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A Level H2 Physics Modern Physics Quiz
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Questions
A-Level Physics H2 Quiz - Modern Physics
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50
Duration: 1 hour 15 minutes Total Marks: 50
Instructions:
- Answer ALL questions in the spaces provided.
- Show all working for calculation questions.
- Use appropriate significant figures in final answers.
- Where explanations are required, use clear and concise language.
- Data: speed of light c = 3.00 × 10⁸ m s⁻¹, Planck constant h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.60 × 10⁻¹⁹ J, 1 u = 931.5 MeV/c²
Section A: Quantum Physics (Questions 1–7)
1. State what is meant by the photoelectric effect.
[2 marks]
2. A metal surface has a work function of 2.30 eV. Calculate the threshold frequency for this metal.
[2 marks]
3. Radiation of wavelength 350 nm is incident on the metal surface in Question 2.
(a) Calculate the maximum kinetic energy, in joules, of the emitted photoelectrons.
[3 marks]
(b) Calculate the stopping potential required to reduce the photoelectric current to zero.
[2 marks]
4. In a photoelectric experiment, the intensity of the incident radiation is kept constant while the wavelength is reduced. State and explain what happens to:
(a) the maximum photoelectric current.
[2 marks]
(b) the stopping potential.
[2 marks]
5. Explain why there is a continuous distribution of wavelengths in the X-ray spectrum produced by an X-ray tube.
[2 marks]
6. Electrons are accelerated through a potential difference of 25 kV in an X-ray tube. Calculate the minimum wavelength of the X-rays produced.
[3 marks]
7. With reference to energy levels, explain why electrons bombarding a target metal must have a minimum energy to produce characteristic X-rays.
[3 marks]
Section B: Nuclear Physics (Questions 8–15)
8. State what is meant by the binding energy of a nucleus.
[2 marks]
9. The mass of a helium-4 nucleus (²⁴He) is 4.002603 u. Given:
- mass of proton = 1.007276 u
- mass of neutron = 1.008665 u
Calculate:
(a) the mass defect of the helium-4 nucleus.
[2 marks]
(b) the binding energy of the helium-4 nucleus in MeV.
[2 marks]
(c) the binding energy per nucleon of helium-4.
[1 mark]
10. Explain why the binding energy per nucleon is a measure of the stability of a nucleus.
[2 marks]
11. A radioactive source has a half-life of 8.0 days. Its initial activity is 480 Bq.
(a) Define the term half-life.
[1 mark]
(b) Calculate the activity of the source after 24 days.
[2 marks]
12. The decay constant λ of a radioactive isotope is 0.0866 day⁻¹. Calculate its half-life in days.
[2 marks]
13. Uranium-238 undergoes alpha decay to form thorium-234.
(a) Write a balanced nuclear equation for this decay.
[2 marks]
(b) The mass of uranium-238 is 238.05079 u, the mass of thorium-234 is 234.04363 u, and the mass of an alpha particle is 4.00260 u. Calculate the total kinetic energy released in this decay.
[3 marks]
14. A sample of ancient wood contains carbon-14 with an activity of 0.15 Bq per gram of carbon. Living wood has an activity of 0.23 Bq per gram of carbon. The half-life of carbon-14 is 5730 years.
Calculate the age of the ancient wood.
[3 marks]
15. Explain the principles behind radioactive dating using carbon-14, and state one assumption made in this method.
[3 marks]
Section C: Lasers and Semiconductors (Questions 16–20)
16. State three properties of laser light that distinguish it from light from a conventional source.
[3 marks]
17. Explain the process of stimulated emission and why it is essential for laser operation.
[3 marks]
18. With the aid of a labelled energy level diagram, explain the principle of population inversion in a laser.
[4 marks]
19. Distinguish between an intrinsic semiconductor and an extrinsic semiconductor.
[2 marks]
20. Explain how a p-n junction diode can function as a rectifier. Include reference to the behaviour of charge carriers under forward and reverse bias.
[4 marks]
END OF QUIZ
Answers
A-Level Physics H2 Quiz - Modern Physics — Answer Key
Total Marks: 50
Section A: Quantum Physics (Questions 1–7)
1. State what is meant by the photoelectric effect. [2 marks]
Answer: The photoelectric effect is the emission of electrons from a metal surface [1 mark] when electromagnetic radiation of sufficiently high frequency (or sufficiently short wavelength) is incident on it [1 mark].
Accept: emission of photoelectrons when light above the threshold frequency strikes a metal surface.
2. Calculate the threshold frequency for a metal with work function 2.30 eV. [2 marks]
Answer: Φ = 2.30 eV = 2.30 × 1.60 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J [1 mark]
f₀ = Φ / h = 3.68 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 5.55 × 10¹⁴ Hz [1 mark]
Award [1] for correct conversion to joules, [1] for correct frequency. Accept 5.6 × 10¹⁴ Hz.
3. Radiation of wavelength 350 nm incident on the metal in Q2. [5 marks total]
(a) Calculate maximum kinetic energy in joules. [3 marks]
Answer: E = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (350 × 10⁻⁹) = 5.68 × 10⁻¹⁹ J [1 mark]
K_max = E - Φ = 5.68 × 10⁻¹⁹ - 3.68 × 10⁻¹⁹ = 2.00 × 10⁻¹⁹ J [1 mark]
Award [1] for photon energy calculation, [1] for subtraction, [1] for correct answer with unit. Accept 2.0 × 10⁻¹⁹ J.
(b) Calculate the stopping potential. [2 marks]
Answer: eV_s = K_max [1 mark]
V_s = K_max / e = 2.00 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.25 V [1 mark]
Accept 1.3 V.
4. Intensity constant, wavelength reduced. [4 marks total]
(a) Maximum photoelectric current. [2 marks]
Answer: The maximum photoelectric current remains approximately constant (or decreases slightly) [1 mark]. Since intensity is constant, the number of photons incident per second remains constant (or decreases slightly as each photon has higher energy). Each photon still liberates at most one electron, so the number of photoelectrons emitted per second is unchanged (or slightly reduced) [1 mark].
(b) Stopping potential. [2 marks]
Answer: The stopping potential increases [1 mark]. Reducing wavelength increases frequency, so photon energy hf increases. Since K_max = hf - Φ, the maximum kinetic energy increases, requiring a larger stopping potential to reduce the current to zero [1 mark].
5. Explain continuous distribution of wavelengths in X-ray spectrum. [2 marks]
Answer: High-speed electrons collide with target atoms and are decelerated [1 mark]. Different electrons lose different amounts of kinetic energy in these collisions, producing photons with a range of energies and therefore a continuous range of wavelengths [1 mark].
Accept reference to Bremsstrahlung (braking radiation).
6. Calculate minimum wavelength for 25 kV X-ray tube. [3 marks]
Answer: Energy of electrons = eV = 1.60 × 10⁻¹⁹ × 25 × 10³ = 4.00 × 10⁻¹⁵ J [1 mark]
For minimum wavelength, all electron energy converted to one photon: E = hc / λ_min [1 mark]
λ_min = hc / E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (4.00 × 10⁻¹⁵) = 4.97 × 10⁻¹¹ m [1 mark]
Accept 5.0 × 10⁻¹¹ m or 0.050 nm.
7. Explain minimum energy for characteristic X-rays. [3 marks]
Answer: Characteristic X-rays are produced when incident electrons knock out inner shell electrons (e.g., from the K-shell) of target atoms [1 mark]. The minimum energy required equals the binding energy (ionization energy) of that inner shell electron [1 mark]. When outer electrons fall into the vacancy, photons of specific energies (characteristic of the target element) are emitted [1 mark].
Section B: Nuclear Physics (Questions 8–15)
8. State what is meant by binding energy of a nucleus. [2 marks]
Answer: The binding energy of a nucleus is the energy required to completely separate the nucleus into its constituent protons and neutrons [1 mark]. It is equal to the energy equivalent of the mass defect, given by ΔE = Δmc² [1 mark].
9. Helium-4 nucleus calculations. [5 marks total]
(a) Mass defect. [2 marks]
Answer: Mass of constituents = 2(1.007276) + 2(1.008665) = 2.014552 + 2.017330 = 4.031882 u [1 mark]
Mass defect = 4.031882 - 4.002603 = 0.029279 u [1 mark]
(b) Binding energy in MeV. [2 marks]
Answer: BE = 0.029279 × 931.5 [1 mark] = 27.3 MeV [1 mark]
Accept 27.2–27.3 MeV.
(c) Binding energy per nucleon. [1 mark]
Answer: BE per nucleon = 27.3 / 4 = 6.83 MeV/nucleon [1 mark]
10. Explain why binding energy per nucleon measures stability. [2 marks]
Answer: A higher binding energy per nucleon indicates that more energy is required to remove a nucleon from the nucleus [1 mark]. Therefore, nuclei with higher binding energy per nucleon are more stable against disintegration [1 mark].
11. Radioactive source with half-life 8.0 days, initial activity 480 Bq. [3 marks total]
(a) Define half-life. [1 mark]
Answer: Half-life is the time taken for the activity (or number of radioactive nuclei) of a radioactive source to decrease to half its initial value [1 mark].
(b) Activity after 24 days. [2 marks]
Answer: Number of half-lives = 24 / 8.0 = 3 [1 mark]
Activity = 480 × (½)³ = 480 × ⅛ = 60 Bq [1 mark]
12. Calculate half-life from decay constant λ = 0.0866 day⁻¹. [2 marks]
Answer: t₁/₂ = ln 2 / λ [1 mark] = 0.693 / 0.0866 = 8.00 days [1 mark]
Accept 8.0 days.
13. Alpha decay of uranium-238. [5 marks total]
(a) Balanced nuclear equation. [2 marks]
Answer: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He [2 marks]
Award [1] for correct mass numbers, [1] for correct atomic numbers.
(b) Total kinetic energy released. [3 marks]
Answer: Mass defect = 238.05079 - (234.04363 + 4.00260) = 238.05079 - 238.04623 = 0.00456 u [1 mark]
Energy released = 0.00456 × 931.5 [1 mark] = 4.25 MeV [1 mark]
Accept 4.2–4.3 MeV.
14. Calculate age of ancient wood. [3 marks]
Answer: A = A₀ e^(-λt) or A = A₀ (½)^(t/t₁/₂) [1 mark]
0.15 = 0.23 × (½)^(t/5730)
(½)^(t/5730) = 0.15/0.23 = 0.6522 [1 mark]
t/5730 = log(0.6522) / log(0.5) = -0.1856 / -0.3010 = 0.6167
t = 0.6167 × 5730 = 3530 years [1 mark]
Accept 3500–3600 years depending on rounding. Award method marks for correct approach.
15. Explain carbon-14 dating principles and one assumption. [3 marks]
Answer: Carbon-14 is produced in the upper atmosphere by cosmic rays and is absorbed by living organisms, maintaining a constant ratio of C-14 to C-12 while alive [1 mark]. When the organism dies, C-14 intake stops and the existing C-14 decays with a half-life of 5730 years. By measuring the remaining C-14 activity and comparing it to the activity in living organisms, the time since death can be calculated [1 mark].
Assumption: The concentration of C-14 in the atmosphere has remained constant over time (or the initial activity in the sample was the same as in living organisms today) [1 mark].
Accept other valid assumptions: no contamination of the sample, closed system since death.
Section C: Lasers and Semiconductors (Questions 16–20)
16. Three properties of laser light. [3 marks]
Answer: Any three from:
- Monochromatic (single wavelength/frequency) [1 mark]
- Coherent (waves are in phase) [1 mark]
- Highly directional (low divergence / collimated beam) [1 mark]
- High intensity (concentrated energy) [1 mark]
Award [1] each for any three distinct properties.
17. Explain stimulated emission and its importance. [3 marks]
Answer: Stimulated emission occurs when an incident photon of energy equal to the energy difference between two levels interacts with an atom in the excited state, causing it to drop to the lower energy level and emit a second photon [1 mark]. The emitted photon has the same energy, phase, direction, and polarization as the incident photon [1 mark]. This process is essential for laser operation because it produces amplification of light (Light Amplification by Stimulated Emission of Radiation), creating a coherent, intense beam [1 mark].
18. Population inversion with energy level diagram. [4 marks]
Answer: Diagram should show:
- Ground state and at least one excited state (or metastable state) [1 mark]
- Higher population in the upper (metastable) state than in the lower state [1 mark]
Explanation: Population inversion is a condition where more atoms exist in an excited (metastable) state than in the ground (or lower) state [1 mark]. This is achieved by pumping (optical or electrical) atoms to a higher energy level, from which they rapidly decay to a metastable state. The metastable state has a relatively long lifetime, allowing population to build up. When population inversion is achieved, stimulated emission dominates over absorption, enabling laser amplification [1 mark].
Award marks for correct diagram with labels and clear explanation of non-equilibrium condition.
19. Distinguish between intrinsic and extrinsic semiconductors. [2 marks]
Answer: An intrinsic semiconductor is a pure semiconductor material (e.g., pure silicon or germanium) where electrical conductivity is due to thermally generated electron-hole pairs [1 mark]. An extrinsic semiconductor is a semiconductor that has been doped with impurity atoms to increase its conductivity; it can be n-type (doped with donor atoms providing extra electrons) or p-type (doped with acceptor atoms creating holes) [1 mark].
20. Explain p-n junction diode as a rectifier. [4 marks]
Answer: A p-n junction diode allows current to flow in only one direction [1 mark].
Under forward bias (p-side connected to positive terminal, n-side to negative): The external voltage reduces the depletion region width. Holes from the p-side and electrons from the n-side are pushed toward the junction, where they recombine, allowing a significant current to flow [1 mark].
Under reverse bias (p-side negative, n-side positive): The external voltage widens the depletion region. Charge carriers are pulled away from the junction, and only a very small leakage current flows due to minority carriers [1 mark].
This one-way conduction property allows the diode to convert alternating current (AC) to direct current (DC), functioning as a rectifier [1 mark].
Award marks for clear explanation of both bias conditions and the rectification principle.
END OF ANSWER KEY