher.

Every measurement in physics is a number with a unit and an uncertainty. This topic sets up the tools used everywhere else: SI base units, checking equations, estimating, combining uncertainties, and working with vectors.

Physical quantities and estimates

A physical quantity has a numerical magnitude and a unit, such as 9.81 m s−2. Paper 1 often asks you to pick a sensible estimate, so learn some reference values:

QuantityTypical valueQuantityTypical value
mass of an adult70 kgwalking speed1.5 m s−1
mass of an apple0.1 kgspeed of sound in air330 m s−1
height of a room3 mpower of a kettle2 kW
diameter of an atom10−10 matmospheric pressure1 × 105 Pa
wavelength of visible light400 to 700 nmdensity of water1000 kg m−3

SI units

All units are built from six SI base units used in this course:

Base quantityUnitBase quantityUnit
masskilogram, kgelectric currentampere, A
lengthmetre, mthermodynamic temperaturekelvin, K
timesecond, samount of substancemole, mol

Every other unit is a derived unit. To express one in base units, start from a defining equation:

UnitFromIn base units
newton, NF = makg m s−2
joule, JW = Fskg m2 s−2
watt, WP = W/tkg m2 s−3
pascal, Pap = F/Akg m−1 s−2
coulomb, CQ = ItA s
volt, VV = W/Qkg m2 s−3 A−1
ohm, ΩR = V/Ikg m2 s−3 A−2

Homogeneity: an equation is homogeneous if every term has the same base units. A homogeneous equation is not necessarily correct, because pure numbers such as ½ have no units, but an equation that is not homogeneous must be wrong.

Prefixes

pico pnano nmicro μmilli mcenti cdeci dkilo kmega Mgiga Gtera T
10−1210−910−610−310−210−11031061091012

Errors and uncertainties

Systematic error
Makes every reading too high or too low by the same amount or proportion, such as a zero error on a meter or a stretched tape. Repeating readings does not reduce it; it is removed by correcting the instrument or the method.
Random error
Makes readings scatter unpredictably either side of the true value, such as reaction time when using a stopwatch. Reduced by repeating readings and averaging.
Accuracy
How close a measurement is to the true value. Affected by systematic error.
Precision
How close repeated measurements are to each other (small scatter). Affected by random error.

If a length is x = 25.0 ± 0.5 cm, the absolute uncertainty is 0.5 cm, the fractional uncertainty is 0.5/25.0 = 0.02, and the percentage uncertainty is 2%.

CalculationRule
Adding or subtracting, y = a ± bAdd the absolute uncertainties: Δy = Δa + Δb
Multiplying or dividing, y = ab or a/bAdd the fractional (or percentage) uncertainties: Δyy = Δaa + Δbb
Powers, y = anMultiply the fractional uncertainty by the power: Δyy = nΔaa

Worked example

A block has mass m = 25.0 ± 0.1 g and volume V = 10.0 ± 0.2 cm3. Find its density with its uncertainty.

  1. ρ = m/V = 25.0/10.0 = 2.50 g cm−3
  2. Percentage uncertainties: m: 0.4%, V: 2.0%. Total = 2.4%.
  3. 2.4% of 2.50 = 0.06 g cm−3

ρ = 2.50 ± 0.06 g cm−3

Exam tipGive the uncertainty to one (or at most two) significant figures, and round the value to the same decimal place as the uncertainty.

Scalars and vectors

A scalar has magnitude only (mass, speed, distance, energy, time, temperature). A vector has magnitude and direction (displacement, velocity, acceleration, force, momentum, weight).

To add vectors, draw them tip to tail; the resultant goes from the start of the first to the end of the last. For two perpendicular vectors, use Pythagoras and trigonometry. To resolve a vector F at angle θ to the horizontal into two perpendicular components:

θ F F cos θ F sin θ
Horizontal componentFx = Fcos θLearn it
Vertical componentFy = Fsin θLearn it

Exam-style questions

1Paper 1 style

Which of these is an SI base unit?

  1. newton
  2. coulomb
  3. kelvin
  4. volt

The six base units in the course are kg, m, s, A, K and mol. The newton, coulomb and volt are derived units.

2Paper 1 style

The volume of a cylinder is V = πr2l. The radius is measured with an uncertainty of 2% and the length with an uncertainty of 1%. What is the percentage uncertainty in V?

  1. 2%
  2. 3%
  3. 5%
  4. 6%

Because r is squared, its percentage uncertainty is doubled: 2 × 2% + 1% = 5%. The constant π has no uncertainty.

3Paper 1 style

Which estimate is the most reasonable?

  1. The mass of an apple is 1 kg.
  2. The power of an electric kettle is 2 kW.
  3. The speed of a sprinter is 40 m s−1.
  4. The thickness of a sheet of paper is 1 mm.

Kettles are rated at about 2 to 3 kW. An apple is about 0.1 kg, a sprinter runs at about 10 m s−1, and paper is about 0.1 mm thick.

4Paper 2 style
(a)
Express the volt in SI base units.
[2]
(b)
Show that the equation Δp = ρgΔh is homogeneous.
[2]
  • V = W/Q (or P/I), with J = kg m2 s−2 and C = A s
  • kg m2 s−3 A−1
  • Units of ρgΔh: kg m−3 × m s−2 × m = kg m−1 s−2
  • Units of Δp: Pa = N m−2 = kg m s−2 m−2 = kg m−1 s−2, so both sides match
5Paper 2 style

The resistivity of a wire is found using ρ = Rπd24L. The measurements are: resistance R = 2.4 ± 0.1 Ω, diameter d = 0.50 ± 0.01 mm, length L = 1.200 ± 0.005 m.

(a)
Calculate the resistivity.
[2]
(b)
Determine the absolute uncertainty in the resistivity, and state the resistivity with its uncertainty.
[3]
  • ρ = 2.4 × π × (0.50 × 10−3)24 × 1.200
  • ρ = 3.9 × 10−7 Ω m (3.93 × 10−7)
  • Percentage uncertainties: R 4.2%, d 2 × 2.0% = 4.0%, L 0.4%
  • Total ≈ 8.6%, so Δρ = 0.086 × 3.93 × 10−7 = 0.34 × 10−7 Ω m
  • ρ = (3.9 ± 0.3) × 10−7 Ω m

The uncertainty in d counts twice because d is squared.

6Paper 2 style
(a)
A rope pulls a sledge with a force of 50 N at 30° above the horizontal. Calculate the horizontal and vertical components of the force.
[2]
(b)
A boat heads straight across a river at 4.0 m s−1 relative to the water. The river flows at 3.0 m s−1. Calculate the magnitude of the boat's resultant velocity and its angle to the direction the boat is heading.
[3]
  • Horizontal: 50 cos 30° = 43 N
  • Vertical: 50 sin 30° = 25 N
  • v = √4.02 + 3.02
  • v = 5.0 m s−1
  • θ = tan −1(3.0/4.0) = 37° downstream of the heading
7Paper 2 style
(a)
Distinguish between systematic errors and random errors.
[2]
(b)
Distinguish between precision and accuracy.
[2]
  • Systematic error: readings are all above (or all below) the true value by the same amount or proportion
  • Random error: readings are scattered about the true value (in either direction)
  • Precision: how close the readings are to each other (small spread)
  • Accuracy: how close a reading (or the mean) is to the true value

Kinematics describes motion without asking what causes it. You need precise definitions, the meaning of the gradient and area of motion graphs, the four equations of uniformly accelerated motion, and projectile motion.

Definitions

Distance
The total length of path travelled. A scalar.
Displacement
The distance in a straight line from the starting point, in a stated direction. A vector.
Speed
Distance travelled per unit time. A scalar.
Velocity
The rate of change of displacement. A vector.
Acceleration
The rate of change of velocity. A vector, in m s−2.

Motion graphs

  • The gradient of a displacement–time graph is the velocity.
  • The gradient of a velocity–time graph is the acceleration.
  • The area under a velocity–time graph is the displacement.
u v t time velocity area = displacement gradient = a
The four equations of motion follow from this graph: the gradient gives v = u + at, and the area of the trapezium gives s = 12(u + v)t.

Equations of uniformly accelerated motion

These apply only when the acceleration is constant and the motion is in a straight line. Choose a positive direction and keep to it: a deceleration is a negative acceleration.

Velocity after time tv = u + atLearn it
Displacement from average velocitys = 12(u + v)tLearn it
Displacement after time ts = ut + 12at2On formulae page
Without timev2 = u2 + 2asOn formulae page

Worked example

A car accelerates uniformly from 12 m s−1 to 30 m s−1 in 6.0 s. Find its acceleration and the distance travelled.

  1. a = (v − u)/t = (30 − 12)/6.0 = 3.0 m s−2
  2. s = 12(u + v)t = 12(12 + 30) × 6.0 = 126 m

a = 3.0 m s−2, s = 130 m (to 2 s.f.)

Free fall and measuring g

Near the Earth's surface, with no air resistance, every object falls with the same acceleration g = 9.81 m s−2.

Experiment: drop a steel ball from rest through a measured height h, timing the fall with an electromagnet release and a trapdoor (or with light gates) connected to an electronic timer. Repeat for several heights. Since h = 12gt2, plot h against t2: it is a straight line through the origin with gradient g/2, so g = 2 × gradient. Using a graph also reveals a systematic delay in the release (an intercept).

Projectile motion

Without air resistance, the horizontal and vertical motions are independent:

  • Horizontally: no force, so constant velocity ucos θ. Use s = vt.
  • Vertically: constant acceleration g downwards, starting at usin θ. Use the equations of motion.
  • The time is the same for both, and links them.

Worked example

A ball is thrown horizontally at 15 m s−1 from the top of a cliff 20 m high. How far from the foot of the cliff does it land?

  1. Vertically: u = 0, s = 20 m, a = 9.81 m s−2, so 20 = 12(9.81)t2 and t = 2.02 s
  2. Horizontally: distance = 15 × 2.02 = 30.3 m

The ball lands 30 m from the foot of the cliff.

With air resistance, the projectile reaches a lower maximum height and a shorter range, and its path is no longer symmetrical: the descent is steeper than the ascent.

Exam-style questions

1Paper 1 style

A stone is dropped from rest and falls freely for 3.0 s. Air resistance is negligible. How far does it fall?

  1. 14.7 m
  2. 29.4 m
  3. 44.1 m
  4. 88.3 m

s = ut + 12at2 = 0 + 12 × 9.81 × 3.02 = 44.1 m. Option B is the speed (29.4 m s−1), a common slip.

2Paper 1 style

A projectile is launched at an angle to the horizontal. Air resistance is negligible. Which statement about its motion at the highest point is correct?

  1. Its velocity is zero.
  2. Its acceleration is zero.
  3. Its velocity and acceleration are both zero.
  4. Its velocity is horizontal and its acceleration is g downwards.

Only the vertical component of velocity is zero at the top; the horizontal component is unchanged. The acceleration is g downwards throughout the flight.

3Paper 2 style

A car travelling at 25 m s−1 brakes with constant deceleration and stops in a distance of 50 m.

(a)
Calculate the deceleration.
[2]
(b)
Calculate the braking time.
[1]
(c)
The driver's reaction time before braking is 0.70 s. Calculate the total stopping distance.
[2]
  • 0 = 252 + 2a(50)
  • deceleration = 6.25 m s−2 (6.3)
  • t = 25/6.25 = 4.0 s
  • thinking distance = 25 × 0.70 = 17.5 m
  • total = 17.5 + 50 = 67.5 m (68 m)
4Paper 2 style

A ball is kicked from level ground with a velocity of 18 m s−1 at 35° above the horizontal. Air resistance is negligible.

(a)
Calculate the horizontal and vertical components of the initial velocity.
[2]
(b)
Calculate the maximum height reached.
[2]
(c)
Calculate the horizontal distance travelled before the ball lands.
[3]
  • horizontal = 18 cos 35° = 14.7 m s−1
  • vertical = 18 sin 35° = 10.3 m s−1
  • 0 = 10.32 − 2(9.81)h
  • h = 5.4 m
  • time to top = 10.3/9.81 = 1.05 s, so time of flight = 2.10 s
  • distance = 14.7 × 2.10
  • = 31 m

Check this with the projectile simulation: set 18 m s⁻¹ and 35°.

5Paper 2 style
Describe an experiment to determine the acceleration of free fall g using a falling object. Include how the measurements are used to find g.
[4]
  • Object released from rest; height of fall h measured with a metre rule
  • Time of fall t measured with light gates or an electromagnet and trapdoor connected to an electronic timer
  • Repeat for different heights (and repeat each to find a mean time)
  • Plot h against t2; the gradient is g/2, so g = 2 × gradient

Dynamics explains why motion changes. The central idea is momentum: a resultant force changes it, and in a collision it is conserved. Air resistance adds the idea of terminal velocity.

Mass, weight and momentum

Mass
The property of an object that resists changes in its motion (its inertia). Measured in kg, the same everywhere.
Weight
The gravitational force on an object: W = mg. It depends on the gravitational field strength.
Linear momentum
The product of mass and velocity, p = mv. A vector, in kg m s−1 or N s.

Newton's laws of motion

  1. First law: an object stays at rest or moves with constant velocity unless a resultant force acts on it.
  2. Second law: the resultant force on an object is proportional to its rate of change of momentum, and acts in the same direction: F = ΔpΔt. For constant mass this becomes F = ma.
  3. Third law: if body A exerts a force on body B, then B exerts a force on A that is equal in magnitude and opposite in direction. The two forces are of the same type and act on different bodies.
Momentump = mvLearn it
Newton's second lawF = ΔpΔtLearn it
Constant massF = maLearn it
WeightW = mgLearn it
Third-law pairs

The weight of a book on a table and the table's contact force on it are not a third-law pair: both act on the book. The pair to the book's weight is the book's gravitational pull on the Earth.

Non-uniform motion: drag and terminal velocity

Drag (air resistance or a viscous force in a liquid) acts opposite to the velocity and increases with speed. For an object falling from rest:

  1. At first the speed is zero, so there is no drag; the resultant force is the weight and the acceleration is g.
  2. As speed increases, drag increases, so the resultant force (weight − drag) and the acceleration decrease.
  3. Eventually drag equals weight. The resultant force is zero, so the object moves at a constant terminal velocity.
terminalvelocity initial gradient = g time velocity

Conservation of momentum

Principle of conservation of momentum: for a system of interacting objects, the total momentum in any direction stays constant, provided no resultant external force acts on the system.

  • In every collision, momentum is conserved (and total energy is conserved).
  • In a perfectly elastic collision, kinetic energy is also conserved. In this case the relative speed of approach equals the relative speed of separation.
  • In an inelastic collision, some kinetic energy becomes other forms (thermal energy, sound, deformation).
  • In two dimensions, momentum is conserved separately in each of two perpendicular directions.

Worked example

A trolley of mass 2.0 kg moving at 3.0 m s−1 hits a stationary trolley of mass 1.0 kg. They stick together. Find their common velocity and the kinetic energy lost.

  1. Momentum before = 2.0 × 3.0 = 6.0 kg m s−1 = momentum after = 3.0 v, so v = 2.0 m s−1
  2. KE before = ½ × 2.0 × 3.0² = 9.0 J; KE after = ½ × 3.0 × 2.0² = 6.0 J

v = 2.0 m s−1; 3.0 J of kinetic energy is transferred to other forms, so the collision is inelastic.

Exam-style questions

1Paper 1 style

Which unit is equivalent to the unit of momentum?

  1. N s−1
  2. N s
  3. kg m s−2
  4. J s

From F = Δp/Δt, momentum has the unit of force × time, N s, which equals kg m s−1.

2Paper 1 style

A ball of mass 0.20 kg hits a wall at 5.0 m s−1 and rebounds along the same line at 3.0 m s−1. It is in contact with the wall for 0.040 s. What is the average force on the ball?

  1. 10 N
  2. 20 N
  3. 40 N
  4. 50 N

Velocity changes from +5.0 to −3.0 m s−1, a change of 8.0 m s−1. Δp = 0.20 × 8.0 = 1.6 N s, and F = 1.6/0.040 = 40 N. Option B comes from forgetting that the direction reverses.

3Paper 2 style
(a)
State Newton's second law of motion.
[1]
(b)
A skydiver jumps from a stationary balloon. Describe and explain the variation of her velocity from the moment she jumps until she reaches terminal velocity.
[4]
  • (a) The resultant force is (proportional to) the rate of change of momentum (and in the same direction)
  • (b) Initially the only force is her weight, so her acceleration is g
  • As her speed increases, air resistance increases
  • So the resultant force decreases and her acceleration decreases
  • When air resistance equals weight, the resultant force is zero and she moves at constant (terminal) velocity
4Paper 2 style

Sphere A of mass 0.30 kg moves at 4.0 m s−1 and collides head-on with a stationary sphere B of mass 0.50 kg. After the collision, A moves backwards along its original line at 1.0 m s−1.

(a)
State the principle of conservation of momentum.
[2]
(b)
Calculate the velocity of B after the collision.
[2]
(c)
Show that the collision is perfectly elastic.
[2]
  • (a) The total momentum of a system remains constant
  • (a) provided no resultant external force acts on the system
  • (b) 0.30 × 4.0 = 0.30 × (−1.0) + 0.50v
  • (b) v = 3.0 m s−1 in A's original direction
  • (c) Relative speed of approach = 4.0 m s−1; relative speed of separation = 3.0 − (−1.0) = 4.0 m s−1
  • (c) They are equal, so the collision is elastic (or: KE before = 2.4 J = KE after, 0.15 + 2.25 J)

Forces can make things turn as well as accelerate. This topic covers moments and couples, the two conditions for equilibrium, and pressure in fluids, including why things float.

Types of force

  • Weight: acts through the centre of gravity, the point where the whole weight of an object appears to act.
  • Normal contact (reaction) force: perpendicular to a surface.
  • Friction: acts along a surface, opposing relative motion.
  • Drag (viscous force, air resistance): opposes motion through a fluid and increases with speed.
  • Upthrust: the upward force a fluid exerts on an object in it.

Turning effects

Moment of a force
The product of the force and the perpendicular distance from the pivot to the line of action of the force. Unit: N m.
Couple
A pair of equal and opposite forces whose lines of action are parallel but do not coincide. A couple produces rotation only, with no resultant force.
Torque of a couple
The product of one of the forces and the perpendicular distance between their lines of action.
Principle of moments
For an object in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point.
Moment (d = perpendicular distance to pivot)moment = F × dLearn it
Torque of a couple (d = separation of the forces)torque = F × dLearn it

Equilibrium

An object is in equilibrium when both conditions hold:

  1. the resultant force is zero (in every direction), and
  2. the resultant moment (torque) about any point is zero.

If three forces keep an object in equilibrium, they can be drawn tip to tail to form a closed triangle (a triangle of forces).

Worked example

A uniform plank 4.0 m long and of weight 200 N rests on supports at its ends A and B. A person of weight 600 N stands 1.0 m from A. Find the force from each support.

  1. Take moments about A (this removes the unknown force at A): RB × 4.0 = 600 × 1.0 + 200 × 2.0
  2. RB = 1000/4.0 = 250 N
  3. Vertical forces balance: RA = 600 + 200 − 250 = 550 N

RA = 550 N, RB = 250 N

Exam tipTake moments about the point where an unknown force acts. That force then has zero moment and drops out of the equation.

Density and pressure

Densityρ = mVLearn it
Pressurep = FALearn it
Hydrostatic pressureΔp = ρgΔhOn formulae page
UpthrustF = ρgVOn formulae page

Pressure is the normal force per unit area, in pascals (Pa = N m−2). In a fluid, pressure increases with depth: Δp = ρgΔh, where ρ is the density of the fluid. The total pressure at a depth also includes the atmospheric pressure on the surface.

Upthrust arises because the pressure on the bottom surface of a submerged object is greater than on its top surface (the bottom is deeper). The resulting upward force equals the weight of fluid displaced, F = ρgV, where ρ is the fluid's density and V the volume submerged (Archimedes' principle). A floating object displaces its own weight of fluid.

Exam-style questions

1Paper 1 style

A diver is 12 m below the surface of water of density 1000 kg m−3. What is the pressure on the diver due to the water alone?

  1. 1.2 × 104 Pa
  2. 1.2 × 105 Pa
  3. 2.2 × 105 Pa
  4. 1.2 × 106 Pa

Δp = ρgΔh = 1000 × 9.81 × 12 = 1.18 × 105 Pa. Option C adds atmospheric pressure, which the question excludes.

2Paper 1 style

A steering wheel is turned by two forces of 15 N, equal and opposite, acting at opposite ends of a diameter of 0.40 m. What is the torque?

  1. 3.0 N m
  2. 6.0 N m
  3. 12 N m
  4. 30 N m

Torque of a couple = one force × perpendicular separation = 15 × 0.40 = 6.0 N m.

3Paper 2 style
(a)
State the two conditions for an object to be in equilibrium.
[2]
(b)
A uniform beam of length 3.0 m and weight 120 N is pivoted at one end, P, and held horizontal by a vertical string attached 2.5 m from P. A load of 80 N hangs from the other end of the beam. Calculate the tension in the string.
[3]
  • (a) Resultant force is zero
  • (a) Resultant moment (torque) about any point is zero
  • (b) Moments about P: T × 2.5 = 120 × 1.5 + 80 × 3.0
  • (b) 2.5T = 180 + 240 = 420
  • (b) T = 168 N (170 N)
4Paper 2 style
(a)
Explain the origin of the upthrust on an object immersed in a liquid.
[2]
(b)
A wooden block of density 600 kg m−3 and volume 5.0 × 10−4 m3 floats in water of density 1000 kg m−3. Calculate the weight of the block and the volume of the block below the water surface.
[3]
  • (a) Pressure in a liquid increases with depth
  • (a) So the (upward) pressure on the bottom face is greater than the (downward) pressure on the top face, giving a resultant upward force
  • (b) Weight = 600 × 5.0 × 10−4 × 9.81 = 2.9 N (2.94)
  • (b) Floating, so upthrust = weight: 1000 × 9.81 × V = 2.94
  • (b) V = 3.0 × 10−4 m3 (60% of the block)
5Paper 2 style
A lamp of weight 40 N hangs at rest from two cables. Each cable makes an angle of 30° with the horizontal. Calculate the tension in each cable.
[2]
  • Vertical components balance the weight: 2Tsin 30° = 40
  • T = 40 N

Energy is never created or destroyed, only transferred. Work is how a force transfers energy, power is how fast it happens, and efficiency tells you how much of it is useful.

Work done

Work done by a force is the product of the force and the displacement in the direction of the force. One joule is the work done when a force of 1 N moves its point of application 1 m in the direction of the force. If the force is at angle θ to the displacement, W = Fscos θ.

A gas expanding against a constant external pressure p does work W = pΔV, where ΔV is the increase in volume.

Energy conservation and efficiency

Principle of conservation of energy: energy cannot be created or destroyed; it can only be transferred from one form to another. The total energy of a closed system is constant.

Work doneW = FsLearn it
Work done by an expanding gasW = pΔVLearn it
PowerP = WtLearn it
Power at constant velocityP = FvLearn it
Efficiencyefficiency = useful energy outputtotal energy input × 100%Learn it

Power is the rate of doing work (or transferring energy), in watts: 1 W = 1 J s−1. When a force F moves an object at constant velocity v, P = Fs/t = Fv.

Worked example

A car's engine delivers a useful power of 30 kW while the car travels at a constant 25 m s−1. What is the total resistive force on the car?

  1. At constant velocity, driving force = resistive force
  2. F = P/v = 30 000/25 = 1200 N

The resistive force is 1200 N.

Potential and kinetic energy

Change in gravitational potential energyΔEp = mgΔhLearn it
Kinetic energyEk = 12mv2Learn it

Gravitational potential energy is the energy an object has because of its position in a gravitational field. Kinetic energy is the energy it has because of its motion. You may be asked to derive both:

  • ΔEp = mgΔh: lifting a mass at constant speed needs a force equal to its weight, mg. The work done is force × distance = mgΔh, and this equals the gain in potential energy. It applies only in a uniform field, near the Earth's surface.
  • Ek = 12mv2: a constant force F accelerates a mass from rest through distance s. Work done W = Fs = mas. From v2 = u2 + 2as with u = 0, as = 12v2, so W = 12mv2, the kinetic energy gained.

Exam-style questions

1Paper 1 style

A crane lifts a load of mass 500 kg vertically at a constant speed of 0.40 m s−1. What useful power does the crane provide?

  1. 200 W
  2. 2.0 kW
  3. 4.9 kW
  4. 20 kW

At constant speed the lifting force equals the weight, 500 × 9.81 = 4905 N. P = Fv = 4905 × 0.40 = 1962 W ≈ 2.0 kW.

2Paper 1 style

A ball is dropped from rest from a height of 5.0 m. Air resistance is negligible. What is its speed just before it hits the ground?

  1. 7.0 m s−1
  2. 9.9 m s−1
  3. 49 m s−1
  4. 98 m s−1

Loss of Ep = gain of Ek: mgh = 12mv2, so v = √2gh = √2 × 9.81 × 5.0 = 9.9 m s−1. The mass cancels.

3Paper 2 style
Use the definition of work done and an equation of motion to show that the kinetic energy of a mass m moving at speed v is 12mv2.
[3]
  • Work done by the force: W = Fs and F = ma, so W = mas
  • From v2 = u2 + 2as with u = 0: as = 12v2
  • So W = 12mv2, which equals the kinetic energy gained
4Paper 2 style

A roller-coaster car of mass 600 kg starts from rest and descends a vertical height of 35 m along a track of length 120 m. At the bottom its speed is 24 m s−1.

(a)
Calculate the loss in gravitational potential energy.
[1]
(b)
Calculate the gain in kinetic energy.
[1]
(c)
Calculate the average resistive force on the car along the track.
[3]
  • (a) 600 × 9.81 × 35 = 2.06 × 105 J
  • (b) 12 × 600 × 242 = 1.73 × 105 J
  • (c) Work done against resistance = 2.06 × 105 − 1.73 × 105 = 3.3 × 104 J
  • (c) F × 120 = 3.3 × 104
  • (c) F = 280 N
5Paper 2 style
An electric motor lifts a mass of 20 kg vertically through 3.0 m in 4.0 s at constant speed. The electrical power input to the motor is 250 W. Calculate the efficiency of the motor.
[3]
  • Useful work = 20 × 9.81 × 3.0 = 589 J
  • Useful power = 589/4.0 = 147 W (or energy input = 250 × 4.0 = 1000 J)
  • Efficiency = 147/250 = 0.59, or 59%

When forces stretch or squash a solid, it changes shape. Stress and strain describe this independently of the sample's size, and the Young modulus is the property of the material that links them.

Hooke's law

Tensile forces stretch an object; compressive forces squash it. Hooke's law: the extension of a spring (or wire) is proportional to the applied force, provided the limit of proportionality is not exceeded.

Hooke's law (k = spring constant)F = kxLearn it
Stressσ = FALearn it
Strainε = xLLearn it
Young modulusE = σεLearn it
Elastic potential energyEp = 12Fx = 12kx2Learn it

Stress, strain and the Young modulus

Stress
The force per unit cross-sectional area, σ = F/A. Unit: Pa.
Strain
The extension per unit original length, ε = x/L. No unit.
Young modulus
Stress divided by strain, for a material obeying Hooke's law. Unit: Pa. A large Young modulus means a stiff material.

Worked example

A steel wire is 2.0 m long with diameter 0.50 mm. The Young modulus of steel is 2.0 × 1011 Pa. Find the extension produced by a load of 50 N.

  1. A = π(0.25 × 10−3)2 = 1.96 × 10−7 m2
  2. σ = 50/1.96 × 10−7 = 2.55 × 108 Pa
  3. ε = σ/E = 1.27 × 10−3, so x = εL = 2.5 × 10−3 m

The extension is 2.5 mm.

Measuring the Young modulus of a metal wire

  1. Use a long, thin wire (so the extension is measurable), clamped at one end and passing over a pulley, with a marker attached near the loaded end against a fixed millimetre scale. (Alternatively, use a vernier on a Searle's apparatus with a reference wire.)
  2. Measure the original length L from the clamp to the marker with a metre rule.
  3. Measure the diameter at several places with a micrometer screw gauge and take the mean; calculate A = πd2/4.
  4. Add masses in steps, recording the extension x for each load F. Check by removing the loads that the wire returns to its original length (it stayed elastic).
  5. Plot F against x. In the straight-line region, E = gradient × LA.

Elastic and plastic behaviour

limit of proportionality elastic limit area = ½Fx extension x force F
Up to the elastic limit the deformation is elastic. Beyond it, the deformation is plastic. The area under a force–extension graph is the work done in stretching (the elastic potential energy, if the deformation is elastic).
Elastic deformation
The object returns to its original shape and size when the load is removed.
Plastic deformation
The object is permanently deformed; it does not return to its original shape when the load is removed.
Limit of proportionality
The point beyond which force is no longer proportional to extension.
Elastic limit
The point beyond which the deformation becomes plastic.

Elastic potential energy (strain energy) is the energy stored in a deformed object. It equals the area under the force–extension graph. In the Hooke's law region this area is a triangle: Ep = 12Fx = 12kx2.

Exam-style questions

1Paper 1 style

What is the SI unit of the Young modulus?

  1. N m−1
  2. N m−2
  3. N m
  4. It has no unit.

Young modulus = stress/strain. Strain has no unit, so the unit is that of stress, N m−2 (Pa). N m−1 is the unit of the spring constant.

2Paper 1 style

A spring with spring constant 200 N m−1 is stretched elastically by 5.0 cm. How much elastic potential energy is stored?

  1. 0.25 J
  2. 0.50 J
  3. 5.0 J
  4. 10 J

Ep = 12kx2 = 12 × 200 × 0.0502 = 0.25 J. Remember to convert cm to m.

3Paper 2 style

A copper wire has length 1.8 m and cross-sectional area 4.0 × 10−7 m2. A tensile force of 60 N extends it by 2.0 mm. The wire obeys Hooke's law.

(a)
Calculate the stress and the strain in the wire.
[2]
(b)
Calculate the Young modulus of copper.
[1]
(c)
Calculate the elastic potential energy stored in the wire.
[2]
  • (a) stress = 60/4.0 × 10−7 = 1.5 × 108 Pa
  • (a) strain = 2.0 × 10−3/1.8 = 1.1 × 10−3
  • (b) E = 1.5 × 108/1.11 × 10−3 = 1.35 × 1011 Pa
  • (c) Ep = 12Fx = 12 × 60 × 2.0 × 10−3
  • (c) Ep = 0.060 J
4Paper 2 style
Describe how to determine the Young modulus of a metal in the form of a wire. Include the measurements, the instruments used, and how the result is calculated.
[5]
  • Long wire clamped at one end, loaded over a pulley (or hanging vertically), with a marker against a scale
  • Original length measured with a metre rule (or tape)
  • Diameter measured with a micrometer at several points; area = πd2/4
  • Extension measured with the scale (or vernier) for increasing loads F = mg
  • Plot F against x (or stress against strain); E = gradient × L/A (or the gradient of stress–strain)

A progressive wave transfers energy from one place to another without transferring matter. This topic covers the vocabulary of waves, the wave equation, the Doppler effect, the electromagnetic spectrum and polarisation.

Describing a wave

amplitude A wavelength λ distance
Displacement
The distance of a point on the wave from its equilibrium position.
Amplitude
The maximum displacement from the equilibrium position.
Wavelength
The distance between two adjacent points on the wave that are in phase (such as two crests).
Period
The time for one complete oscillation of a point on the wave.
Frequency
The number of oscillations per unit time. Unit: hertz (Hz).
Phase difference
How far one oscillation is ahead of another, as a fraction of a cycle, in degrees or radians. Points one wavelength apart are in phase; points half a wavelength apart are in antiphase (180°).
Wave speed
The distance travelled by the wave energy per unit time.
Intensity
The power transmitted per unit area at right angles to the wave direction. Intensity ∝ amplitude2.
Wave equationv = fλLearn it
Frequency and periodf = 1TLearn it
IntensityI = PA, and I ∝ A2Learn it

Deriving v = fλ: in one period T the wave moves forward one wavelength, so v = λ/T = fλ.

Using an oscilloscope: the time-base setting (such as 0.5 ms per division) converts horizontal divisions into time. Measure the number of divisions for one cycle to find T, then f = 1/T. The vertical (Y-gain) setting converts divisions into voltage, giving the amplitude.

Transverse and longitudinal waves

TransverseLongitudinal
The particles (or fields) oscillate perpendicular to the direction of energy transfer.The particles oscillate parallel to the direction of energy transfer, forming compressions and rarefactions.
Examples: all electromagnetic waves, waves on a string, water ripples.Examples: sound, ultrasound, some seismic (P) waves.
Can be polarised.Cannot be polarised.

The Doppler effect

When a source of sound moves relative to a stationary observer, the observed frequency differs from the emitted frequency. As the source approaches, the wavefronts in front of it are squashed together (shorter wavelength), so the observer hears a higher frequency; as it recedes, the wavefronts are spread out and the frequency is lower.

Doppler effect (source moving; − when approaching, + when receding)fo = fsvv ± vsOn formulae page

Worked example

An ambulance siren emits 700 Hz and approaches a stationary observer at 30 m s−1. The speed of sound is 340 m s−1. What frequency is heard as it approaches, and after it has passed?

  1. Approaching: fo = 700 × 340/(340 − 30) = 768 Hz
  2. Receding: fo = 700 × 340/(340 + 30) = 643 Hz

The electromagnetic spectrum

All electromagnetic waves are transverse, can travel through a vacuum, and travel at the same speed in a vacuum, c = 3.00 × 108 m s−1. They differ in wavelength and frequency.

RegionApproximate wavelength
Radio waveslonger than about 10−1 m
Microwaves10−3 m to 10−1 m
Infrared7 × 10−7 m to 10−3 m
Visible light400 nm (violet) to 700 nm (red)
Ultraviolet10−8 m to 4 × 10−7 m
X-rays10−12 m to 10−8 m
Gamma raysshorter than about 10−11 m (overlaps with X-rays)

Polarisation

In plane-polarised light, the oscillations are in one plane only. Only transverse waves can be polarised. When plane-polarised light of intensity I0 passes through a polarising filter whose transmission axis is at angle θ to the plane of polarisation, the transmitted intensity is given by Malus's law:

Malus's lawI = I0cos 2θLearn it

Exam-style questions

1Paper 1 style

Plane-polarised light passes through a polarising filter whose axis is at 60° to the plane of polarisation. What fraction of the intensity is transmitted?

  1. 0.25
  2. 0.50
  3. 0.75
  4. 0.87

I/I0 = cos 260° = 0.52 = 0.25. Option B forgets to square the cosine.

2Paper 1 style

Which waves cannot be polarised?

  1. light waves
  2. microwaves
  3. sound waves in air
  4. radio waves

Sound in air is longitudinal. Only transverse waves, such as all electromagnetic waves, can be polarised.

3Paper 1 style

The amplitude of a wave is doubled. Its frequency is unchanged. By what factor does its intensity change?

  1. ½
  2. √2
  3. 2
  4. 4

Intensity is proportional to amplitude squared, so it increases by a factor of 22 = 4.

4Paper 2 style

A wave on a stretched string has wavelength 0.80 m and travels at 240 m s−1.

(a)
Calculate the frequency and the period of the wave.
[2]
(b)
Calculate the phase difference between two points on the string 0.20 m apart along the direction of travel.
[1]
  • (a) f = 240/0.80 = 300 Hz
  • (a) T = 1/300 = 3.3 × 10−3 s
  • (b) 0.20/0.80 = ¼ of a wavelength, so 90° (π/2 rad)
5Paper 2 style
(a)
A sound wave is displayed on an oscilloscope with a time-base setting of 0.50 ms per division. One complete cycle occupies 4.0 divisions. Calculate the frequency of the sound.
[2]
(b)
A car horn emits sound of frequency 400 Hz. The car moves directly away from a stationary observer at 25 m s−1. The speed of sound is 340 m s−1. Calculate the frequency heard by the observer.
[2]
  • (a) T = 4.0 × 0.50 = 2.0 ms
  • (a) f = 1/2.0 × 10−3 = 500 Hz
  • (b) fo = 400 × 340/(340 + 25)
  • (b) fo = 373 Hz (370 Hz)

When two waves meet, their displacements add. This principle of superposition explains stationary waves, interference fringes and the sharp bright lines of a diffraction grating.

The principle of superposition

When two or more waves meet at a point, the resultant displacement is the sum of the individual displacements at that point.

Stationary waves

A stationary (standing) wave forms when two progressive waves of the same frequency (and similar amplitude), travelling at the same speed in opposite directions, superpose. Usually one wave is a reflection of the other.

  • Nodes are points of zero amplitude; antinodes are points of maximum amplitude.
  • The distance between adjacent nodes (or adjacent antinodes) is λ/2.
  • All points between two adjacent nodes oscillate in phase; points either side of a node are in antiphase.
  • A stationary wave stores energy but does not transfer it.
Progressive waveStationary wave
EnergyTransferredNot transferred
AmplitudeSame for all pointsVaries from zero at nodes to maximum at antinodes
PhaseChanges continuously along the waveSame between adjacent nodes; reverses across a node

Experiments

  • Stretched string: a string is driven by a vibration generator and passes over a pulley to a hanging mass. Adjust the frequency until a stationary wave forms. Measure the node-to-node distance with a metre rule, so λ = 2 × node separation and v = fλ.
  • Air column: a tuning fork of known frequency is held over a tube closed at one end (for example by water). Change the length of the air column until the sound is loudest (resonance). The first two resonant lengths differ by λ/2, so λ = 2(L2 − L1) and the speed of sound v = fλ. Using the difference removes the end correction.
  • Microwaves: a microwave transmitter faces a metal reflector. A detector moved along the line between them finds maxima and minima; adjacent minima (nodes) are λ/2 apart.

Diffraction

Diffraction is the spreading of a wave as it passes through a gap or around the edge of an obstacle. It is most noticeable when the gap width is similar to the wavelength. The wavelength and speed do not change.

Interference

Two sources are coherent if they have a constant phase difference (which requires the same frequency). Interference fringes can only be observed with coherent sources. For two sources in phase:

  • Constructive interference (maximum) where the path difference is a whole number of wavelengths, nλ.
  • Destructive interference (minimum) where the path difference is (n + 12)λ.

Two-source interference can be shown with water waves in a ripple tank, with microwaves from one transmitter passing through two slits, and with light through a double slit (Young's experiment).

Double-slit fringe spacing (a = slit separation, D = slit-to-screen distance)λ = axDLearn it
Diffraction grating (d = line spacing)dsin θ = nλLearn it

Worked example

In a double-slit experiment the slits are 0.50 mm apart and the screen is 1.20 m away. The fringe separation is 1.4 mm. Find the wavelength.

  1. λ = ax/D = (0.50 × 10−3)(1.4 × 10−3)/1.20

λ = 5.8 × 10−7 m (580 nm)

The diffraction grating

A grating has a very large number of equally spaced parallel slits. Light diffracts through each slit, and the waves interfere constructively only at precise angles where the path difference between adjacent slits is a whole number of wavelengths, giving sharp, bright maxima. If the grating has N lines per metre, d = 1/N. The highest order seen is the largest integer n for which nλ/d ≤ 1.

Exam-style questions

1Paper 1 style

What is the distance between two adjacent nodes in a stationary wave of wavelength λ?

  1. λ/4
  2. λ/2
  3. λ
  4. 2λ

Adjacent nodes are half a wavelength apart. A node and the next antinode are λ/4 apart.

2Paper 1 style

Two coherent sources emit waves in phase. At which path difference is there destructive interference?

  1. λ
  2. 2λ
  3. 1.5λ
  4. 3λ

Destructive interference needs a path difference of a whole number plus half a wavelength. 1.5λ fits; the others give constructive interference.

3Paper 2 style
(a)
Explain how a stationary wave is formed on a stretched string.
[2]
(b)
A tuning fork of frequency 512 Hz is held above a tube closed at one end. Resonance is heard when the air column is 0.165 m long and again when it is 0.495 m long. Calculate the speed of sound.
[3]
  • (a) The wave travels along the string and is reflected at the fixed end
  • (a) The incident and reflected waves (same frequency and speed, opposite directions) superpose
  • (b) L2 − L1 = λ/2 = 0.330 m
  • (b) λ = 0.660 m
  • (b) v = 512 × 0.660 = 338 m s−1
4Paper 2 style

Light of wavelength 600 nm is incident normally on a diffraction grating with 500 lines per millimetre.

(a)
Calculate the angle of the first-order maximum.
[2]
(b)
Determine the highest order of maximum that can be observed.
[2]
(c)
State what is meant by coherent sources.
[1]
  • (a) d = 1/500 000 = 2.0 × 10−6 m
  • (a) sin θ = 600 × 10−9/2.0 × 10−6 = 0.30, so θ = 17.5°
  • (b) n ≤ d/λ = 3.3
  • (b) highest order = 3
  • (c) Sources with a constant phase difference (and the same frequency)

Try it in the grating simulation: set 500 lines per mm and 600 nm.

Electric current is a flow of charge, potential difference measures the energy each coulomb transfers, and resistance links the two. Learn the precise definitions: they are asked for again and again.

Electric current and charge

Electric current is the rate of flow of charge. In metals the charge carriers are free electrons; in electrolytes they are ions. Charge is quantised: any charge is a whole-number multiple of the elementary charge e = 1.60 × 10−19 C. One coulomb is the charge passing a point when a current of 1 A flows for 1 s.

ChargeQ = ItLearn it
Current and drift velocity (n = number density of carriers, q = charge of each)I = AnvqOn formulae page
Potential differenceV = WQLearn it
PowerP = VI = I2R = V2RLearn it
ResistanceR = VILearn it
ResistivityR = ρLALearn it

Worked example

A copper wire of cross-sectional area 1.0 mm2 carries a current of 2.0 A. Copper has 8.5 × 1028 free electrons per m3. Find the drift velocity.

  1. v = IAnq = 2.0(1.0 × 10−6)(8.5 × 1028)(1.60 × 10−19)

v = 1.5 × 10−4 m s−1: the electrons drift very slowly, even though the signal travels almost at the speed of light.

Potential difference and power

The potential difference across a component is the energy transferred from electrical to other forms per unit charge passing through it. One volt is one joule per coulomb.

Resistance and I–V characteristics

Resistance is the potential difference across a component divided by the current through it. One ohm is one volt per ampere. Ohm's law: for a metallic conductor at constant temperature, the current is proportional to the potential difference.

VIMetallic conductor
at constant temperature: a straight line through the origin, so R is constant
VIFilament lamp
the gradient falls as V increases: the filament gets hotter and its resistance increases
VISemiconductor diode
almost no current in reverse; conducts in the forward direction above about 0.6 V

Why a filament's resistance rises: a larger current heats the metal, so its ions vibrate with greater amplitude and the electrons collide with them more often.

  • The resistance of a light-dependent resistor (LDR) decreases as light intensity increases.
  • The resistance of an NTC thermistor decreases as its temperature increases.

Resistivity ρ is a property of the material: the resistance of a 1 m length with 1 m2 cross-section. Unit: Ω m. A wire's resistance increases with length and decreases with cross-sectional area.

Exam-style questions

1Paper 1 style

A lamp carries a current of 0.25 A for 2.0 minutes. How much charge passes through it?

  1. 0.50 C
  2. 30 C
  3. 60 C
  4. 480 C

Q = It = 0.25 × 120 = 30 C. Convert minutes to seconds first.

2Paper 1 style

A wire has resistance R. A second wire of the same material has twice the length and twice the diameter. What is its resistance?

  1. R/4
  2. R/2
  3. R
  4. 2R

Doubling the diameter multiplies the area by 4. R ∝ L/A, so the new resistance is R × 2/4 = R/2.

3Paper 2 style
(a)
Define the ohm.
[1]
(b)
Describe and explain the I–V characteristic of a filament lamp.
[3]
  • (a) The resistance when a p.d. of one volt produces a current of one ampere (1 Ω = 1 V A−1)
  • (b) A curve through the origin, symmetrical for negative values, with the gradient decreasing as V increases
  • (b) Larger current means higher temperature of the filament
  • (b) So the resistance increases (V/I increases)
4Paper 2 style

A heater is rated at 2.0 kW when connected to a 230 V supply.

(a)
Calculate the current in the heater and its resistance.
[2]
(b)
Calculate the energy transferred in 15 minutes.
[1]
(c)
The heater element is a nichrome wire of resistivity 1.1 × 10−6 Ω m and cross-sectional area 2.0 × 10−7 m2. Calculate its length.
[2]
  • (a) I = 2000/230 = 8.7 A
  • (a) R = 230/8.7 = 26 Ω (or V2/P = 26.5 Ω)
  • (b) E = 2000 × 900 = 1.8 × 106 J
  • (c) L = RA/ρ = 26.5 × 2.0 × 10−7/1.1 × 10−6
  • (c) L = 4.8 m

Real sources have internal resistance, Kirchhoff's two laws let you solve any network, and the potential divider is the basis of every sensor circuit.

E.m.f. and internal resistance

Electromotive force (e.m.f.)
The energy transferred from other forms (such as chemical) to electrical energy per unit charge, by the source, when charge passes round a complete circuit.
Potential difference
The energy transferred from electrical to other forms per unit charge, in a component.
Internal resistance
The resistance of the source itself. Some energy is transferred to thermal energy inside the source.
Source with internal resistance rE = I(R + r) = V + IrLearn it
Resistors in seriesR = R1 + R2 + …On formulae page
Resistors in parallel1R = 1R1 + 1R2 + …On formulae page

The terminal p.d. V is less than the e.m.f. by the "lost volts" Ir. The more current drawn, the lower the terminal p.d. A graph of V against I is a straight line with y-intercept E and gradient −r.

Worked example

A battery of e.m.f. 12 V and internal resistance 0.50 Ω is connected to a 5.5 Ω resistor. Find the current and the terminal p.d.

  1. I = E/(R + r) = 12/6.0 = 2.0 A
  2. V = IR = 2.0 × 5.5 = 11 V (lost volts = 1.0 V)

Kirchhoff's laws

First law
The sum of the currents entering a junction equals the sum of the currents leaving it. This is a consequence of conservation of charge.
Second law
Around any closed loop in a circuit, the sum of the e.m.f.s equals the sum of the p.d.s. This is a consequence of conservation of energy.

The series and parallel formulas follow from these laws. For series resistors the current is the same, so V = IR1 + IR2. For parallel resistors the p.d. is the same, so I = V/R1 + V/R2.

Potential dividers

Two resistors in series across a supply share its p.d. in the ratio of their resistances:

Potential divider outputVout = R2R1 + R2 × VinLearn it

Replacing one resistor with a thermistor or LDR makes a sensor: for example, with an LDR as R2, more light means a lower R2 and so a lower output p.d.

The potentiometer

A potentiometer is a uniform resistance wire with a steady current from a driver cell, so the p.d. along it is proportional to length. To compare two p.d.s, connect each in turn (through a galvanometer) to a sliding contact and find the balance (null) point where the galvanometer reads zero. Then V1/V2 = L1/L2. At balance no current is drawn from the cell being measured, so its e.m.f. is measured without any lost volts.

Exam-style questions

1Paper 1 style

What is the combined resistance of a 6.0 Ω resistor and a 3.0 Ω resistor connected in parallel?

  1. 0.50 Ω
  2. 2.0 Ω
  3. 4.5 Ω
  4. 9.0 Ω

1/R = 1/6 + 1/3 = 1/2, so R = 2.0 Ω. The combination is always smaller than the smallest resistor.

2Paper 1 style

A 9.0 V supply of negligible internal resistance is connected across a 4.0 kΩ resistor in series with a 2.0 kΩ resistor. What is the p.d. across the 2.0 kΩ resistor?

  1. 3.0 V
  2. 4.5 V
  3. 6.0 V
  4. 9.0 V

V = 9.0 × 2.0/(4.0 + 2.0) = 3.0 V.

3Paper 1 style

On a potentiometer, a standard cell of e.m.f. 1.50 V balances at a length of 60.0 cm. A second cell balances at 80.0 cm. What is the e.m.f. of the second cell?

  1. 1.13 V
  2. 1.50 V
  3. 2.00 V
  4. 2.40 V

The e.m.f. is proportional to balance length: 1.50 × 80.0/60.0 = 2.00 V.

4Paper 2 style
(a)
Distinguish between electromotive force and potential difference, in terms of energy.
[2]
(b)
A cell of e.m.f. 6.0 V is connected to a 10 Ω resistor. The p.d. across the resistor is 5.0 V. Calculate the internal resistance of the cell.
[3]
  • (a) E.m.f.: energy transferred from other forms to electrical energy per unit charge (in the source)
  • (a) P.d.: energy transferred from electrical to other forms per unit charge (in a component)
  • (b) I = 5.0/10 = 0.50 A
  • (b) lost volts = 6.0 − 5.0 = 1.0 V
  • (b) r = 1.0/0.50 = 2.0 Ω
5Paper 2 style

A 12 V battery of negligible internal resistance is connected to a 4.0 Ω resistor in series with a parallel combination of a 6.0 Ω and a 12 Ω resistor.

(a)
State Kirchhoff's first law.
[1]
(b)
Calculate the current from the battery and the current in the 6.0 Ω resistor.
[4]
  • (a) The sum of currents into a junction equals the sum of currents out of it
  • (b) Parallel combination: 1/R = 1/6.0 + 1/12, so R = 4.0 Ω
  • (b) Total = 8.0 Ω, so battery current = 12/8.0 = 1.5 A
  • (b) p.d. across the parallel section = 1.5 × 4.0 = 6.0 V
  • (b) Current in 6.0 Ω = 6.0/6.0 = 1.0 A
6Paper 2 style
A potential divider consists of a fixed resistor in series with an NTC thermistor, connected to a 9.0 V supply. The output is taken across the fixed resistor. Explain what happens to the output p.d. as the temperature of the thermistor increases.
[3]
  • The resistance of the thermistor decreases as temperature increases
  • The total resistance decreases so the current increases (or: the fixed resistor takes a larger share of the supply p.d.)
  • So the p.d. across the fixed resistor (the output) increases

The atom has a tiny, dense nucleus; nuclei can be unstable and emit radiation; and protons and neutrons are themselves made of quarks. This topic links all three.

The nuclear atom

In the α-particle scattering experiment, a thin gold foil in a vacuum was bombarded with α-particles:

ObservationConclusion
Most α-particles passed straight through, undeflected.Most of the atom is empty space.
A few were deflected through small angles.The nucleus is positively charged (it repels the positive α-particles).
A very small number were deflected through more than 90°.Most of the mass is concentrated in a very small nucleus.

A nuclide is written AZX, where A is the nucleon number (protons + neutrons) and Z the proton number. Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. A nucleus has a diameter of about 10−15 to 10−14 m, compared with about 10−10 m for the atom.

Radioactive decay

In every nuclear process, nucleon number and charge are conserved.

αβ−β+γ
Naturehelium nucleus, 2p + 2nelectronpositron (antielectron)electromagnetic radiation
Charge+2e−e+e0
Mass4 uabout 1/2000 uabout 1/2000 u0
Ionisingstronglymoderatelymoderatelyweakly
Stopped bypaper, a few cm of aira few mm of aluminium(annihilates with an electron)reduced by thick lead or concrete
Energiesdiscretecontinuous rangecontinuous rangediscrete

β particles have a continuous range of energies because the energy released in each decay is shared with a second particle: an antineutrino in β− decay and a neutrino in β+ decay. These particles have almost no mass and no charge.

Example decay equations:

  • α: 23892U → 23490Th + 42α
  • β−: 146C → 147N + 0−1e + ν
  • β+: 116C → 115B + 0+1e + ν

Fundamental particles

Quarks are fundamental particles. You need three of the six flavours. Each quark has an antiquark with the opposite charge.

QuarkSymbolChargeAntiquark charge
upu+⅔ e−⅔ e
downd−⅓ e+⅓ e
stranges−⅓ e+⅓ e

The other three are charm, top and bottom.

Hadrons
Particles made of quarks, which feel the strong force. Baryons are three quarks (proton = uud, neutron = udd); mesons are a quark and an antiquark.
Leptons
Fundamental particles that are not made of quarks, such as the electron and the neutrino.
Antiparticles
Every particle has an antiparticle with the same mass and opposite charge (the positron is the antiparticle of the electron).

Quark changes in β decay: in β− decay a down quark changes into an up quark (a neutron becomes a proton): d → u + e− + ν. In β+ decay an up quark changes into a down quark (a proton becomes a neutron): u → d + e+ + ν. Both are caused by the weak interaction.

Exam-style questions

1Paper 1 style

What is the quark composition of a neutron?

  1. uud
  2. udd
  3. uuu
  4. ddd

udd has charge ⅔ − ⅓ − ⅓ = 0. The proton is uud, with charge +1.

2Paper 1 style

Which quark change takes place in β− decay?

  1. up to down
  2. down to up
  3. strange to up
  4. up to strange

A neutron (udd) becomes a proton (uud), so one down quark becomes an up quark.

3Paper 1 style

The nucleus 22688Ra emits an α-particle. What are the nucleon number and proton number of the nucleus formed?

  1. 226 and 86
  2. 224 and 86
  3. 222 and 88
  4. 222 and 86

An α-particle carries away 4 nucleons and 2 protons: 226 − 4 = 222 and 88 − 2 = 86 (radon).

4Paper 2 style
Describe the results of the α-particle scattering experiment and explain what each result shows about the structure of the atom.
[4]
  • Most α-particles pass straight through (undeflected)
  • so most of the atom is empty space
  • A very small fraction are deflected through large angles (more than 90°)
  • so the nucleus is very small and contains most of the mass, and is positively charged
5Paper 2 style
(a)
Carbon-14 (proton number 6) decays by β− emission to nitrogen. Write the nuclear equation for this decay.
[2]
(b)
Explain why the β− particles emitted have a range of kinetic energies.
[2]
(c)
A particle has the quark composition ud. State whether it is a baryon or a meson, and determine its charge.
[2]
  • (a) 146C → 147N + 0−1e
  • (a) + antineutrino (ν), with all numbers correct
  • (b) The energy released in each decay is shared between the β− particle and an antineutrino
  • (b) in varying proportions, so the β particle's energy varies (up to a maximum)
  • (c) Meson (quark–antiquark)
  • (c) Charge = +⅔e + ⅓e = +e

A Level Physics, one topic at a time

Notes for the fourteen A Level topics of the Cambridge 9702 syllabus: fields, thermal physics, oscillations, electromagnetism, quantum and nuclear physics, medical physics and astronomy. Each has worked examples, simulations and exam-style questions with mark schemes.

Paper 4: A Level Structured Questions

2 h100 marks38.5% of A Level

Structured questions on the A Level topics, which can draw on AS knowledge.

Paper 5: Planning, Analysis and Evaluation

1 h 15 min30 marks11.5% of A Level

Planning an experiment, and analysing and evaluating data, including logarithmic graphs and uncertainties.

The AS papers make up the other half of the A Level: Paper 1 is 15.5%, Paper 2 is 23% and Paper 3 is 11.5%.

Topics

Revising A Level physics

  • Learn definitions word for word. Gravitational potential, electric field strength, the tesla, simple harmonic motion and binding energy are asked for again and again, and each phrase earns a mark.
  • Spot the shared maths. Capacitor discharge and radioactive decay are both exponential decays; gravitational and electric fields follow the same inverse-square pattern.
  • Practise with logarithms. Paper 5 expects you to linearise relationships such as x = x0e−kt by plotting ln x against t.
About the questions

The questions here are written in the style of 9702 papers. They are not copies of official papers. For genuine past papers and mark schemes, use the Cambridge School Support Hub or ask your teacher.

An object moving in a circle at constant speed is accelerating, because its direction keeps changing. The acceleration, and the resultant force that causes it, point towards the centre.

Angles and angular speed

Radian
The angle subtended at the centre of a circle by an arc equal in length to the radius. A full circle is 2π rad.
Angular displacement
The angle through which an object has moved around a circle, in radians.
Angular speed ω
The rate of change of angular displacement, in rad s−1.
Angular speedω = 2πT = 2πfLearn it
Linear and angular speedv = rωLearn it
Centripetal accelerationa = rω2 = v2rLearn it
Centripetal forceF = mrω2 = mv2rLearn it

Centripetal acceleration and force

Velocity is a vector. Moving in a circle at constant speed, the direction of the velocity changes continuously, so the object is accelerating. This centripetal acceleration is directed towards the centre of the circle, perpendicular to the velocity, so it changes the direction but not the speed.

By Newton's second law, a resultant force towards the centre is needed: the centripetal force. It is not a new kind of force; it is provided by an existing one:

  • a car on a flat bend: friction between the tyres and the road;
  • a satellite in orbit: gravitational force;
  • a ball whirled on a string: tension (plus a component of the weight in a vertical circle);
  • an electron orbiting in a magnetic field: the magnetic force.

Because the force is perpendicular to the velocity, it does no work, and the kinetic energy stays constant.

Worked example

A car of mass 1200 kg travels round a flat bend of radius 50 m at 15 m s−1. What friction force is needed?

  1. F = mv2/r = 1200 × 152/50

F = 5400 N, directed towards the centre of the bend.

Vertical circles

For a mass on a string in a vertical circle, at the top both tension and weight point to the centre: T + mg = mv2/r. At the bottom the tension points up and the weight down: T − mg = mv2/r. So the tension is greatest at the bottom, and the string is most likely to go slack at the top.

Exam-style questions

1Paper 4 practice

What is the angular speed of the second hand of a clock?

  1. 0.105 rad s−1
  2. 0.0167 rad s−1
  3. 6.28 rad s−1
  4. 1.75 × 10−3 rad s−1

It turns 2π rad in 60 s: ω = 2π/60 = 0.105 rad s−1. Option D is the minute hand.

2Paper 4 practice

A puck on a frictionless horizontal table moves in a circle, held by a string. Its speed is doubled while the radius stays the same. By what factor does the tension change?

  1. ½
  2. √2
  3. 2
  4. 4

T = mv2/r, so doubling v multiplies the tension by 4.

3Paper 4 style
(a)
Define the radian.
[1]
(b)
Explain why an object moving in a circle at constant speed is accelerating.
[2]
  • (a) The angle subtended at the centre of a circle by an arc of length equal to the radius
  • (b) The direction of motion (velocity) is continuously changing
  • (b) Velocity is a vector, so its velocity changes and it accelerates (towards the centre)
4Paper 4 style

A ball of mass 0.20 kg is whirled in a vertical circle of radius 0.80 m on a light string. At the top of the circle its speed is 4.0 m s−1.

(a)
Calculate the tension in the string at the top of the circle.
[2]
(b)
Use conservation of energy to find the speed at the bottom, and hence calculate the tension at the bottom.
[3]
  • (a) T + mg = mv2/r: T = 0.20 × 4.02/0.80 − 0.20 × 9.81
  • (a) T = 4.0 − 1.96 = 2.0 N
  • (b) v2 = 4.02 + 2 × 9.81 × 1.60 = 47.4, so v = 6.9 m s−1
  • (b) T − mg = mv2/r: T = 0.20 × 47.4/0.80 + 1.96
  • (b) T = 13.8 N (14 N)

Every mass attracts every other mass. This topic describes that attraction as a field, uses it to explain orbits, and introduces gravitational potential, which is always negative.

The gravitational field

A gravitational field is a region of space where a mass experiences a force. Gravitational field strength g at a point is the gravitational force per unit mass at that point, in N kg−1. Field lines show the direction of the force on a mass. Close to the Earth's surface the field is approximately uniform: parallel, equally spaced lines, with g = 9.81 N kg−1.

Field strengthg = FmLearn it
Newton's law of gravitationF = GMmr2Learn it
Field strength of a point massg = GMr2Learn it
Gravitational potentialφ = −GMrOn formulae page
Gravitational potential energyEp = −GMmrOn formulae page

Newton's law of gravitation

The gravitational force between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation. For a uniform sphere, outside the sphere it acts as if all its mass were concentrated at its centre, so r is measured from the centre.

Orbits

For a satellite in a circular orbit, gravity provides the centripetal force:

GMmr2 = mv2r, so v2 = GMr. Using v = 2πr/T gives T2 = 4π2GMr3, so T2 ∝ r3 (Kepler's third law).

A geostationary satellite stays above the same point on the Earth. It must have a period of 24 hours, orbit above the equator, and move in the same direction as the Earth rotates (west to east). This fixes its orbital radius at about 4.2 × 107 m. Ground aerials do not need to track it.

Gravitational potential

The gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point. It is zero at infinity and negative everywhere else, because gravity is attractive: work is done by the field as the mass moves in. The potential energy of a mass m at that point is Ep = mφ.

When a mass moves from radius r1 to r2, the change in potential energy is ΔEp = GMm(1r1 − 1r2). Near the Earth's surface this reduces to mgΔh.

The field strength is the negative of the potential gradient: g = −ΔφΔr.

Exam-style questions

1Paper 4 practice

The gravitational field strength at the Earth's surface is 9.8 N kg−1. What is it at a height above the surface equal to the Earth's radius?

  1. 9.8 N kg−1
  2. 4.9 N kg−1
  3. 2.5 N kg−1
  4. 1.1 N kg−1

The distance from the centre doubles, so g falls by a factor of 22 = 4: 9.8/4 = 2.45 N kg−1.

2Paper 4 practice

Why is gravitational potential always negative?

  1. Because mass is always positive.
  2. Because the force is attractive and the potential is defined as zero at infinity.
  3. Because the field lines point away from the mass.
  4. Because potential energy decreases with height.

With zero at infinity, moving a mass inward means work is done by the attractive field, so the potential falls below zero everywhere at a finite distance.

3Paper 4 style
(a)
Define gravitational potential at a point.
[2]
(b)
A satellite of mass m is in a circular orbit of radius r about a planet of mass M. Show that the period T is given by T2 = 4π2r3GM.
[3]
  • (a) Work done per unit mass
  • (a) in bringing a small (test) mass from infinity to the point
  • (b) Gravitational force provides the centripetal force: GMm/r2 = mv2/r (or mrω2)
  • (b) v = 2πr/T (or ω = 2π/T)
  • (b) Substituting and rearranging gives T2 = 4π2r3/GM
4Paper 4 style

The Earth has mass 5.97 × 1024 kg and radius 6.37 × 106 m. A satellite of mass 1000 kg orbits 400 km above the surface.

(a)
Calculate the orbital speed and the period of the satellite.
[4]
(b)
The satellite is moved to a geostationary orbit of radius 4.22 × 107 m. Calculate the increase in its gravitational potential energy.
[3]
  • (a) r = 6.37 × 106 + 0.40 × 106 = 6.77 × 106 m
  • (a) v = √GM/r = 7.67 × 103 m s−1
  • (a) T = 2πr/v
  • (a) T = 5.5 × 103 s (about 92 minutes)
  • (b) ΔEp = GMm(1r1 − 1r2)
  • (b) = 6.67 × 10−11 × 5.97 × 1024 × 1000 × (16.77 × 106 − 14.22 × 107)
  • (b) ΔEp = 4.9 × 1010 J

Temperature decides which way thermal energy flows. This topic covers the absolute temperature scale and how much energy it takes to warm a substance or change its state.

Thermal equilibrium

Thermal energy is transferred from a region of higher temperature to a region of lower temperature. Objects in thermal equilibrium are at the same temperature, and there is no net transfer of thermal energy between them.

Temperature scales

A practical thermometer relies on a physical property that varies with temperature, such as the resistance of a thermistor, the e.m.f. of a thermocouple, or the expansion of a liquid. The thermodynamic (Kelvin) scale does not depend on the property of any particular substance. Its zero is absolute zero, the temperature at which a substance has minimum internal energy.

Kelvin and CelsiusT/K = θ/°C + 273.15Learn it
Specific heat capacityE = mcΔθLearn it
Specific latent heatE = mLLearn it

A temperature change of 1 K is the same as a change of 1 °C.

Specific heat capacity and specific latent heat

Specific heat capacity
The energy needed per unit mass to raise the temperature of a substance by one degree (1 K). Unit: J kg−1 K−1.
Specific latent heat
The energy needed per unit mass to change the state of a substance without any change in temperature. Unit: J kg−1. Of fusion for melting, of vaporisation for boiling.

During a change of state the temperature stays constant: the energy supplied increases the potential energy of the molecules (breaking bonds), not their kinetic energy. The specific latent heat of vaporisation is much larger than that of fusion because, on boiling, the molecules must be separated completely and work is done pushing back the atmosphere as the volume increases greatly.

Measuring specific heat capacity: the electrical method

  1. Place an electrical heater and a thermometer in holes in a metal block (or in a liquid), and insulate it.
  2. Measure the mass with a balance. Record the starting temperature.
  3. Switch on the heater for a measured time t, recording the p.d. V and current I (or use a joulemeter).
  4. Record the highest temperature reached. Energy supplied = VIt = mcΔθ, so c = VItmΔθ.
  5. Energy losses to the surroundings make the value of c too high. To reduce them: insulate, or start below room temperature and finish the same amount above it.

Worked example

How long does a 50 W heater take to warm 0.50 kg of water from 20 °C to 30 °C? Water has c = 4200 J kg−1 K−1. Assume no energy is lost.

  1. E = mcΔθ = 0.50 × 4200 × 10 = 21 000 J
  2. t = E/P = 21 000/50

t = 420 s (7.0 minutes)

Exam-style questions

1Paper 4 practice

What is a temperature of 300 K on the Celsius scale (to the nearest degree)?

  1. 27 °C
  2. −27 °C
  3. 573 °C
  4. 300 °C

θ = 300 − 273 = 27 °C.

2Paper 4 practice

Why does the temperature of ice remain at 0 °C while it is melting?

  1. The energy supplied is lost to the surroundings.
  2. The kinetic energy of the molecules increases.
  3. The energy supplied increases the potential energy of the molecules.
  4. The specific heat capacity of ice is zero at 0 °C.

During melting, energy breaks the bonds holding the molecules in place, increasing their potential energy. The mean kinetic energy, and so the temperature, stays the same.

3Paper 4 style
(a)
Define specific heat capacity.
[1]
(b)
Describe an electrical method to determine the specific heat capacity of a metal block. Explain how the result is calculated and state one source of error.
[4]
  • (a) Energy required per unit mass to raise the temperature by one degree
  • (b) Heater and thermometer in the block; block insulated; mass measured with a balance
  • (b) Measure V, I and heating time t (or joulemeter reading), and the temperature rise
  • (b) c = VIt/(mΔθ)
  • (b) Energy lost to the surroundings (so c is too high); or thermal energy absorbed by the heater and thermometer
4Paper 4 style

A drink contains 0.20 kg of ice at 0 °C. Calculate the energy needed to melt all the ice and warm the water to 15 °C. Specific latent heat of fusion of ice = 3.34 × 105 J kg−1; specific heat capacity of water = 4200 J kg−1 K−1.

[3]
  • To melt: 0.20 × 3.34 × 105 = 6.68 × 104 J
  • To warm: 0.20 × 4200 × 15 = 1.26 × 104 J
  • Total = 7.9 × 104 J

A gas is a huge number of molecules moving at random. The ideal gas equation describes its large-scale behaviour, and kinetic theory explains that behaviour from the motion of the molecules.

The mole

The mole is the SI unit of amount of substance. One mole contains NA = 6.02 × 1023 particles (the Avogadro constant). For n moles, the number of molecules is N = nNA.

The equation of state

An ideal gas is one that obeys pV ∝ T exactly, at all pressures, volumes and temperatures, where T is the thermodynamic temperature in kelvin. Real gases behave nearly ideally at low pressure and high temperature.

Ideal gas equation (n moles)pV = nRTLearn it
Ideal gas equation (N molecules)pV = NkTLearn it
Kinetic theory pressurep = 13NmV⟨c2⟩On formulae page
Mean kinetic energy of a molecule12m⟨c2⟩ = 32kTLearn it

The Boltzmann constant is k = R/NA. The root-mean-square speed is crms = √⟨c2⟩.

Kinetic theory

Assumptions

  • A gas contains a very large number of molecules moving randomly.
  • The volume of the molecules is negligible compared with the volume of the container.
  • There are no forces between molecules except during collisions.
  • Collisions between molecules and with the walls are perfectly elastic.
  • The time spent in collisions is negligible compared with the time between collisions.

Deriving the pressure

  1. A molecule of mass m moving with velocity component cx hits a wall and rebounds: its change of momentum is 2mcx.
  2. In a cube of side l, it returns to the same wall every 2l/cx, so the force on the wall is 2mcx2l/cx = mcx2l.
  3. For N molecules, total force = Nm⟨cx2⟩l, and pressure = forcel2 = Nm⟨cx2⟩V.
  4. Motion is random, so ⟨cx2⟩ = ⟨cy2⟩ = ⟨cz2⟩ = 13⟨c2⟩, giving pV = 13Nm⟨c2⟩.

Linking to temperature: comparing pV = 13Nm⟨c2⟩ with pV = NkT gives 13m⟨c2⟩ = kT, so the mean kinetic energy of a molecule is 12m⟨c2⟩ = 32kT. The thermodynamic temperature is proportional to the mean kinetic energy of the molecules.

Worked example

Find the r.m.s. speed of nitrogen molecules (molecular mass 28 u) at 300 K.

  1. m = 28 × 1.66 × 10−27 = 4.65 × 10−26 kg
  2. ⟨c2⟩ = 3kT/m = 3 × 1.38 × 10−23 × 300/4.65 × 10−26 = 2.67 × 105 m2 s−2

crms = 520 m s−1

Exam-style questions

1Paper 4 practice

The thermodynamic temperature of an ideal gas is doubled. By what factor does the r.m.s. speed of its molecules change?

  1. 2
  2. √2
  3. 4
  4. 1/√2

⟨c2⟩ ∝ T, so crms ∝ √T and doubling T multiplies crms by √2.

2Paper 4 practice

A container of volume 2.0 m3 holds an ideal gas at a pressure of 1.0 × 105 Pa and a temperature of 300 K. How many moles of gas does it contain?

  1. 8.0
  2. 27
  3. 80
  4. 240

n = pV/RT = (1.0 × 105 × 2.0)/(8.31 × 300) = 80 mol.

3Paper 4 style
State four assumptions of the kinetic theory of gases.
[4]
  • Large number of molecules in random motion
  • Volume of molecules negligible compared with the volume of the container
  • No intermolecular forces (except during collisions)
  • Collisions are elastic; time of collision negligible compared with time between collisions
4Paper 4 style

A cylinder of volume 3.0 × 10−3 m3 contains an ideal gas at a pressure of 2.0 × 105 Pa and a temperature of 27 °C.

(a)
Calculate the amount of gas, in moles, and the number of molecules.
[3]
(b)
The gas is heated to 127 °C at constant volume. Calculate the new pressure.
[2]
  • (a) T = 300 K
  • (a) n = pV/RT = (2.0 × 105 × 3.0 × 10−3)/(8.31 × 300) = 0.24 mol
  • (a) N = 0.241 × 6.02 × 1023 = 1.4 × 1023
  • (b) p ∝ T: p = 2.0 × 105 × 400/300
  • (b) p = 2.7 × 105 Pa

Using 27 instead of 300 K, or 127 instead of 400 K, loses the marks.

5Paper 4 style
Use the equations pV = 13Nm⟨c2⟩ and pV = NkT to show that the mean kinetic energy of a molecule of an ideal gas is proportional to its thermodynamic temperature.
[2]
  • Equating: 13Nm⟨c2⟩ = NkT, so m⟨c2⟩ = 3kT
  • Mean KE = 12m⟨c2⟩ = 32kT, and k is a constant, so mean KE ∝ T

Internal energy is the total energy of the molecules in a system. The first law of thermodynamics is conservation of energy applied to it: heat it or do work on it, and its internal energy rises.

Internal energy

The internal energy of a system is the sum of a random distribution of the kinetic and potential energies of its molecules. A rise in temperature increases the kinetic energy of the molecules, so it increases the internal energy.

In an ideal gas there are no intermolecular forces, so the molecules have no potential energy. The internal energy is purely kinetic, U = N × 32kT, and so it is proportional to the thermodynamic temperature.

The first law of thermodynamics

First lawΔU = q + WLearn it
Work done on a gas at constant pressureW = −pΔVLearn it
ΔU
The increase in internal energy of the system.
q
The energy transferred to the system by heating (negative if the system loses thermal energy).
W
The work done on the system. When a gas is compressed, work is done on it and W is positive; when it expands, it does work on the surroundings and W is negative.
Watch the signs

9702 uses "work done on the system". Some textbooks use work done by the system, with the opposite sign. Always say in words which you mean.

Process (ideal gas)What happens
Constant temperature (isothermal)ΔU = 0, so q = −W. If the gas is compressed, the work done on it is lost as thermal energy.
No heat transfer (rapid compression or expansion)q = 0, so ΔU = W. Rapid compression makes the gas hotter; rapid expansion makes it cooler.
Constant volumeW = 0, so ΔU = q. All the heat supplied raises the internal energy.

Worked example

A gas absorbs 500 J of thermal energy and expands, doing 200 J of work on its surroundings. What is the change in its internal energy?

  1. q = +500 J; work done on the gas W = −200 J
  2. ΔU = 500 + (−200) = 300 J

Exam-style questions

1Paper 4 practice

An ideal gas is compressed slowly at constant temperature. Which statement is correct?

  1. Its internal energy increases.
  2. No work is done on the gas.
  3. It transfers thermal energy to the surroundings.
  4. Its internal energy decreases.

At constant temperature, ΔU = 0. Work is done on the gas (W > 0), so q = −W is negative: the gas loses thermal energy to the surroundings.

2Paper 4 style
(a)
Define the internal energy of a system.
[2]
(b)
Explain why the internal energy of an ideal gas is proportional to its thermodynamic temperature.
[2]
  • (a) Sum of the kinetic and potential energies of the molecules
  • (a) (due to their) random distribution
  • (b) There are no intermolecular forces, so no potential energy: internal energy is the total kinetic energy
  • (b) Mean kinetic energy is proportional to thermodynamic temperature (32kT)
3Paper 4 style

A gas at a constant pressure of 1.0 × 105 Pa expands from a volume of 2.0 × 10−3 m3 to 3.5 × 10−3 m3. During this, 400 J of thermal energy is supplied to it.

(a)
Calculate the work done on the gas.
[2]
(b)
Calculate the change in internal energy of the gas.
[1]
  • (a) pΔV = 1.0 × 105 × 1.5 × 10−3 = 150 J
  • (a) Work done on the gas = −150 J (the gas expands)
  • (b) ΔU = 400 − 150 = +250 J

Simple harmonic motion describes pendulums, masses on springs and vibrating molecules. Its defining equation leads to sinusoidal graphs, a constant total energy, and the dramatic effect of resonance.

Describing oscillations

Displacement x
The distance from the equilibrium position in a given direction.
Amplitude x₀
The maximum displacement.
Period T
The time for one complete oscillation.
Frequency f
The number of oscillations per unit time.
Angular frequency ω
ω = 2πf = 2π/T, in rad s−1.
Phase difference
The fraction of a cycle by which one oscillation leads another, in radians or degrees.

Simple harmonic motion

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from a fixed point and is always directed towards that point.

Defining equationa = −ω2xOn formulae page
Displacement (starting at the centre)x = x0sin ωtLearn it
Velocityv = v0cos ωtOn formulae page
Velocity at displacement xv = ±ω√x02 − x2On formulae page
Maximum speedv0 = ωx0Learn it
Total energyE = 12mω2x02Learn it
  • The velocity is maximum at the equilibrium position and zero at the extremes.
  • The acceleration is zero at equilibrium and maximum (magnitude ω2x0) at the extremes.
  • Velocity leads displacement by a quarter cycle (π/2); acceleration is in antiphase with displacement (π).
  • A graph of a against x is a straight line through the origin with gradient −ω2.

Energy in s.h.m.

Energy is continuously exchanged between kinetic and potential energy, while the total stays constant (if there is no damping). Kinetic energy = 12mω2(x02 − x2); potential energy = 12mω2x2. Both graphs against displacement are parabolas; their sum is a horizontal line.

Worked example

A mass oscillates in s.h.m. with amplitude 5.0 cm and frequency 2.0 Hz. Find its maximum speed, its maximum acceleration, and its speed when the displacement is 3.0 cm.

  1. ω = 2π × 2.0 = 12.6 rad s−1
  2. v0 = ωx0 = 12.6 × 0.050 = 0.63 m s−1; amax = ω2x0 = 7.9 m s−2
  3. v = ω√x02 − x2 = 12.6√0.0502 − 0.0302 = 12.6 × 0.040 = 0.50 m s−1

Damping and resonance

Damping is the loss of energy from an oscillating system, usually by resistive forces, so the amplitude decreases.

  • Light damping: the system oscillates with amplitude decreasing gradually (exponentially) over many cycles; the period stays almost constant.
  • Critical damping: the system returns to equilibrium in the shortest possible time without oscillating (as in car suspension).
  • Heavy damping: the system returns to equilibrium very slowly, without oscillating.

In forced oscillations, a periodic driving force makes a system oscillate at the driving frequency. Resonance is when the amplitude of the forced oscillation is a maximum, which happens when the driving frequency equals the natural frequency of the system. At resonance, energy is transferred most efficiently from the driver.

f0 driving frequency amplitude light damping more damping heavy damping
Increasing the damping lowers the peak amplitude, makes the peak broader, and moves it to a slightly lower frequency.

Resonance can be useful (tuning a radio circuit, magnetic resonance imaging, musical instruments) or destructive (bridges and buildings driven by wind or earthquakes, rattling vehicle parts), which is why structures are damped.

Exam-style questions

1Paper 4 practice

A body moves with simple harmonic motion. At the moment its displacement is maximum, which quantity is also at its maximum magnitude?

  1. speed
  2. kinetic energy
  3. momentum
  4. acceleration

a = −ω2x, so the acceleration is largest where the displacement is largest. The speed, kinetic energy and momentum are all zero there.

2Paper 4 style
(a)
Define simple harmonic motion.
[2]
(b)
A mass of 0.40 kg oscillates in s.h.m. with amplitude 0.10 m and angular frequency 5.0 rad s−1. Calculate its total energy and its kinetic energy when its displacement is 0.060 m.
[3]
  • (a) Acceleration is (directly) proportional to displacement (from a fixed point)
  • (a) and is always directed towards that fixed point (opposite to displacement)
  • (b) E = 12 × 0.40 × 5.02 × 0.102 = 0.050 J
  • (b) Ek = 12 × 0.40 × 5.02 × (0.102 − 0.0602)
  • (b) Ek = 0.032 J
3Paper 4 style
(a)
Explain what is meant by resonance.
[2]
(b)
Describe the effect of increasing the damping on the resonance curve of a system.
[2]
  • (a) The amplitude of a forced oscillation is maximum
  • (a) when the driving frequency equals the natural frequency of the system
  • (b) The amplitude at every frequency decreases, especially the peak
  • (b) The peak becomes flatter (broader) and occurs at a slightly lower frequency

Electric fields follow the same inverse-square pattern as gravitational fields, with one big difference: charges can repel as well as attract. This topic covers uniform fields between plates, Coulomb's law and electric potential.

Electric fields and field lines

An electric field is a region of space where a charge experiences a force. Electric field strength is the force per unit positive charge, in N C−1 (equivalently V m−1). Field lines show the direction of the force on a positive charge: away from positive charges, towards negative charges.

Field strengthE = FqLearn it
Uniform field between parallel platesE = ΔVΔdLearn it
Coulomb's lawF = Qq4πε0r2Learn it
Field of a point chargeE = Q4πε0r2Learn it
Electric potentialV = Q4πε0rOn formulae page
Electric potential energyEp = Qq4πε0rOn formulae page

Uniform fields

Between two parallel plates with a p.d. across them, the field is uniform (except near the edges): field lines are parallel, equally spaced and perpendicular to the plates. A charged particle entering the field at right angles moves in a parabola, like a projectile: constant velocity parallel to the plates, and constant acceleration a = qE/m towards one plate.

Worked example

Two parallel plates 2.0 cm apart have a p.d. of 500 V across them. Find the force on an electron between them, and its acceleration.

  1. E = 500/0.020 = 2.5 × 104 V m−1
  2. F = Ee = 2.5 × 104 × 1.60 × 10−19 = 4.0 × 10−15 N
  3. a = F/m = 4.0 × 10−15/9.11 × 10−31 = 4.4 × 1015 m s−2

Point charges

Coulomb's law: the force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation. For a charged conducting sphere, outside the sphere the charge acts as if concentrated at its centre.

Electric potential

The electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point. It is positive near a positive charge and negative near a negative charge. The field strength equals the negative of the potential gradient: E = −ΔVΔr.

Gravitational fieldElectric field
Acts onmasscharge
Force lawF = GMm/r2F = Qq/4πε0r2
Directionalways attractiveattractive or repulsive
Potentialalways negativepositive or negative
SimilaritiesBoth are inverse-square laws for point sources; field strength is the negative potential gradient; potential is zero at infinity.

Exam-style questions

1Paper 4 practice

The separation of two point charges is doubled. What happens to the force between them?

  1. It becomes a quarter.
  2. It halves.
  3. It doubles.
  4. It becomes four times larger.

F ∝ 1/r2, so doubling r divides the force by 4.

2Paper 4 style

An electron enters the uniform field between two horizontal plates, moving horizontally at 3.0 × 107 m s−1. The plates are 4.0 cm apart with a p.d. of 600 V between them, and they are 5.0 cm long.

(a)
Define electric field strength.
[1]
(b)
Calculate the vertical acceleration of the electron.
[2]
(c)
Calculate the vertical deflection of the electron as it leaves the plates.
[3]
  • (a) Force per unit positive charge
  • (b) E = 600/0.040 = 1.5 × 104 V m−1
  • (b) a = eE/m = 1.60 × 10−19 × 1.5 × 104/9.11 × 10−31 = 2.6 × 1015 m s−2
  • (c) time between plates = 0.050/3.0 × 107 = 1.67 × 10−9 s
  • (c) y = 12at2 = 12 × 2.63 × 1015 × (1.67 × 10−9)2
  • (c) y = 3.7 × 10−3 m (3.7 mm)
3Paper 4 style

An isolated metal sphere of radius 0.10 m carries a charge of +4.0 × 10−8 C.

(a)
Define electric potential at a point.
[2]
(b)
Calculate the electric potential at the surface of the sphere.
[2]
(c)
Calculate the work done in moving a charge of +2.0 × 10−9 C from infinity to the surface.
[1]
  • (a) Work done per unit positive charge
  • (a) in bringing a small test charge from infinity to the point
  • (b) V = 8.99 × 109 × 4.0 × 10−8/0.10
  • (b) V = 3.6 × 103 V
  • (c) W = qV = 2.0 × 10−9 × 3.6 × 103 = 7.2 × 10−6 J

A capacitor stores charge, and with it energy. When it discharges through a resistor the charge, current and p.d. all fall exponentially, governed by one number: the time constant RC.

Capacitance

Capacitance is the charge stored per unit potential difference. A capacitor stores equal and opposite charges on two plates separated by an insulator (dielectric); the charge Q means the charge on one plate. One farad is one coulomb per volt. The idea also applies to an isolated conducting sphere.

CapacitanceC = QVLearn it
Capacitors in parallelC = C1 + C2 + …On formulae page
Capacitors in series1C = 1C1 + 1C2 + …On formulae page
Energy storedW = 12QV = 12CV2Learn it
Discharge (x = Q, V or I)x = x0e−t/RCOn formulae page
Time constantτ = RCLearn it

The series and parallel formulas are the reverse of those for resistors. In parallel, the p.d. is the same and the charges add. In series, the charge on each is the same and the p.d.s add.

Energy stored

On a graph of p.d. V against charge Q for a capacitor, the line is straight through the origin (since V = Q/C). The work done to charge it is the area under this graph, a triangle: W = 12QV. Substituting Q = CV gives W = 12CV2. When a capacitor is charged from a battery, the battery supplies energy QV; only half is stored, and half is dissipated in the resistance of the circuit.

Discharging a capacitor

When a charged capacitor discharges through a resistor R, the current is proportional to the p.d., which is proportional to the remaining charge. So the rate of loss of charge is proportional to the charge: an exponential decay. The time constant τ = RC is the time for the charge (or p.d., or current) to fall to 1/e (about 37%) of its initial value. It takes 0.69RC to halve.

Taking natural logs: ln Q = ln Q0 − tRC, so a graph of ln Q (or ln V) against t is a straight line of gradient −1RC.

Worked example

A 470 μF capacitor is charged to 12 V and then discharged through a 10 kΩ resistor. Find the initial charge, the energy stored, the time constant, and the p.d. after 5.0 s.

  1. Q = CV = 470 × 10−6 × 12 = 5.6 × 10−3 C
  2. W = 12CV2 = 12 × 470 × 10−6 × 144 = 0.034 J
  3. τ = RC = 10 × 103 × 470 × 10−6 = 4.7 s
  4. V = 12e−5.0/4.7 = 4.1 V

Exam-style questions

1Paper 4 practice

Two 6.0 μF capacitors are connected in series. What is their combined capacitance?

  1. 2.0 μF
  2. 3.0 μF
  3. 6.0 μF
  4. 12 μF

1/C = 1/6.0 + 1/6.0, so C = 3.0 μF. In parallel they would give 12 μF.

2Paper 4 practice

The p.d. across a capacitor is doubled. By what factor does the energy stored change?

  1. ½
  2. √2
  3. 2
  4. 4

W = 12CV2, so doubling V multiplies the energy by 4.

3Paper 4 style
(a)
Define capacitance.
[1]
(b)
Use a graph of p.d. against charge to show that the energy stored in a capacitor is 12QV.
[2]
(c)
A 2200 μF capacitor is charged to 9.0 V. Calculate the energy stored.
[1]
  • (a) Charge (stored on one plate) per unit potential difference
  • (b) The graph of V against Q is a straight line through the origin; the work done (energy stored) is the area under the graph
  • (b) The area is a triangle, so W = 12QV
  • (c) W = 12 × 2200 × 10−6 × 9.02 = 0.089 J
4Paper 4 style

A capacitor of capacitance 100 μF is charged to 6.0 V and then discharged through a resistor of resistance 47 kΩ.

(a)
Calculate the time constant of the circuit.
[1]
(b)
Calculate the p.d. across the capacitor 10 s after the discharge starts.
[2]
(c)
Calculate the time for the p.d. to fall to 1.0 V.
[2]
  • (a) τ = 47 × 103 × 100 × 10−6 = 4.7 s
  • (b) V = 6.0e−10/4.7
  • (b) V = 0.72 V
  • (c) 1.0 = 6.0e−t/4.7, so t = 4.7ln 6.0
  • (c) t = 8.4 s

Magnetic fields push on moving charges and on currents, and changing magnetic flux induces an e.m.f. Together these explain motors, particle beams, Hall probes and generators.

Magnetic fields

A magnetic field is a region of space where a magnet, a current-carrying conductor or a moving charge experiences a force. Field lines run from north to south outside a magnet; closer lines mean a stronger field.

Fields due to currents

  • Long straight wire: concentric circles around the wire, getting further apart with distance. Direction given by the right-hand grip rule (thumb along the current, fingers curl with the field).
  • Flat circular coil: like that of a short bar magnet, with field lines through the centre of the coil.
  • Long solenoid: a strong, uniform field inside, like a bar magnet outside. Inserting a ferrous (iron) core greatly increases the field strength.

Two parallel wires carrying currents in the same direction attract; in opposite directions they repel. Each wire sits in the other's field.

Force on a current-carrying conductor

Force on a conductorF = BILsin θLearn it
Force on a moving chargeF = BQvsin θLearn it
Hall voltage (t = thickness of the slice)VH = BIntqOn formulae page
Magnetic fluxΦ = BALearn it
Faraday's lawE = −Δ(NΦ)ΔtLearn it

The direction of the force is given by Fleming's left-hand rule: First finger = Field, seCond finger = Current, thuMb = Motion (force). There is no force when the current is parallel to the field.

Magnetic flux density B is the force per unit length per unit current on a conductor placed at right angles to the field. One tesla is the flux density when a wire carrying 1 A at right angles to the field experiences a force of 1 N per metre of its length (1 T = 1 N A−1 m−1).

Force on a moving charge

A charge moving at right angles to a uniform magnetic field feels a force perpendicular to its velocity, so it moves in a circle: BQv = mv2r, giving r = mvBQ. Faster or heavier particles make bigger circles; a stronger field makes smaller ones. The force does no work, so the speed is constant.

Velocity selection: crossed electric and magnetic fields exert opposite forces. Only particles with EQ = BQv, that is v = E/B, pass through undeflected.

The Hall effect: when a current flows through a thin slice of conductor in a magnetic field perpendicular to it, the magnetic force pushes the charge carriers towards one side. Charge builds up until the electric force from it balances the magnetic force, producing a steady Hall voltage across the slice. Since VH ∝ B, a Hall probe measures flux density. Semiconductors give a larger Hall voltage than metals because their number density of charge carriers n is much smaller.

Worked example

A proton moves at 2.0 × 106 m s−1 at right angles to a magnetic field of 0.50 T. Find the radius of its path.

  1. r = mvBQ = 1.67 × 10−27 × 2.0 × 1060.50 × 1.60 × 10−19

r = 0.042 m (4.2 cm)

Electromagnetic induction

Magnetic flux
The product of the flux density and the area perpendicular to the field, Φ = BA. Unit: weber (Wb).
Magnetic flux linkage
The product of the flux through a coil and its number of turns, NΦ.
Faraday's law
The induced e.m.f. is proportional to the rate of change of magnetic flux linkage.
Lenz's law
The direction of the induced e.m.f. is such that it tends to oppose the change producing it. This is a consequence of conservation of energy: work must be done to produce the induced current.

An e.m.f. is induced whenever the flux linking a circuit changes: moving a magnet into a coil, moving a wire across a field, rotating a coil in a field, or changing the current in a nearby coil. Experiments show the e.m.f. is larger if the change is faster, the field stronger, or the coil has more turns.

Exam-style questions

1Paper 4 practice

A charged particle moves parallel to the field lines of a uniform magnetic field. What is the magnetic force on it?

  1. along its direction of motion
  2. opposite to its direction of motion
  3. perpendicular to its direction of motion
  4. zero

F = BQvsin θ and θ = 0, so the force is zero.

2Paper 4 style
(a)
Define the tesla.
[2]
(b)
A straight wire of length 0.15 m carries a current of 3.0 A at right angles to a uniform magnetic field of 0.20 T. Calculate the force on the wire.
[1]
  • (a) The (uniform) magnetic flux density that produces a force of 1 N per metre
  • (a) on a wire carrying a current of 1 A at right angles to the field
  • (b) F = BIL = 0.20 × 3.0 × 0.15 = 0.090 N
3Paper 4 style

An electron is accelerated to a speed of 4.0 × 106 m s−1 and enters a uniform magnetic field of 2.5 mT at right angles to the field.

(a)
Explain why the electron moves in a circle.
[2]
(b)
Show that the radius of the circle is given by r = mv/(Be), and calculate it.
[3]
  • (a) The magnetic force is always perpendicular to the velocity
  • (a) so it provides the centripetal force (constant magnitude, speed unchanged)
  • (b) Bev = mv2/r, so r = mv/(Be)
  • (b) r = 9.11 × 10−31 × 4.0 × 106/(2.5 × 10−3 × 1.60 × 10−19)
  • (b) r = 9.1 × 10−3 m
4Paper 4 style

A coil of 200 turns and area 4.0 × 10−3 m2 is perpendicular to a magnetic field of 0.30 T. The field is reduced steadily to zero in 0.15 s.

(a)
State Faraday's law and Lenz's law.
[2]
(b)
Calculate the e.m.f. induced in the coil.
[2]
  • (a) Faraday: the induced e.m.f. is proportional to the rate of change of magnetic flux linkage
  • (a) Lenz: the induced e.m.f. acts in a direction to oppose the change causing it
  • (b) change in flux linkage = 200 × 0.30 × 4.0 × 10−3 = 0.24 Wb
  • (b) E = 0.24/0.15 = 1.6 V
5Paper 4 style
Explain how a Hall voltage is produced across a thin slice of semiconductor carrying a current in a magnetic field, and why it is larger than in a metal slice of the same size.
[4]
  • The moving charge carriers experience a magnetic force perpendicular to their velocity and to the field
  • Charge accumulates on one face, creating an electric field (p.d.) across the slice
  • A steady Hall voltage is reached when the electric force balances the magnetic force
  • VH = BI/ntq; a semiconductor has a much smaller number density n, so VH is larger

Mains electricity is alternating: the current reverses direction many times a second. This topic covers how to describe an a.c. supply, why we quote r.m.s. values, and how diodes and a capacitor turn a.c. into steady d.c.

Describing alternating current

Period
The time for one complete cycle.
Frequency
The number of cycles per unit time.
Peak value
The maximum value of the current or p.d., I0 or V0.
Root-mean-square (r.m.s.) value
The value of the direct current (or p.d.) that would produce the same mean power in a resistor as the alternating current (or p.d.).
Sinusoidal current or p.d.x = x0sin ωtOn formulae page
R.m.s. currentIrms = I0√2Learn it
R.m.s. p.d.Vrms = V0√2Learn it
Mean power in a resistive load⟨P⟩ = Irms2R = 12P0Learn it

For a sinusoidal supply, the mean power in a resistor is half the peak power, because the power I2R varies between 0 and I02R and the mean of sin 2 is ½. Mains voltages are quoted as r.m.s. values: a 230 V supply has a peak value of 230√2 = 325 V.

Rectification and smoothing

a.c. input
Half-wave rectified
one diode: only half of each cycle passes
Full-wave rectified
bridge of four diodes: both halves pass, in one direction
Smoothed
a capacitor across the load reduces the ripple
  • Half-wave rectification uses a single diode in series with the load. The current flows only during the half-cycles when the diode is forward-biased.
  • Full-wave rectification uses a bridge rectifier of four diodes. In each half-cycle a different pair of diodes conducts, so the current always passes through the load in the same direction.
  • Smoothing: a capacitor connected in parallel with the load charges up as the p.d. rises, then, as the supply p.d. falls, discharges through the load, keeping the output p.d. fairly steady. The remaining variation is the ripple. A larger time constant RC (larger capacitor or larger load resistance) gives less ripple.

Exam-style questions

1Paper 4 practice

A sinusoidal alternating current has a peak value of 5.0 A. What is its r.m.s. value?

  1. 2.5 A
  2. 3.5 A
  3. 5.0 A
  4. 7.1 A

Irms = 5.0/√2 = 3.5 A. Option D multiplies instead of dividing.

2Paper 4 style

An alternating p.d. is given by V = 340sin (314t), where V is in volts and t in seconds. It is applied across a 100 Ω resistor.

(a)
Define the root-mean-square value of an alternating current.
[2]
(b)
Determine the frequency of the supply and the r.m.s. p.d.
[2]
(c)
Calculate the mean power dissipated in the resistor.
[2]
  • (a) The value of the direct current
  • (a) that produces the same (mean) power in a resistor
  • (b) f = 314/2π = 50 Hz
  • (b) Vrms = 340/√2 = 240 V
  • (c) P = Vrms2/R = 2402/100
  • (c) P = 580 W
3Paper 4 style
Describe how a bridge rectifier and a capacitor are used to produce a smoothed direct p.d. across a load resistor, and state how the ripple could be reduced.
[4]
  • Four diodes in a bridge; in each half-cycle two diodes conduct (opposite pairs)
  • So the current through the load is always in the same direction (full-wave rectification)
  • The capacitor in parallel with the load charges as the p.d. rises and discharges through the load as the p.d. falls
  • Ripple is reduced by a larger capacitance (or larger load resistance), so a larger time constant

Light behaves as particles called photons, and electrons behave as waves. The photoelectric effect, electron diffraction and atomic line spectra are the three pieces of evidence you need to know.

Photons

Electromagnetic radiation has a particulate nature: it is emitted and absorbed in discrete packets of energy called photons. A photon's energy depends only on its frequency.

Photon energyE = hf = hcλLearn it
Photon momentump = EcLearn it
Photoelectric equationhf = Φ + 12mvmax2Learn it
de Broglie wavelengthλ = hpLearn it
Photon emitted or absorbedhf = E1 − E2Learn it

The electronvolt is the energy gained by an electron accelerated through a p.d. of 1 V: 1 eV = 1.60 × 10−19 J. A photon of wavelength 500 nm has energy hc/λ = 3.98 × 10−19 J = 2.49 eV.

The photoelectric effect

When electromagnetic radiation of high enough frequency falls on a metal surface, electrons are emitted. The observations:

  • No electrons are emitted below a threshold frequency, however intense the radiation.
  • Above it, emission is instantaneous, even for very weak radiation.
  • The maximum kinetic energy of the electrons depends on the frequency, not on the intensity.
  • The number of electrons emitted per second (the photoelectric current) is proportional to the intensity.

Explanation: each electron absorbs the energy of one photon. It needs at least the work function Φ (the minimum energy needed to release an electron from the surface) to escape, so the photon energy hf must be at least Φ: the threshold frequency is f0 = Φ/h. Any extra energy becomes kinetic energy; electrons from deeper in the metal lose some on the way out, which is why the equation gives the maximum kinetic energy. Greater intensity means more photons per second, so more electrons, but not more energy per electron.

The wave theory fails: a wave's energy depends on its intensity, so it predicts that intense light of any frequency would eventually release electrons, with energy increasing with intensity, and with a time delay at low intensity. None of these happen.

Wave-particle duality

Electron diffraction is evidence that particles have a wave nature. A beam of electrons passed through a thin film of polycrystalline graphite produces a pattern of concentric rings on a fluorescent screen, which is a diffraction pattern. Increasing the accelerating p.d. increases the electrons' momentum, so their wavelength decreases and the rings get smaller.

Any particle of momentum p has a de Broglie wavelength λ = h/p. For an electron accelerated from rest through p.d. V: 12mv2 = eV, so p = √2meV.

Worked example

Find the de Broglie wavelength of an electron accelerated through 100 V.

  1. p = √2 × 9.11 × 10−31 × 1.60 × 10−19 × 100 = 5.40 × 10−24 kg m s−1
  2. λ = h/p = 6.63 × 10−34/5.40 × 10−24

λ = 1.2 × 10−10 m, similar to the spacing of atoms in a crystal, which is why diffraction is seen.

Energy levels and line spectra

Electrons in an isolated atom can only have certain discrete energy levels. When an electron falls from a higher level E1 to a lower level E2, a photon of energy hf = E1 − E2 is emitted. Because only certain differences are possible, only certain frequencies appear: an emission line spectrum of bright lines. When white light passes through a cool gas, photons with exactly these energies are absorbed to excite electrons, giving an absorption line spectrum: dark lines on a continuous spectrum.

0 (ionisation)−0.85 eV, n = 4−1.51 eV, n = 3−3.40 eV, n = 2−13.6 eV, n = 1 (ground state) ultravioletvisible
Hydrogen: transitions to the ground state emit ultraviolet photons; transitions to n = 2 emit visible light. The energies are negative because energy must be supplied to free the electron.

Exam-style questions

1Paper 4 practice

In a photoelectric experiment, the intensity of the light is increased while its frequency stays the same, above the threshold. What increases?

  1. the work function of the metal
  2. the maximum kinetic energy of the electrons
  3. the number of electrons emitted per second
  4. the threshold frequency

More intense light means more photons per second, so more electrons are emitted each second. Each photon's energy, and so the maximum kinetic energy, is unchanged.

2Paper 4 style

Ultraviolet radiation of wavelength 250 nm falls on a zinc surface. The work function of zinc is 4.3 eV.

(a)
Explain what is meant by the work function.
[1]
(b)
Calculate the maximum kinetic energy of the emitted electrons, in eV.
[3]
(c)
Explain why no electrons are emitted when red light falls on zinc, however bright it is.
[2]
  • (a) The minimum energy needed to remove an electron from the surface of the metal
  • (b) E = hc/λ = 6.63 × 10−34 × 3.00 × 108/250 × 10−9 = 7.96 × 10−19 J
  • (b) = 4.97 eV
  • (b) Ek, max = 4.97 − 4.3 = 0.67 eV
  • (c) Each electron absorbs one photon; a red photon's energy is less than the work function
  • (c) Brighter light only gives more photons, not more energy per photon
3Paper 4 style
(a)
Describe the evidence that electrons can behave as waves.
[2]
(b)
Calculate the de Broglie wavelength of an electron moving at 3.0 × 106 m s−1.
[2]
  • (a) Electrons passed through (polycrystalline) graphite produce rings on a screen
  • (a) This is a diffraction pattern, and diffraction is a wave property
  • (b) p = 9.11 × 10−31 × 3.0 × 106 = 2.73 × 10−24 kg m s−1
  • (b) λ = 6.63 × 10−34/2.73 × 10−24 = 2.4 × 10−10 m
4Paper 4 style
(a)
Explain how the existence of discrete electron energy levels in atoms leads to line spectra.
[2]
(b)
An electron in a hydrogen atom moves from the −3.40 eV level to the −13.6 eV level. Calculate the wavelength of the emitted photon.
[3]
  • (a) Only certain energy differences between levels are possible, and each transition emits (or absorbs) one photon of energy hf = E1 − E2
  • (a) so only certain discrete frequencies (wavelengths) appear
  • (b) ΔE = 10.2 eV = 1.63 × 10−18 J
  • (b) λ = hc/ΔE = 6.63 × 10−34 × 3.00 × 108/1.63 × 10−18
  • (b) λ = 1.2 × 10−7 m (ultraviolet)

A nucleus has less mass than the protons and neutrons it is made from; the missing mass is the binding energy. This explains why fission and fusion release energy. The second half of the topic treats radioactive decay mathematically.

Mass and energy

Mass–energy equivalenceE = mc2Learn it
ActivityA = λNLearn it
Radioactive decay (x = A, N or count rate)x = x0e−λtOn formulae page
Decay constant and half-lifeλ = 0.693t1/2On formulae page
Mass defect
The difference between the total mass of the separate nucleons and the mass of the nucleus.
Binding energy
The minimum energy needed to separate a nucleus into its individual protons and neutrons. Binding energy = mass defect × c2.
Binding energy per nucleon
The binding energy divided by the nucleon number. A higher value means a more stable nucleus.
02468 50100150200 iron-56: most stable helium-4 uranium-235 fusion fission nucleon number A binding energy per nucleon / MeV
Light nuclei can release energy by fusing; heavy nuclei by splitting. Either way the products have a higher binding energy per nucleon, and the difference in total binding energy is released.

Fission is the splitting of a large nucleus (such as uranium-235, after absorbing a neutron) into two smaller nuclei and a few neutrons. Fusion is the joining of two light nuclei (such as hydrogen isotopes) to form a heavier one; it needs very high temperatures to overcome the electrostatic repulsion. In both, the total mass of the products is less than the total mass before, and the mass difference appears as energy, ΔE = Δmc2.

Worked example

A helium-4 nucleus has mass 4.001506 u. A proton has mass 1.007276 u and a neutron 1.008665 u. Find the binding energy per nucleon in MeV. (1 u = 1.66 × 10−27 kg)

  1. Mass of separate nucleons = 2(1.007276) + 2(1.008665) = 4.031882 u
  2. Mass defect = 0.030376 u = 5.04 × 10−29 kg
  3. Binding energy = Δmc2 = 5.04 × 10−29 × (3.00 × 108)2 = 4.54 × 10−12 J = 28.4 MeV

Binding energy per nucleon = 28.4/4 = 7.1 MeV

Radioactive decay

Radioactive decay is spontaneous (not affected by external factors such as temperature or pressure) and random (it is impossible to predict which nucleus will decay next, or when; each nucleus has a constant probability of decaying per unit time). Evidence for randomness: the count rate from a source fluctuates.

Activity
The number of nuclei decaying per unit time, in becquerels (1 Bq = 1 decay per second).
Decay constant λ
The probability that a nucleus decays per unit time.
Half-life
The mean time taken for half of the active nuclei in a sample to decay.

Because A = λN, the number of nuclei, the activity and the count rate all decay exponentially with the same half-life. After n half-lives, the fraction remaining is (12)n. Plotting ln A against t gives a straight line of gradient −λ.

Exam-style questions

1Paper 4 practice

A radioactive sample has a half-life of 8.0 days. What fraction of the original nuclei remains undecayed after 24 days?

  1. 1/3
  2. 1/4
  3. 1/8
  4. 1/24

24 days is three half-lives: (12)3 = 18.

2Paper 4 practice

Which nucleus has the highest binding energy per nucleon?

  1. helium-4
  2. iron-56
  3. uranium-235
  4. hydrogen-2

The binding energy per nucleon peaks near iron-56 (and nickel-62), making these the most stable nuclei.

3Paper 4 style
(a)
Define nuclear binding energy.
[1]
(b)
Use the idea of binding energy per nucleon to explain why the fission of uranium-235 releases energy.
[3]
  • (a) The minimum energy required to separate a nucleus into its constituent protons and neutrons
  • (b) The fission products (middle-sized nuclei) have a higher binding energy per nucleon than uranium-235
  • (b) So the total binding energy of the products is greater than that of the uranium nucleus
  • (b) The difference is released as energy (the products have less total mass)
4Paper 4 style

A sample of cobalt-60 contains 2.0 × 1020 nuclei. The half-life of cobalt-60 is 5.27 years. (1 year = 3.16 × 107 s)

(a)
State what is meant by the decay constant.
[1]
(b)
Calculate the decay constant in s−1 and the activity of the sample.
[3]
(c)
Calculate the activity after 2.0 years.
[2]
  • (a) The probability of decay of a nucleus per unit time
  • (b) λ = 0.693/(5.27 × 3.16 × 107) = 4.16 × 10−9 s−1
  • (b) A = λN = 4.16 × 10−9 × 2.0 × 1020
  • (b) A = 8.3 × 1011 Bq
  • (c) A = 8.3 × 1011e−0.693 × 2.0/5.27
  • (c) A = 6.4 × 1011 Bq

Three ways of seeing inside the body: ultrasound reflects from boundaries between tissues, X-rays are absorbed differently by bone and soft tissue, and PET detects gamma photons from a radioactive tracer.

Ultrasound

Ultrasound is sound with a frequency above the range of human hearing (above 20 kHz); medical scans use about 1 to 10 MHz.

The piezoelectric transducer

  • Generating: an alternating p.d. applied across a piezoelectric crystal makes it change shape (expand and contract) at the same frequency, so it vibrates and emits ultrasound. The crystal's thickness is chosen so it resonates at the required frequency.
  • Detecting: when reflected ultrasound hits the crystal, it is made to vibrate, and the changing shape produces an alternating p.d. across it, which is processed.

Reflection at boundaries

Specific acoustic impedanceZ = ρcLearn it
Intensity reflection coefficientIRI0 = (Z2 − Z1)2(Z2 + Z1)2On formulae page
Attenuation (μ = linear attenuation coefficient)I = I0e−μxLearn it
Minimum X-ray wavelength (accelerating p.d. V)λmin = hceVLearn it

The specific acoustic impedance Z of a medium is its density × the speed of sound in it, unit kg m−2 s−1. When ultrasound meets a boundary, the fraction of intensity reflected depends on the difference in Z. A large difference means strong reflection.

Why gel is used: air and skin have very different impedances, so almost all ultrasound would be reflected at the skin. A gel with an impedance close to skin's fills the gap and removes the air layer, so most of the ultrasound enters the body.

In an A-scan, the time between sending a pulse and receiving each echo gives the depth of each boundary: depth = 12ct. A B-scan combines many A-scans taken at different angles into a two-dimensional image.

Worked example

Fat has Z = 1.38 × 106 kg m−2 s−1 and muscle Z = 1.70 × 106 kg m−2 s−1. What fraction of the intensity is reflected at a fat–muscle boundary?

  1. IRI0 = (1.70 − 1.38)2(1.70 + 1.38)2 = 0.3223.082

IR/I0 = 0.011 (about 1%), so most of the ultrasound passes on to deeper boundaries.

X-rays

In an X-ray tube, electrons emitted from a heated cathode are accelerated through a large p.d. (tens of kilovolts) and strike a metal target (such as tungsten). Most of their energy becomes thermal energy, so the anode is cooled, often by rotating it. About 1% becomes X-rays:

  • Braking radiation: electrons decelerating in the target emit a continuous spectrum of X-ray photons. The highest-energy photon carries all of one electron's kinetic energy eV, which sets a minimum wavelength λmin = hc/eV.
  • Characteristic lines: sharp peaks at wavelengths set by electron energy levels in the target atoms.

A higher accelerating p.d. gives "harder" X-rays (higher photon energies, more penetrating). Increasing the tube current gives more photons, so a more intense beam, but does not change the photon energies.

Image quality. Sharpness improves with a smaller target area (a near point source) and with a lead grid that stops scattered X-rays. Contrast depends on the difference in attenuation between tissues: bone attenuates far more than soft tissue. Contrast media such as barium (high attenuation) are used to show soft tissues such as the digestive tract.

CT scanning: an X-ray tube and detectors rotate around the patient, taking images of a thin slice from many angles. A computer combines them to find the attenuation in each small volume element (voxel) of the slice. Repeating for many slices builds a three-dimensional image that can be viewed from any angle.

PET scanning

  1. A tracer is injected: a substance used by the body (such as a glucose analogue, FDG) labelled with a β+-emitting isotope such as fluorine-18, with a short half-life.
  2. The tracer collects where it is used most, for example in active tumours.
  3. Each emitted positron travels a very short distance before meeting an electron. They annihilate, turning their mass into two gamma photons.
  4. To conserve momentum, the two photons travel in opposite directions, each with energy mec2 = 0.51 MeV (8.2 × 10−14 J).
  5. A ring of detectors around the patient detects both photons. The line between the two detectors, and the small difference in arrival times, locate where the annihilation happened. Many events build up an image of tracer concentration.

Exam-style questions

1Paper 4 practice

Why is a gel applied between an ultrasound transducer and the skin?

  1. to cool the transducer
  2. to remove air, whose acoustic impedance is very different from skin's
  3. to increase the frequency of the ultrasound
  4. to absorb reflected ultrasound

Air and skin have very different acoustic impedances, so nearly all the ultrasound would be reflected at an air gap. The gel matches the impedance of skin, so most of the ultrasound is transmitted into the body.

2Paper 4 style
(a)
Explain how a piezoelectric crystal is used to generate ultrasound.
[3]
(b)
The specific acoustic impedance of soft tissue is 1.63 × 106 kg m−2 s−1 and that of bone is 7.80 × 106 kg m−2 s−1. Calculate the intensity reflection coefficient at the boundary.
[2]
  • (a) An alternating p.d. is applied across the crystal
  • (a) causing the crystal to change shape (vibrate) at the same frequency
  • (a) at its resonant frequency (in the ultrasound range), producing ultrasound waves
  • (b) (7.80 − 1.63)2(7.80 + 1.63)2
  • (b) = 0.43
3Paper 4 style
(a)
An X-ray tube operates with an accelerating p.d. of 80 kV. Calculate the minimum wavelength of the X-rays produced.
[2]
(b)
A beam of X-rays passes through 5.0 mm of a material with linear attenuation coefficient 0.20 mm−1. Calculate the fraction of the intensity transmitted.
[2]
  • (a) λ = hc/eV = 6.63 × 10−34 × 3.00 × 108/(1.60 × 10−19 × 80 × 103)
  • (a) λ = 1.6 × 10−11 m
  • (b) I/I0 = e−0.20 × 5.0 = e−1.0
  • (b) = 0.37
4Paper 4 style
(a)
Explain how the photons detected in PET scanning are produced, and why they travel in opposite directions.
[3]
(b)
Calculate the energy of each photon.
[2]
  • (a) The tracer emits positrons (β+)
  • (a) Each positron annihilates with an electron, producing two gamma photons
  • (a) The total momentum before is (approximately) zero, so the photons travel in opposite directions
  • (b) Each photon's energy = mec2 = 9.11 × 10−31 × (3.00 × 108)2
  • (b) = 8.2 × 10−14 J

How do we measure the size of a star we can't resolve, or the distance to a galaxy? Luminosity, the inverse-square law, the laws of black-body radiation and redshift together answer both questions, and point to an expanding universe.

Luminosity and standard candles

Luminosity
The total power of the electromagnetic radiation emitted by a star, in watts.
Radiant flux intensity
The radiant power passing normally through unit area, in W m−2. This is what we measure from Earth.
Standard candle
An astronomical object of known luminosity. Examples: Cepheid variable stars (whose period of brightness variation is linked to their luminosity) and type 1a supernovae (which all reach about the same peak luminosity).
Inverse-square law for fluxF = L4πd2Learn it
Wien's displacement lawλmax ∝ 1TLearn it
Stefan–Boltzmann lawL = 4πσr2T4On formulae page
Hubble's lawv ≈ H0dLearn it
Doppler redshiftΔλλ ≈ Δff ≈ vcOn formulae page

Radiation spreads out equally in all directions over the surface of a sphere of area 4πd2. If we know a standard candle's luminosity and measure its flux, we can find its distance: d = √L/4πF.

Stellar radii

A star behaves approximately as a black body. Wien's displacement law: the wavelength at which the intensity of radiation is greatest is inversely proportional to the thermodynamic temperature, λmaxT = constant (about 2.9 × 10−3 m K). So a star's colour gives its surface temperature: hotter stars are bluer.

The Stefan–Boltzmann law links luminosity to surface temperature and radius. Knowing L (from a standard candle or flux and distance) and T (from Wien's law), the radius is r = √L4πσT4.

Worked example

The Sun's luminosity is 3.83 × 1026 W and its spectrum peaks at about 500 nm. Estimate its surface temperature and radius. (Wien's constant = 2.9 × 10−3 m K)

  1. T = 2.9 × 10−3/500 × 10−9 = 5800 K
  2. r = √3.83 × 10264π × 5.67 × 10−8 × 58004

r = 6.9 × 108 m

Hubble's law and the Big Bang

Spectral lines in light from distant galaxies are shifted to longer wavelengths: a redshift. This is the Doppler effect for light from a source moving away. The fractional shift gives the recession speed: v ≈ cΔλ/λ (for speeds much less than c).

Hubble's law: the speed at which a galaxy is receding is proportional to its distance from us, v ≈ H0d, where H0 is the Hubble constant (about 2.2 × 10−18 s−1).

Since more distant galaxies recede faster, and in every direction, the universe is expanding. Running the expansion backwards suggests it began from an extremely small, hot, dense state: the Big Bang. A rough age for the universe is 1/H0.

Exam-style questions

1Paper 4 practice

A star is moved twice as far from an observer. By what factor does the radiant flux intensity received change?

  1. 2
  2. ½
  3. 4
  4. ¼

F ∝ 1/d2, so doubling the distance reduces the flux to a quarter.

2Paper 4 practice

Star X has twice the surface temperature of star Y and the same radius. What is the ratio of their luminosities, LX/LY?

  1. 16
  2. 8
  3. 4
  4. 2

L ∝ r2T4, so with equal radii LX/LY = 24 = 16.

3Paper 4 style
(a)
Explain what is meant by a standard candle.
[1]
(b)
A Cepheid variable star has luminosity 4.0 × 1028 W. The radiant flux intensity received at Earth is 2.0 × 10−9 W m−2. Calculate its distance from Earth.
[2]
  • (a) An object of known luminosity
  • (b) d = √L/(4πF) = √4.0 × 1028/(4π × 2.0 × 10−9)
  • (b) d = 1.3 × 1018 m
4Paper 4 style

A star has luminosity 6.0 × 1027 W. The peak wavelength of its spectrum is 360 nm. Wien's constant is 2.9 × 10−3 m K.

(a)
State Wien's displacement law.
[1]
(b)
Calculate the surface temperature of the star.
[1]
(c)
Calculate the radius of the star.
[2]
  • (a) The wavelength at which the emitted intensity is maximum is inversely proportional to the thermodynamic temperature
  • (b) T = 2.9 × 10−3/360 × 10−9 = 8.1 × 103 K
  • (c) r = √L/(4πσT4) = √6.0 × 1027/(4π × 5.67 × 10−8 × (8.06 × 103)4)
  • (c) r = 1.4 × 109 m
5Paper 4 style

A hydrogen line has wavelength 656.3 nm in the laboratory. In the spectrum of a distant galaxy it is observed at 662.9 nm.

(a)
Calculate the speed at which the galaxy is receding.
[2]
(b)
Taking the Hubble constant as 2.2 × 10−18 s−1, estimate the distance to the galaxy.
[1]
(c)
Explain how Hubble's law supports the Big Bang theory.
[2]
  • (a) v = cΔλ/λ = 3.00 × 108 × 6.6/656.3
  • (a) v = 3.0 × 106 m s−1
  • (b) d = v/H0 = 1.4 × 1024 m
  • (c) Galaxies are receding, faster the further away they are, so the universe is expanding
  • (c) so in the past it was smaller (denser and hotter), and began from a single point (or very small region)