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A Level H1 Physics Modern Physics Quiz

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A Level H1 Physics AI Generated Generated by Qwen3.6 Plus Updated 2026-08-17

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Questions

A-Level Physics H1 Quiz - Modern Physics

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 45

Duration: 45 minutes
Total Marks: 45

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working clearly. Marks are awarded for correct reasoning and steps, not just the final answer.
Use $g = 9.81 \text{ m s}^{-2}$ , $c = 3.00 \times 10^8 \text{ m s}^{-1}$ , $h = 6.63 \times 10^{-34} \text{ J s}$ , and $e = 1.60 \times 10^{-19} \text{ C}$ where appropriate.
The data booklet may be consulted.

Section A: Photoelectric Effect (Questions 1–8)

1. Define the term work function of a metal.
[1]

2. State one experimental observation of the photoelectric effect that cannot be explained by the classical wave theory of light.
[1]

3. A metal surface has a work function of $3.2 \times 10^{-19} \text{ J}$ . Calculate the threshold frequency for this metal.
[2]

Answer space

4. Ultraviolet light of wavelength $250 \text{ nm}$ is incident on the metal surface in Question 3.
(a) Calculate the energy of a single photon of this light in Joules.
[2]

Answer space

(b) Determine the maximum kinetic energy of the emitted photoelectrons.
[2]

Answer space

5. Explain why increasing the intensity of the incident light (while keeping the frequency constant) increases the photoelectric current but does not change the maximum kinetic energy of the photoelectrons.
[3]

6. In a photoelectric experiment, the stopping potential $V_s$ is measured for different frequencies $f$ of incident light. A graph of $V_s$ against $f$ is plotted.
(a) State the physical significance of the gradient of this graph.
[1]

(b) State the physical significance of the intercept on the frequency axis ( $x$ -intercept).
[1]

7. The graph of $V_s$ against $f$ for a specific metal has a gradient of $4.14 \times 10^{-15} \text{ V s}$ .
Calculate the value of Planck’s constant $h$ derived from this gradient.
[2]

Answer space

8. Two different metals, X and Y, are illuminated with light of the same frequency $f$ , where $f$ is greater than the threshold frequency for both metals. Metal X has a larger work function than Metal Y.
Compare the maximum kinetic energy of photoelectrons emitted from X and Y. Explain your answer.
[2]

Section B: Atomic Energy Levels (Questions 9–14)

9. Explain what is meant by the term ionisation energy of an atom.
[1]

10. The diagram below shows three energy levels of a hydrogen atom:

Image pending generation for this question.

$n=3$ : $-1.51 \text{ eV}$
$n=2$ : $-3.40 \text{ eV}$
$n=1$ : $-13.6 \text{ eV}$

(a) Calculate the wavelength of the photon emitted when an electron transitions from $n=3$ to $n=2$ .
[3]

Answer space

(b) State which region of the electromagnetic spectrum this photon belongs to.
[1]

11. An electron in the ground state ( $n=1$ ) of a hydrogen atom absorbs a photon with energy $12.1 \text{ eV}$ .
(a) Determine the final energy level ( $n$ ) of the electron.
[2]

Answer space

(b) Explain why a photon with energy $11.0 \text{ eV}$ would not be absorbed by the ground-state electron.
[1]

12. Describe the difference between excitation and ionisation of an atom.
[2]

13. A gas discharge tube emits a line spectrum rather than a continuous spectrum.
Explain how the existence of line spectra provides evidence for discrete atomic energy levels.
[3]

14. Calculate the minimum frequency of radiation required to ionise a hydrogen atom initially in the $n=2$ state. (Energy of $n=2$ state is $-3.40 \text{ eV}$ ).
[2]

Answer space

Section C: Nuclear Physics (Questions 15–20)

15. Define the term isotope.
[1]

16. A radioactive isotope has a half-life of $12.0 \text{ hours}$ . The initial activity of a sample is $8000 \text{ Bq}$ .
Calculate the activity of the sample after $48.0 \text{ hours}$ .
[2]

Answer space

17. Explain why the mass of a stable nucleus is less than the sum of the masses of its constituent protons and neutrons.
[2]

18. Consider the following nuclear fission reaction: $^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3^{1}_{0}\text{n}$ (a) Verify that the nucleon number (mass number) is conserved in this reaction.
[1]

(b) Explain why energy is released in this reaction.
[2]

19. The binding energy per nucleon for $^{56}\text{Fe}$ is approximately $8.8 \text{ MeV}$ , while for $^{235}\text{U}$ it is approximately $7.6 \text{ MeV}$ .
Using these values, explain why energy is released when Uranium-235 undergoes fission to form lighter nuclei like Iron (conceptually, though Fe is not the direct product, it represents higher stability).
[2]

20. A student suggests that because radioactive decay is random, it is impossible to predict when a specific nucleus will decay, and therefore impossible to predict the activity of a large sample.
Evaluate this statement.
[2]

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Answers

A-Level Physics H1 Quiz - Modern Physics (Answer Key)

Total Marks: 45

Section A: Photoelectric Effect

1. [1 mark]

The minimum energy required to remove an electron from the surface of a metal.
Accept: Energy required to release an electron from the metal surface.

2. [1 mark]

Any one of the following:
- Existence of a threshold frequency (no emission below $f_0$ regardless of intensity).
- Immediate emission of electrons (no time lag).
- Maximum kinetic energy depends on frequency, not intensity.

3. [2 marks]

Formula: $\Phi = h f_0$ [M1]
Calculation: $f_0 = \frac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.83 \times 10^{14} \text{ Hz}$ [A1]

4. (a) [2 marks]

Formula: $E = \frac{hc}{\lambda}$ [M1]
Calculation: $E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{250 \times 10^{-9}} = 7.96 \times 10^{-19} \text{ J}$ [A1]

(b) [2 marks]

Formula: $K_{max} = E_{photon} - \Phi$ [M1]
Calculation: $K_{max} = 7.96 \times 10^{-19} - 3.2 \times 10^{-19} = 4.76 \times 10^{-19} \text{ J}$ [A1]

5. [3 marks]

Intensity is proportional to the number of photons incident per unit time. [B1]
One photon interacts with one electron (1:1 interaction). More photons mean more electrons emitted per second, hence higher current. [B1]
The energy of each photon ( $hf$ ) remains unchanged, so the energy transferred to each electron is unchanged. Thus, $K_{max}$ remains constant. [B1]

6. (a) [1 mark]

$h/e$ (Planck’s constant divided by elementary charge).

(b) [1 mark]

Threshold frequency ( $f_0$ ).

7. [2 marks]

Gradient $= h/e$ [M1]
$h = \text{Gradient} \times e = 4.14 \times 10^{-15} \times 1.60 \times 10^{-19} = 6.62 \times 10^{-34} \text{ J s}$ [A1]
Note: Accept $6.6 \times 10^{-34}$ to $6.63 \times 10^{-34}$ .

8. [2 marks]

$K_{max} = hf - \Phi$ . [B1]
Since $hf$ is constant and $\Phi_X > \Phi_Y$ , then $K_{max}(X) < K_{max}(Y)$ . The electrons from X have lower maximum kinetic energy. [B1]

Section B: Atomic Energy Levels

9. [1 mark]

The minimum energy required to remove an electron from the ground state of an atom to infinity (completely free).

10. (a) [3 marks]

Energy difference: $\Delta E = E_3 - E_2 = -1.51 - (-3.40) = 1.89 \text{ eV}$ [M1]
Convert to Joules: $1.89 \times 1.60 \times 10^{-19} = 3.024 \times 10^{-19} \text{ J}$ [M1]
Wavelength: $\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.024 \times 10^{-19}} = 6.59 \times 10^{-7} \text{ m}$ ( $659 \text{ nm}$ ) [A1]

(b) [1 mark]

Visible (Red).

11. (a) [2 marks]

Energy of level $n$ : $E_n = E_1 + 12.1 = -13.6 + 12.1 = -1.5 \text{ eV}$ [M1]
This corresponds to $n=3$ (since $E_3 = -1.51 \text{ eV} \approx -1.5 \text{ eV}$ ). [A1]

(b) [1 mark]

Energy levels are discrete/quantised. There is no energy level at $-13.6 + 11.0 = -2.6 \text{ eV}$ . The photon energy does not match any transition difference.

12. [2 marks]

Excitation: Electron moves to a higher bound energy level within the atom. [B1]
Ionisation: Electron is removed completely from the atom (moves to $E \ge 0$ ). [B1]

13. [3 marks]

Electrons can only exist in specific, discrete energy levels. [B1]
Photons are emitted only when electrons transition between these specific levels. [B1]
Therefore, only photons with specific energies (and thus specific frequencies/wavelengths) are emitted, creating lines rather than a continuous range. [B1]

14. [2 marks]

Energy required to ionise from $n=2$ : $\Delta E = 0 - (-3.40) = 3.40 \text{ eV}$ . [M1]
$f = \frac{E}{h} = \frac{3.40 \times 1.60 \times 10^{-19}}{6.63 \times 10^{-34}} = 8.17 \times 10^{14} \text{ Hz}$ [A1]

Section C: Nuclear Physics

15. [1 mark]

Atoms of the same element (same proton number) with different numbers of neutrons (different nucleon numbers).

16. [2 marks]

Number of half-lives: $n = \frac{48.0}{12.0} = 4$ . [M1]
Activity: $A = \frac{A_0}{2^n} = \frac{8000}{2^4} = \frac{8000}{16} = 500 \text{ Bq}$ . [A1]

17. [2 marks]

When nucleons combine to form a nucleus, energy is released (binding energy). [B1]
By mass-energy equivalence ( $E=mc^2$ ), this loss of energy corresponds to a loss of mass (mass defect). [B1]

18. (a) [1 mark]

LHS: $235 + 1 = 236$ . RHS: $141 + 92 + 3(1) = 236$ . Conserved.

(b) [2 marks]

The total mass of the products is less than the total mass of the reactants. [B1]
This mass difference (mass defect) is converted into energy according to $E=mc^2$ . [B1]

19. [2 marks]

Fe-56 has a higher binding energy per nucleon than U-235, meaning it is more stable. [B1]
When U-235 splits into lighter, more tightly bound nuclei, the total binding energy of the system increases. This increase in binding energy is released as kinetic energy/radiation. [B1]

20. [2 marks]

The first part is correct: Decay of a single nucleus is random and unpredictable. [B1]
The second part is incorrect: For a large sample, the statistical behavior is predictable. The activity follows the exponential decay law ( $A = A_0 e^{-\lambda t}$ ) reliably. [B1]