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A Level H1 Physics Modern Physics Quiz

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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]

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]

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

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]

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:

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

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

(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]

(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]

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]

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]

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]