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A Level H1 Physics Modern Physics Quiz

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Questions

A-Level Physics H1 Quiz - Modern Physics

Name: __________________________
Class: __________________________
Date: __________________________
Score: _______ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working clearly. Numerical answers should be given to an appropriate number of significant figures.
The use of an approved scientific calculator is expected.

Section A: Multiple Choice & Short Concepts (Questions 1-5)

1. Which of the following statements correctly describes the photon model of light?
[1]

A. Light energy is distributed continuously across the wavefront.
B. The energy of a photon is proportional to the square of the frequency.
C. Light interacts with matter in discrete packets of energy.
D. The intensity of light determines the kinetic energy of emitted electrons.

2. A metal surface has a work function of $4.0 \text{ eV}$ . Which of the following photons will not cause photoelectric emission from this surface?
[1]

A. A photon with energy $4.5 \text{ eV}$
B. A photon with wavelength $250 \text{ nm}$
C. A photon with frequency $1.0 \times 10^{15} \text{ Hz}$
D. A photon with energy $3.8 \text{ eV}$

(Use $h = 6.63 \times 10^{-34} \text{ J s}$ , $c = 3.00 \times 10^8 \text{ m s}^{-1}$ , $1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$ )

3. In a photoelectric experiment, the intensity of the incident monochromatic light is doubled while keeping the frequency constant. Which of the following quantities remains unchanged?
[1]

A. The photocurrent
B. The number of photoelectrons emitted per second
C. The maximum kinetic energy of the photoelectrons
D. The total energy incident on the surface per second

4. Define the term work function.
[1]

5. State the significance of the threshold frequency in the photoelectric effect.
[1]

Section B: Core Concepts & Calculations (Questions 6-10)

6. Explain why the classical wave theory of light fails to explain the existence of a threshold frequency.
[2]

7. Calculate the energy of a photon of green light with a wavelength of $550 \text{ nm}$ . Give your answer in Joules.
[2]

8. The graph below shows the variation of maximum kinetic energy $E_{max}$ of photoelectrons with the frequency $f$ of incident radiation for a specific metal.

(Imagine a linear graph starting from $f_0$ on the x-axis, with positive slope)

State what the gradient of this graph represents.
[1]

9. A zinc plate has a work function of $4.3 \text{ eV}$ . Ultraviolet light of wavelength $200 \text{ nm}$ is incident on the plate. Calculate the frequency of the incident ultraviolet light.
[2]

10. Using the data from Question 9, determine the maximum kinetic energy of the emitted photoelectrons in Joules.
[3]

Section C: Structured Problems & Nuclear Physics (Questions 11-15)

11. If the intensity of the ultraviolet light in Question 9 is increased, state and explain the effect on the maximum kinetic energy of the photoelectrons.
[1]

12. In an experiment to determine Planck’s constant, a student measures the stopping potential $V_s$ for different frequencies $f$ of incident light. Write down the equation relating $V_s$ , $f$ , the work function $\Phi$ , and fundamental constants.
[2]

13. Explain how Planck’s constant $h$ can be determined from the gradient of a graph of stopping potential $V_s$ against frequency $f$ .
[2]

14. The graph of $V_s$ against $f$ intersects the frequency axis at $f_0 = 5.5 \times 10^{14} \text{ Hz}$ . Calculate the work function of the metal in electron-volts (eV).
[3]

15. Consider the nuclear reaction where a neutron collides with a Uranium-235 nucleus:

$^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3^{1}_{0}\text{n} + \text{energy}$

Name this type of nuclear reaction.
[1]

Section D: Data Analysis & Applications (Questions 16-20)

16. Explain, in terms of binding energy per nucleon, why energy is released in the reaction described in Question 15.
[2]

17. The mass defect for the reaction in Question 15 is $3.0 \times 10^{-28} \text{ kg}$ . Calculate the energy released in Joules.
[2]

18. A radioactive source emits $\alpha$ -particles. Describe the composition of an $\alpha$ -particle and state one property regarding its ionizing ability or penetrating power.
[2]

Composition: ______________________________________________________________
Property: __________________________________________________________________

19. The table below shows the results of a photoelectric effect experiment using a sodium surface.

Frequency $f$ ( $10^{14} \text{ Hz}$ )	Stopping Potential $V_s$ (V)
5.0	0.0
6.0	0.4
7.0	0.8
8.0	1.2
9.0	1.6

Use the data to determine the threshold frequency $f_0$ for sodium.
[1]

$f_0 =$ __________________________ Hz

20. A certain isotope has a half-life of 10 days. If the initial activity is $800 \text{ Bq}$ , calculate the activity after 30 days.
[2]

End of Quiz

Answers

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

Total Marks: 40

Section A: Multiple Choice & Short Concepts

1. C [1]
Reasoning: The photon model posits that light energy is quantized into discrete packets (photons). A is wave theory, B is incorrect ( $E=hf$ ), D is incorrect (frequency determines KE).

2. D [1]
Reasoning: For emission, Photon Energy $E \ge \Phi$ .
$\Phi = 4.0 \text{ eV}$ .
A: $4.5 > 4.0$ (Emits).
B: $E = \frac{hc}{\lambda} = \frac{1240 \text{ eV nm}}{250 \text{ nm}} \approx 4.96 \text{ eV} > 4.0$ (Emits).
C: $E = hf = (4.14 \times 10^{-15} \text{ eV s})(1.0 \times 10^{15} \text{ Hz}) = 4.14 \text{ eV} > 4.0$ (Emits).
D: $3.8 < 4.0$ (Does not emit).

3. C [1]
Reasoning: $KE_{max} = hf - \Phi$ . It depends only on frequency and work function, not intensity. Intensity affects the number of photons, and thus the photocurrent (A, B) and total energy (D).

4. Definition: [1]
The minimum energy required to remove an electron from the surface of a metal.
(Accept: "Minimum energy needed to liberate an electron from the metal surface.")

5. Significance: [1]
The minimum frequency of incident radiation required to eject electrons from the metal surface. Below this frequency, no emission occurs regardless of intensity.

Section B: Core Concepts & Calculations

6. Wave Theory Failure: [2]
[B1] In the wave model, energy is distributed continuously over the wavefront and accumulates over time.
[B1] Therefore, even low-frequency light should eventually provide enough energy to eject electrons if the intensity is high enough or exposure time is long enough. It cannot explain why there is a specific frequency cutoff below which no emission occurs instantly.

7. Calculation: [2]
$E = \frac{hc}{\lambda}$ [M1]
$E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{550 \times 10^{-9}}$
$E = 3.62 \times 10^{-19} \text{ J}$ [A1]

8. Gradient Meaning: [1]
Planck’s constant ( $h$ ).
(Note: Graph is $E_{max}$ vs $f$ . Equation: $E_{max} = hf - \Phi$ . Gradient = $h$ .)

9. Frequency Calculation: [2]
$f = \frac{c}{\lambda}$ [M1]
$f = \frac{3.00 \times 10^8}{200 \times 10^{-9}} = 1.50 \times 10^{15} \text{ Hz}$ [A1]

10. Max Kinetic Energy: [3]
Work Function $\Phi = 4.3 \text{ eV} = 4.3 \times 1.60 \times 10^{-19} = 6.88 \times 10^{-19} \text{ J}$ .
Photon Energy $E = hf = (6.63 \times 10^{-34})(1.50 \times 10^{15}) = 9.945 \times 10^{-19} \text{ J}$ .
$KE_{max} = E - \Phi$ [M1]
$KE_{max} = 9.945 \times 10^{-19} - 6.88 \times 10^{-19}$
$KE_{max} = 3.065 \times 10^{-19} \text{ J}$ [A1]
(Accept $3.1 \times 10^{-19} \text{ J}$ ) [A1 for unit/sig fig]

Section C: Structured Problems & Nuclear Physics

11. Effect on Max KE: [1]
Remains unchanged. KE depends on frequency ( $hf$ ) and work function, not intensity.

12. Equation: [2]
$eV_s = hf - \Phi$ [M1]
Or $V_s = (\frac{h}{e})f - \frac{\Phi}{e}$ [A1]

13. Gradient Explanation: [2]
[B1] The equation is in the form $y = mx + c$ , where $y=V_s$ and $x=f$ .
[B1] The gradient $m$ is equal to $\frac{h}{e}$ . Therefore, $h = \text{gradient} \times e$ .

14. Work Function Calculation: [3]
At threshold frequency $f_0$ , $KE_{max} = 0$ , so $hf_0 = \Phi$ . [M1]
$\Phi = (6.63 \times 10^{-34})(5.5 \times 10^{14})$
$\Phi = 3.6465 \times 10^{-19} \text{ J}$ [M1]
Convert to eV: $\frac{3.6465 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.28 \text{ eV}$ [A1]
(Accept $2.3 \text{ eV}$ )

15. Name: [1]
Nuclear Fission.

Section D: Data Analysis & Applications

16. Binding Energy Explanation: [2]
[B1] The binding energy per nucleon of the products (Ba and Kr) is greater than that of the reactant (U-235).
[B1] This means the products are more stable, and the difference in binding energy is released as kinetic energy/radiation. (Or: Mass of products < Mass of reactants, mass difference converted to energy).

17. Energy Released: [2]
$E = \Delta m c^2$ [M1]
$E = (3.0 \times 10^{-28})(3.00 \times 10^8)^2$
$E = 2.7 \times 10^{-11} \text{ J}$ [A1]

18. Alpha Radiation: [2]
Composition: 2 protons and 2 neutrons (or a Helium nucleus, $^4_2\text{He}$ ). [B1]
Property: High ionizing ability OR Low penetrating power (stopped by paper/skin). [B1]

19. Threshold Frequency: [1]
From data/table, x-intercept where $V_s = 0$ .
$f_0 = 5.0 \times 10^{14} \text{ Hz}$ .

20. Radioactive Decay: [2]
Number of half-lives $n = \frac{30}{10} = 3$ . [M1]
$A = A_0 (\frac{1}{2})^n = 800 (\frac{1}{2})^3 = 800 \times \frac{1}{8} = 100 \text{ Bq}$ [A1]