A Level Physics 9702 Paper 41, May/June 2026

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Question paper

9702/41 · May/June 2026

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1
Figure 1 of 3 for question 1
Figure 2 of 3 for question 1
Figure 3 of 3 for question 1
a

Define gravitational potential at a point.

[2]
b(i)

A uniform spherical planet has radius R. The gravitational potential at its surface is −Φ. P is a point on a line that passes through the centre of the planet, at variable displacement x from the centre, as shown in Figure 1.1. Figure 1.2 shows the variation with x of the gravitational potential ϕ due to the planet in the region outside the planet close to its surface. Explain why ϕ is negative for all values of x.

[2]
b(ii)

State an expression, in terms of Φ and R, for the mass M of the planet. Identify any other symbols that you use. M=.........................................................

[2]
b(iii)

Determine an expression, in terms of Φ and R, for the gravitational field strength g0 at the surface of the planet. g0=.........................................................

[1]
b(iv)

On Figure 1.3, sketch the variation of the gravitational field g with x between x=−3R and x=3R. Do not include the region inside the sphere between x=−R and x=R.

[3]
2
Figure 1 of 1 for question 2
a

State what is meant by an ideal gas.

[2]
b(i)

Ideal gases may be used to measure the temperature of the environment in a laboratory experiment. In some situations, however, it is preferable to use a thermocouple thermometer. Explain how a thermometer that uses an ideal gas can be advantageous compared with a thermometer that uses a liquid in glass.

[1]
b(ii)

Describe a laboratory situation in which a thermocouple thermometer is preferable to a thermometer that uses an ideal gas.

[1]
c

A very small sample of an ideal gas consists of five molecules, each of mass 3.34×10−27 kg. At a particular instant, the speeds of the molecules are as shown in Figure 2.1. Determine, to three significant figures, the root-mean-square (r.m.s.) speed c0 of the molecules. c0=................................................ ms−1

[2]
d(i)

A larger sample of the gas in 2(c) consists of 5.0×1023 molecules. These molecules have the same r.m.s. speed as the molecules of the gas in 2(c). Calculate the amount of gas. amount of gas=.................................................. mol

[2]
d(ii)

Determine the thermodynamic temperature T0 of the gas. T0=...................................................... K

[2]
d(iii)

The temperature of the gas is now gradually increased. On Figure 2.2, sketch the variation of the r.m.s. speed of the molecules of the gas with thermodynamic temperature.

[3]
3
Figure 1 of 2 for question 3
Figure 2 of 2 for question 3
a

State what is meant by the internal energy of a system.

[2]
b(i)

An ideal gas containing 8.7×1023 molecules is at a temperature of 300 K and a pressure of 2.0×105 Pa. Show that the volume of the gas is 0.018 m3.

[2]
b(ii)

Determine the internal energy of the gas. Explain your reasoning. internal energy=...................................................... J

[3]
c

The gas in 3(b) undergoes a sequence of changes AB, BC and CA in its pressure p and volume V, as shown in Figure 3.1. Determine three conclusions that can be drawn from Figure 3.1 about the individual and overall effects of these changes on the energy of the gas. The conclusions may be qualitative or quantitative, and may refer to any of work done, transfer of thermal energy and change in internal energy. Use the space for any working.

[3]
4

A hydrogen atom consists of a proton and an electron separated by distance 2x, as shown in Figure 4.1. Point P is at the midpoint of the line joining the centres of the proton and the electron.

Figure 1 of 1 for question 4
a(i)

For point P, compare the electric field due to the proton with the electric field due to the electron.

[3]
a(ii)

For point P, compare the electric potential due to the proton with the electric potential due to the electron.

[3]
b(i)

The elementary charge is e and the permittivity of free space is ε0. State an expression, in terms of x, e and ε0, for the magnitude of the electric force F between the proton and the electron. F=.........................................................

[2]
b(ii)

Determine an expression, in terms of F and x, for the electric potential energy EP of the proton and the electron. EP=.........................................................

[1]
5
Figure 1 of 1 for question 5
a

Define the capacitance of a parallel plate capacitor.

[2]
b

Two capacitors X and Y have capacitances CX and CY respectively. The capacitors are connected in parallel to a supply that has electromotive force (e.m.f.) E, as shown in Figure 5.1. Derive the expression for the relationship between CX, CY and the combined capacitance CT of the circuit.

[3]
c

Two capacitors are connected in series to a supply that has e.m.f. 6.00 V. One of the capacitors has capacitance C and the other has capacitance 2C. The total energy stored by the two capacitors is 1.30 mJ. Calculate C in μF. C=.................................................... μF

[3]
6
Figure 1 of 1 for question 6
a

Define magnetic flux density.

[2]
b(i)

Figure 6.1 shows a beam of charged particles in a device known as a mass spectrometer. The shaded region contains a uniform magnetic field that is out of the plane of the page. The particles enter the region of the magnetic field at point X and strike a detector at point Y. Explain whether the sign of the charge on the particles is positive or negative.

[2]
b(ii)

Explain why the path followed by the beam of particles inside the magnetic field is circular.

[2]
c(i)

Each of the particles in 6(b) has a mass of 2.0×10−26 kg and a charge of magnitude 3.2×10−19 C. All of the particles have a speed of 7.3×104 ms−1. The distance XY is 8.2 cm. Show that the centripetal acceleration of one of the particles is 1.3×1011 ms−2.

[2]
c(ii)

Use the information in 6(c)(i) to determine the flux density B of the uniform magnetic field. Give a unit with your answer. B=.................................. unit ...............

[2]
d

The particles in 6(b) are replaced with particles that have half of the charge and a quarter of the mass of the particles in 6(b). The sign of the charge and the speed of the particles are both unchanged. On Figure 6.1, draw the path of this beam of particles.

[2]
7
Figure 1 of 3 for question 7
Figure 2 of 3 for question 7
Figure 3 of 3 for question 7
a(i)

State what is meant by half-wave rectification of an alternating current.

[2]
a(ii)

Figure 7.1 shows an incomplete diagram of the circuit used to achieve half-wave rectification of an alternating input voltage VIN to produce output voltage VOUT. Complete the circuit diagram in Figure 7.1.

[2]
b

Two alternating voltages V1 and V2 have the same peak voltage V0 and the same period T. One is sinusoidal and the other is a square wave, as shown by the variations with time t in Figure 7.2 and Figure 7.3. The alternating voltages are applied in turn to the input terminals in the completed circuit of Figure 7.1. Complete Table 7.1 to show, in terms of V0, expressions for the root-mean-square (r.m.s.) values of the two unrectified and the two half-wave rectified voltages.

[4]
8
a

State what is meant by a photon.

[2]
b(i)

X-rays for use in medical diagnosis are produced by accelerating charged particles towards a metal target. State the name of the charged particles.

[1]
b(ii)

Describe how X-rays are formed at the target.

[2]
b(iii)

For any given accelerating voltage, there is a range of wavelengths of X-rays produced that has a minimum value. Explain, with reference to photons, why there is a range of wavelengths and why this range has a minimum value.

[3]
b(iv)

Calculate the minimum wavelength of X-rays produced when the accelerating voltage is 52.0 kV. minimum wavelength=..................................................... m

[3]
c

X-rays are used to create an image of the internal structure of an object that is made from two types of material of similar thickness. Describe and explain how a property of the materials determines whether or not the image produced has good contrast.

[2]
9
a

Define the mass defect of a nucleus.

[2]
b(i)

Thorium-228 (90228Th) is radioactive. Table 9.1 shows the mass defects of thorium-228 and some other nuclides. Use the data in Table 9.1 to explain why thorium-228 is an α-emitter and not a β− emitter or a β+ emitter.

[4]
b(ii)

Calculate the energy, in MeV, released by the α-decay of one thorium-228 nucleus. Give your answer to three significant figures. energy=................................................. MeV

[3]
10
a

State Wien’s displacement law.

[2]
b(i)

The Sun has a radius of 6.96×108 m and a surface temperature of 5770 K. The Earth is at a distance of 1.52×1011 m from the Sun. Calculate the luminosity of the Sun. luminosity=..................................................... W

[2]
b(ii)

Determine the radiant flux intensity at the Earth of radiation emitted by the Sun. Give a unit with your answer. radiant flux intensity=................................... unit ..............

[3]
b(iii)

Star X has a wavelength of maximum emission of 0.581λs, where λs is the wavelength of maximum emission from the Sun. Calculate the surface temperature of star X. surface temperature=...................................................... K

[1]

Mark scheme

9702/41 · May/June 2026

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1a · 2 marks

  • B1

    work done per unit mass

  • B1

    work done (on mass) moving mass from infinity (to the point)

1b(i) · 2 marks

  • B1

    potential is (defined as) zero at infinity

  • B1

    work is done on (two) masses to move them apart from each other or work is done by (two) masses as they move towards each other

1b(ii) · 2 marks

  • M1

    M = ΦR / G

  • A1

    G is the (universal) gravitational constant

1b(iii) · 1 mark

  • A1

    g0 = Φ / R

1b(iv) · 3 marks

  • B1

    curve in at least one quadrant passing through (±R, ±g0) and (±2R, ± 1/4 g0) and a non-zero value of g at ±3R

  • B1

    curve between –3R and –R of increasing magnitude of gradient and magnitude of g and curve between R and 3R of decreasing magnitude of gradient and magnitude of g

  • B1

    two lines drawn, one in the top left quadrant, the other in the bottom right quadrant

2a · 2 marks

  • M1

    gas which obeys pV ∝ T (under all conditions)

  • A1

    where T is the thermodynamic temperature

2b(i) · 1 mark

  • B1

    measures absolute temperature (unlike liquid in glass) or much greater range of temperature (than liquid in glass) or gives a measurement that does not depend on the properties of a (real) substance

2b(ii) · 1 mark

  • B1

    very small mass of substance or rapidly changing temperatures

2c · 2 marks

  • C1

    357² + 424² + 509² + 286² + 435² (= 837 327)

  • A1

    c0 = √(837327 / 5) = 409 m s⁻¹

2d(i) · 2 marks

  • C1

    N = n × NA

  • A1

    n = (5.0 × 10²³) / (6.02 × 10²³) = 0.83 mol

2d(ii) · 2 marks

  • C1

    ½m <c²> = (3/2) kT

  • A1

    T0 = [(3.34 × 10⁻²⁷) × 409²] / (3 × 1.38 × 10⁻²³) = 13.5 K

2d(iii) · 3 marks

  • B1

    line starting at (T0, c0) with positive gradient at that point

  • B1

    line with decreasing positive gradient

  • B1

    line that passes through (0, 0) when extrapolated back from T = T0

3a · 2 marks

  • B1

    sum of potential energy and kinetic energy

  • B1

    (total) energy of random motion of particles

3b(i) · 2 marks

  • C1

    pV = NkT

  • A1

    V = (8.7 × 10²³ × 1.38 × 10⁻²³ × 300) / (2.0 × 10⁵) = 0.018 m³

3b(ii) · 3 marks

  • B1

    potential energy of molecules is zero

  • C1

    internal energy = (3/2) NkT

  • A1

    = (3/2) × 8.7 × 10²³ × 1.38 × 10⁻²³ × 300 = 5400 J

3c · 3 marks

  • B3

    Any three points from: Up to two points relating to change AB - no change in internal energy - work done = thermal energy transferred - work done on the gas and thermal energy transferred from it Up to two points relating to change BC - internal energy is halved - no work done (on/by gas) - decrease in internal energy = thermal energy removed Up to two points relating to change CA - internal energy is doubled - work done by gas = 1.8 kJ - thermal energy transferred to gas = 4.5 kJ Up to two points relating to the complete cycle ABCA - no overall change in internal energy - overall work is done on gas and thermal energy is removed from it - net work done = net thermal energy transferred

4a(i) · 3 marks

  • B1

    magnitudes (of electric fields) are equal

  • B2

    Any two from: - (electric field) due to proton is to the right - (electric field) due to electron is to the right - directions (of electric fields) are the same

4a(ii) · 3 marks

  • B1

    magnitudes (of electric potentials) are equal

  • B2

    Any two from: - (electric potential) due to proton is positive - (electric potential) due to electron is negative - signs (of electric potentials) are opposite

4b(i) · 2 marks

  • C1

    F = Q1Q2 / 4πε0R²

  • A1

    F = e² / 16πε0x²

4b(ii) · 1 mark

  • A1

    EP = 2Fx

5a · 2 marks

  • M1

    charge / potential (difference)

  • A1

    charge is charge on one plate and potential is p.d. between the plates

5b · 3 marks

  • B1

    E = VX = VY

  • M1

    QT = QX + QY

  • A1

    CTV = CXV + CYV leading to CT = CX + CY or QT / V = QX / V + QY / V leading to CT = CX + CY

5c · 3 marks

  • C1

    E = ½ CV²

  • C1

    CT = (1 / C + 1 / 2C)⁻¹ or 1 / CT = 1 / C + 1 / 2C (CT = 2/3 C)

  • A1

    1.30 × 10⁻³ = ½ × 2/3 C × 6.00² C = 1.08 × 10⁻⁴ F = 108 µF

6a · 2 marks

  • B2

    - force per unit length - force per unit current - length / current perpendicular to field 1 mark for any two bullet points, 2 marks for all three bullet points.

6b(i) · 2 marks

  • B1

    (at X, magnetic) force on particles is to the left

  • B1

    left hand rule tells us the (effective) current is in the direction of movement (of the particles), so charge is positive

6b(ii) · 2 marks

  • B1

    (magnetic) force is always perpendicular to the direction of motion (of the particles)

  • B1

    (magnetic) force changes direction only, and not the speed of the particles or (magnetic) force causes centripetal acceleration

6c(i) · 2 marks

  • C1

    a = v² / r

  • A1

    = (7.3 × 10⁴)² / (0.5 × 8.2 × 10⁻²) = 1.3 × 10¹¹ m s⁻²

6c(ii) · 2 marks

  • C1

    BQv = ma

  • A1

    B = (2.0 × 10⁻²⁶ × 1.3 × 10¹¹) / (3.2 × 10⁻¹⁹ × 7.3 × 10⁴) = 0.11 T

6d · 2 marks

  • B1

    beam shown entering the field at X along same line, and deflecting in same direction in field

  • B1

    circular path with half the radius along the whole length of the curve

7a(i) · 2 marks

  • B1

    allows the current (unhindered) in one direction

  • B1

    blocks the current (completely) in the other direction

7a(ii) · 2 marks

  • B1

    diode shown connecting the wires in one gap, and direct connection across the other gap

  • B1

    correct symbol used for diode

7b · 4 marks

  • B1

    V1 unrectified: r.m.s. voltage shown as V0 / √2

  • B1

    V2 unrectified: r.m.s. voltage shown as V0

  • B1

    V1 rectified: r.m.s. voltage shown as V0 / 2

  • B1

    V2 rectified: r.m.s. voltage shown as V0 / √2

8a · 2 marks

  • M1

    packet / quantum of energy

  • A1

    of electromagnetic radiation

8b(i) · 1 mark

  • B1

    electrons

8b(ii) · 2 marks

  • B1

    electrons are decelerated by target

  • B1

    (some) kinetic energy of electrons becomes energy of photons

8b(iii) · 3 marks

  • B1

    some electron kinetic energy dissipated as thermal energy (in target)

  • B1

    wavelength (of X-rays) is inversely proportional to photon energy

  • B1

    maximum photon energy (and therefore minimum wavelength) when all of the electron kinetic energy becomes the energy of a photon

8b(iv) · 3 marks

  • C1

    hc / λ = qV

  • C1

    λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (1.60 × 10⁻¹⁹ × 52.0 × 10³)

  • A1

    = 2.39 × 10⁻¹¹ m

8c · 2 marks

  • B1

    discussion of differences in (linear) attenuation coefficient

  • B1

    good contrast when the transmitted intensities (through the two materials) are very different

9a · 2 marks

  • M1

    difference between mass of nucleus and mass of constituent nucleons

  • A1

    when separated to infinity

9b(i) · 4 marks

  • B1

    alpha decay identified as producing Ra-224 (plus He-4)

  • B1

    β⁺ decay identified as producing Ac-228 and β⁻ decay as producing Pa-228

  • B1

    alpha decay products calculated to have total mass defect of 1.877217u

  • B1

    reaction only happens if mass defect of products is greater than mass defect of parent nucleus (so decay is alpha and is not beta)

9b(ii) · 3 marks

  • C1

    E = (Δ)mc²

  • C1

    = (1.877217 – 1.871290) × 1.66 × 10⁻²⁷ × (3.00 × 10⁸)²

  • A1

    ( = 8.85 × 10⁻¹³ J) = 5.53 MeV

10a · 2 marks

  • M1

    temperature inversely proportional to wavelength

  • A1

    temperature is thermodynamic temperature of surface, and wavelength is the wavelength at which maximum emission rate occurs

10b(i) · 2 marks

  • C1

    L = 4π × 5.67 × 10⁻⁸ × (6.96 × 10⁸)² × 5770⁴

  • A1

    = 3.83 × 10²⁶ W

10b(ii) · 3 marks

  • C1

    F = L / 4πd²

  • C1

    = 3.83 × 10²⁶ / (4π × (1.52 × 10¹¹)² )

  • A1

    = 1320 W m⁻²

10b(iii) · 1 mark

  • A1

    temperature = 5770 / 0.581 = 9930 K

Questions

  1. 1
    Question 110 marks
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  2. 2
    Question 213 marks
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  3. 3
    Question 310 marks
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  4. 4
    Question 49 marks
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  5. 5
    Question 58 marks
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  6. 6
    Question 612 marks
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  7. 7
    Question 78 marks
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  8. 8
    Question 813 marks
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  9. 9
    Question 99 marks
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  10. 10
    Question 108 marks
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