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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: ______ / 60
Duration: 1 hour 15 minutes
Total Marks: 60
Instructions:
- Answer ALL questions in the spaces provided.
- Show all working clearly for calculation questions.
- Use appropriate units and significant figures.
- Where explanations are required, give clear, concise answers.
- g = 9.81 m s⁻², c = 3.00 × 10⁸ m s⁻¹, h = 6.63 × 10⁻³⁴ J s, e = 1.60 × 10⁻¹⁹ C, 1 u = 931.5 MeV/c²
Section A: Quantum Physics (Questions 1–6)
Total: 18 marks
1. (a) State what is meant by the term work function of a metal.
[2 marks]
(b) Explain why, in the photoelectric effect, electrons are emitted only when the frequency of incident light exceeds a certain threshold value, regardless of the intensity of the light.
[2 marks]
2. A clean sodium surface has a work function of 2.3 eV. It is illuminated with ultraviolet light of wavelength 250 nm.
(a) Calculate the energy, in joules, of a single photon of this ultraviolet light.
[2 marks]
(b) Determine the maximum kinetic energy, in eV, of the photoelectrons emitted from the sodium surface.
[2 marks]
(c) Calculate the stopping potential required to prevent the most energetic photoelectrons from reaching the collector.
[1 mark]
3. In a photoelectric experiment, monochromatic light of frequency 8.0 × 10¹⁴ Hz is incident on a metal surface. The stopping potential is measured to be 0.85 V.
(a) Write down Einstein's photoelectric equation, defining each symbol.
[2 marks]
(b) Use the data to calculate the work function of the metal in eV.
[2 marks]
4. A beam of light of constant intensity and wavelength 450 nm is directed onto a photocell. The wavelength is then reduced to 350 nm while the intensity is kept constant.
State and explain the effect, if any, on:
(a) the maximum kinetic energy of the emitted photoelectrons;
[1 mark]
(b) the photoelectric current.
[2 marks]
Section B: Wave-Particle Duality and Atomic Spectra (Questions 5–10)
Total: 18 marks
5. (a) State the de Broglie relationship, defining each symbol.
[1 mark]
(b) Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 150 V.
[3 marks]
6. In an electron diffraction experiment, a beam of electrons is accelerated through a potential difference of 4.0 kV and directed at a thin polycrystalline graphite film. A pattern of concentric rings is observed on a fluorescent screen.
(a) Explain why the pattern of rings is observed.
[2 marks]
(b) The diameter of the first ring is observed to decrease when the accelerating voltage is increased. Explain this observation.
[2 marks]
7. The diagram below represents some of the energy levels of a hydrogen atom.
Energy/eV
0 ──────────────────── n = ∞
-0.54 ──────────────── n = 5
-0.85 ──────────────── n = 4
-1.51 ──────────────── n = 3
-3.40 ──────────────── n = 2
-13.60 ──────────────── n = 1 (ground state)
(a) Calculate the wavelength of the photon emitted when an electron makes a transition from the n = 4 level to the n = 2 level.
[3 marks]
(b) State the region of the electromagnetic spectrum in which this radiation lies.
[1 mark]
(c) Explain why the energy levels are negative.
[1 mark]
8. An electron in a hydrogen atom is in the ground state. It absorbs a photon of energy 12.75 eV.
(a) Determine the principal quantum number of the excited state to which the electron is promoted.
[2 marks]
(b) List all the possible photon energies that could be emitted when this excited electron returns to the ground state via all possible intermediate transitions.
[3 marks]
Section C: Nuclear Physics (Questions 9–15)
Total: 21 marks
9. (a) Define the term mass defect of a nucleus.
[1 mark]
(b) Explain how the mass defect is related to the binding energy of a nucleus.
[2 marks]
10. The nuclide iron-56 (⁵⁶₂₆Fe) has an atomic mass of 55.934937 u. The mass of a proton is 1.007276 u and the mass of a neutron is 1.008665 u.
(a) Calculate the mass defect of the iron-56 nucleus in atomic mass units.
[2 marks]
(b) Hence calculate the binding energy of the iron-56 nucleus in MeV.
[2 marks]
(c) Determine the binding energy per nucleon of iron-56.
[1 mark]
11. The graph below shows the variation of binding energy per nucleon with nucleon number A for stable nuclides.
(a) Explain why the binding energy per nucleon is approximately constant for nuclides with A > 20.
[2 marks]
(b) State and explain whether energy is released in nuclear fission or nuclear fusion by referring to the binding energy per nucleon curve.
[3 marks]
12. A sample of the radioactive isotope iodine-131 has an initial activity of 2.4 × 10⁵ Bq. The half-life of iodine-131 is 8.0 days.
(a) Define the term half-life.
[1 mark]
(b) Calculate the decay constant of iodine-131 in s⁻¹.
[2 marks]
(c) Determine the activity of the sample after 24 days.
[2 marks]
13. Carbon-14 (¹⁴₆C) is a radioactive isotope with a half-life of 5730 years. It is used in radiocarbon dating. A sample of ancient wood is found to have a ¹⁴C activity that is 15% of the activity of living wood.
Estimate the age of the ancient wood sample.
[3 marks]
Section D: Lasers and Semiconductors (Questions 14–17)
Total: 12 marks
14. (a) State two essential characteristics of laser light that distinguish it from light emitted by a conventional light source such as a filament lamp.
[2 marks]
(b) Explain the process of stimulated emission and state why it is essential for laser operation.
[2 marks]
15. A helium-neon laser produces light of wavelength 632.8 nm with an output power of 2.0 mW.
(a) Calculate the energy of a single photon emitted by this laser.
[2 marks]
(b) Determine the number of photons emitted per second by the laser.
[2 marks]
16. (a) Distinguish between an intrinsic semiconductor and an extrinsic semiconductor.
[2 marks]
(b) Explain, with reference to charge carriers, how doping silicon with phosphorus produces an n-type semiconductor.
[2 marks]
Section E: Integrated Modern Physics (Questions 17–20)
Total: 12 marks
17. In an X-ray tube, electrons are accelerated through a potential difference of 50 kV and strike a tungsten target.
(a) Calculate the minimum wavelength of the X-rays produced.
[2 marks]
(b) Explain why the X-ray spectrum from such a tube consists of a continuous background with sharp peaks superimposed on it.
[3 marks]
18. A radioactive nucleus ²³⁸₉₂U decays by alpha emission to form thorium (Th).
(a) Write a balanced nuclear equation for this decay.
[2 marks]
(b) The masses of the nuclei involved are:
- ²³⁸₉₂U: 238.05079 u
- ²³⁴₉₀Th: 234.04363 u
- ⁴₂He: 4.00260 u
Calculate the energy released in this alpha decay, expressing your answer in MeV.
[3 marks]
19. In a demonstration of the photoelectric effect, ultraviolet light of wavelength 200 nm is incident on a metal surface of work function 4.5 eV.
(a) Determine whether photoelectrons will be emitted. Show your reasoning.
[2 marks]
20. A student suggests that increasing the intensity of the incident light in the photoelectric effect will increase the maximum kinetic energy of the emitted electrons because "more energy is being delivered to the metal surface."
Explain why this statement is incorrect, making reference to the photon model of light.
[2 marks]
END OF QUIZ
Total Marks: 60
Answers
A-Level Physics H2 Quiz - Modern Physics: Answer Key and Marking Scheme
Total Marks: 60
Section A: Quantum Physics (Questions 1–6)
Total: 18 marks
1. (a) State what is meant by the term work function of a metal.
[2 marks]
Answer: The work function is the minimum energy required to remove an electron from the surface of a metal. [1 mark for "minimum energy", 1 mark for "remove an electron from the surface"]
(b) Explain why, in the photoelectric effect, electrons are emitted only when the frequency of incident light exceeds a certain threshold value, regardless of the intensity of the light.
[2 marks]
Answer: According to the photon model, each photon transfers its entire energy hf to a single electron. [1 mark] For emission to occur, the photon energy must be at least equal to the work function Φ of the metal (hf ≥ Φ). If f < f₀ (threshold frequency), individual photons do not have sufficient energy to liberate electrons, regardless of how many photons arrive (intensity). [1 mark for linking to work function and photon energy]
2. (a) Calculate the energy, in joules, of a single photon of this ultraviolet light.
[2 marks]
Answer: E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (250 × 10⁻⁹) [1 mark for correct substitution]
= 7.96 × 10⁻¹⁹ J [1 mark for correct answer with unit]
(b) Determine the maximum kinetic energy, in eV, of the photoelectrons emitted from the sodium surface.
[2 marks]
Answer: Photon energy in eV = 7.96 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 4.97 eV [1 mark for conversion]
K_max = hf - Φ = 4.97 - 2.3 = 2.67 eV [1 mark for correct answer with unit]
(c) Calculate the stopping potential required to prevent the most energetic photoelectrons from reaching the collector.
[1 mark]
Answer: V_s = K_max / e = 2.67 V [1 mark for correct answer with unit]
3. (a) Write down Einstein's photoelectric equation, defining each symbol.
[2 marks]
Answer: hf = Φ + K_max or hf = Φ + ½mv²_max [1 mark for correct equation]
Where: h = Planck's constant, f = frequency of incident light, Φ = work function of the metal, K_max = maximum kinetic energy of emitted electrons [1 mark for defining all symbols]
(b) Use the data to calculate the work function of the metal in eV.
[2 marks]
Answer: Photon energy E = hf = (6.63 × 10⁻³⁴ × 8.0 × 10¹⁴) = 5.30 × 10⁻¹⁹ J = 3.31 eV [1 mark for photon energy]
Φ = hf - eV_s = 3.31 - 0.85 = 2.46 eV [1 mark for correct answer with unit]
4. (a) State and explain the effect, if any, on the maximum kinetic energy of the emitted photoelectrons.
[1 mark]
Answer: The maximum kinetic energy increases. [½ mark] Reducing wavelength increases frequency, so photon energy hf increases. Since K_max = hf - Φ, a larger photon energy results in larger maximum kinetic energy. [½ mark]
(b) State and explain the effect, if any, on the photoelectric current.
[2 marks]
Answer: The photoelectric current decreases. [1 mark] With constant intensity but higher photon energy, fewer photons arrive per second (since intensity = number of photons per second × energy per photon). Fewer photons means fewer photoelectrons emitted per second, so the current decreases. [1 mark for explanation linking intensity, photon number, and current]
Section B: Wave-Particle Duality and Atomic Spectra (Questions 5–10)
Total: 18 marks
5. (a) State the de Broglie relationship, defining each symbol.
[1 mark]
Answer: λ = h/p or λ = h/mv [½ mark] where λ is the de Broglie wavelength, h is Planck's constant, p is momentum (m is mass, v is velocity). [½ mark]
(b) Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 150 V.
[3 marks]
Answer: Kinetic energy gained: K = eV = 1.60 × 10⁻¹⁹ × 150 = 2.40 × 10⁻¹⁷ J [1 mark]
Velocity: v = √(2K/m) = √(2 × 2.40 × 10⁻¹⁷ / 9.11 × 10⁻³¹) = 7.26 × 10⁶ m s⁻¹ [1 mark]
de Broglie wavelength: λ = h/mv = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 7.26 × 10⁶) = 1.00 × 10⁻¹⁰ m [1 mark for correct answer with unit]
Alternative method using λ = h/√(2meV) accepted.
6. (a) Explain why the pattern of rings is observed.
[2 marks]
Answer: The electrons are diffracted by the regularly spaced atoms in the graphite crystals. [1 mark] The polycrystalline nature means many small crystals are randomly oriented, so diffraction occurs in all directions around the incident beam, producing concentric rings rather than discrete spots. [1 mark]
(b) The diameter of the first ring is observed to decrease when the accelerating voltage is increased. Explain this observation.
[2 marks]
Answer: Increasing the accelerating voltage increases the kinetic energy and momentum of the electrons, reducing their de Broglie wavelength (λ = h/p). [1 mark] For a given crystal plane spacing d, the diffraction angle θ is given by nλ = 2d sin θ. A smaller wavelength results in a smaller diffraction angle, so the ring diameter decreases. [1 mark]
7. (a) Calculate the wavelength of the photon emitted when an electron makes a transition from the n = 4 level to the n = 2 level.
[3 marks]
Answer: Energy difference: ΔE = E₄ - E₂ = (-0.85) - (-3.40) = 2.55 eV [1 mark]
Convert to joules: ΔE = 2.55 × 1.60 × 10⁻¹⁹ = 4.08 × 10⁻¹⁹ J [1 mark]
Wavelength: λ = hc/ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (4.08 × 10⁻¹⁹) = 4.88 × 10⁻⁷ m = 488 nm [1 mark for correct answer with unit]
(b) State the region of the electromagnetic spectrum in which this radiation lies.
[1 mark]
Answer: Visible light (blue-green region). [1 mark]
(c) Explain why the energy levels are negative.
[1 mark]
Answer: The negative values indicate that the electron is bound to the nucleus. Zero energy corresponds to the electron being completely free from the atom (at infinity). Energy must be supplied to remove the electron, hence the bound states have negative energy. [1 mark]
8. (a) Determine the principal quantum number of the excited state to which the electron is promoted.
[2 marks]
Answer: Energy of ground state: E₁ = -13.60 eV. Energy after absorbing photon: E_n = -13.60 + 12.75 = -0.85 eV [1 mark]
From the energy level diagram, -0.85 eV corresponds to n = 4. [1 mark]
(b) List all the possible photon energies that could be emitted when this excited electron returns to the ground state via all possible intermediate transitions.
[3 marks]
Answer: Possible transitions from n = 4:
4 → 3: ΔE = (-0.85) - (-1.51) = 0.66 eV
4 → 2: ΔE = (-0.85) - (-3.40) = 2.55 eV
4 → 1: ΔE = (-0.85) - (-13.60) = 12.75 eV
3 → 2: ΔE = (-1.51) - (-3.40) = 1.89 eV
3 → 1: ΔE = (-1.51) - (-13.60) = 12.09 eV
2 → 1: ΔE = (-3.40) - (-13.60) = 10.20 eV
[½ mark each for all six correct energies; award 3 marks for complete correct set]
Section C: Nuclear Physics (Questions 9–15)
Total: 21 marks
9. (a) Define the term mass defect of a nucleus.
[1 mark]
Answer: The mass defect is the difference between the total mass of the individual protons and neutrons that make up a nucleus and the actual mass of the nucleus. [1 mark]
(b) Explain how the mass defect is related to the binding energy of a nucleus.
[2 marks]
Answer: The mass defect Δm represents the mass that is converted into energy when nucleons bind together to form the nucleus. [1 mark] The binding energy is the energy equivalent of the mass defect, given by E = Δmc². It is the energy that would need to be supplied to separate the nucleus into its constituent nucleons. [1 mark]
10. (a) Calculate the mass defect of the iron-56 nucleus in atomic mass units.
[2 marks]
Answer: Iron-56 has 26 protons and (56 - 26) = 30 neutrons.
Total mass of constituents = 26(1.007276) + 30(1.008665) = 26.189176 + 30.25995 = 56.449126 u [1 mark]
Mass defect = 56.449126 - 55.934937 = 0.514189 u [1 mark for correct answer]
(b) Hence calculate the binding energy of the iron-56 nucleus in MeV.
[2 marks]
Answer: Binding energy = Δm × 931.5 MeV/u = 0.514189 × 931.5 [1 mark]
= 479 MeV [1 mark for correct answer with unit; accept 478-480 MeV]
(c) Determine the binding energy per nucleon of iron-56.
[1 mark]
Answer: Binding energy per nucleon = 479 / 56 = 8.55 MeV per nucleon [1 mark]
11. (a) Explain why the binding energy per nucleon is approximately constant for nuclides with A > 20.
[2 marks]
Answer: The nuclear force is short-range and saturates, meaning each nucleon only interacts strongly with its nearest neighbours. [1 mark] For larger nuclei (A > 20), nucleons in the interior are surrounded by approximately the same number of neighbours, so the binding energy contributed per nucleon is roughly constant. [1 mark]
(b) State and explain whether energy is released in nuclear fission or nuclear fusion by referring to the binding energy per nucleon curve.
[3 marks]
Answer: Energy is released in both nuclear fission and nuclear fusion. [1 mark]
In fission, a heavy nucleus (e.g., A ≈ 240) splits into two medium-mass nuclei (A ≈ 120). The binding energy per nucleon increases from about 7.6 MeV to about 8.5 MeV, so the products are more tightly bound. The increase in total binding energy corresponds to released energy. [1 mark]
In fusion, two light nuclei (e.g., A = 2 or 3) combine to form a heavier nucleus (e.g., A = 4). The binding energy per nucleon increases significantly (e.g., from about 1.1 MeV to about 7.1 MeV), releasing energy. [1 mark]
12. (a) Define the term half-life.
[1 mark]
Answer: The half-life of a radioactive isotope is the time taken for the number of undecayed nuclei (or the activity) of a sample to decrease to half its initial value. [1 mark]
(b) Calculate the decay constant of iodine-131 in s⁻¹.
[2 marks]
Answer: T₁/₂ = 8.0 days = 8.0 × 24 × 3600 = 6.912 × 10⁵ s [1 mark for conversion]
λ = ln 2 / T₁/₂ = 0.693 / (6.912 × 10⁵) = 1.00 × 10⁻⁶ s⁻¹ [1 mark for correct answer with unit]
(c) Determine the activity of the sample after 24 days.
[2 marks]
Answer: Number of half-lives = 24 / 8.0 = 3 [1 mark]
Activity after 3 half-lives = A₀ / 2³ = 2.4 × 10⁵ / 8 = 3.0 × 10⁴ Bq [1 mark for correct answer with unit]
Alternative method using A = A₀e^(-λt) accepted.
13. Estimate the age of the ancient wood sample.
[3 marks]
Answer: A = A₀e^(-λt) → 0.15 = e^(-λt) [1 mark for setting up equation]
λ = ln 2 / 5730 = 1.21 × 10⁻⁴ year⁻¹ [1 mark for decay constant]
ln(0.15) = -λt → t = -ln(0.15) / λ = 1.897 / (1.21 × 10⁻⁴) = 15,700 years [1 mark for correct answer with unit; accept 15,600-15,800 years]
Section D: Lasers and Semiconductors (Questions 14–17)
Total: 12 marks
14. (a) State two essential characteristics of laser light that distinguish it from light emitted by a conventional light source such as a filament lamp.
[2 marks]
Answer: Any two from: [1 mark each]
- Laser light is highly monochromatic (single wavelength/very narrow range of wavelengths).
- Laser light is coherent (waves are in phase both temporally and spatially).
- Laser light is highly directional (very low divergence/collimated beam).
- Laser light can achieve very high intensity.
(b) Explain the process of stimulated emission and state why it is essential for laser operation.
[2 marks]
Answer: Stimulated emission occurs when an incoming photon of energy exactly equal to the energy difference between an excited state and a lower energy state interacts with an excited atom, causing it to emit a second photon. [1 mark] The emitted photon is identical to the incoming photon in energy, phase, direction, and polarization. This process produces coherent amplification of light, which is the fundamental mechanism of laser operation (Light Amplification by Stimulated Emission of Radiation). [1 mark]
15. (a) Calculate the energy of a single photon emitted by this laser.
[2 marks]
Answer: E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (632.8 × 10⁻⁹) [1 mark]
= 3.14 × 10⁻¹⁹ J [1 mark for correct answer with unit]
(b) Determine the number of photons emitted per second by the laser.
[2 marks]
Answer: Power = energy per second = N × E, where N is number of photons per second.
N = P/E = (2.0 × 10⁻³) / (3.14 × 10⁻¹⁹) [1 mark]
= 6.37 × 10¹⁵ photons per second [1 mark for correct answer with unit]
16. (a) Distinguish between an intrinsic semiconductor and an extrinsic semiconductor.
[2 marks]
Answer: An intrinsic semiconductor is a pure semiconductor material (e.g., pure silicon) where the number of free electrons equals the number of holes, and conductivity depends solely on thermally generated electron-hole pairs. [1 mark] An extrinsic semiconductor is a semiconductor that has been doped with impurity atoms to increase its conductivity by introducing additional charge carriers (either excess electrons or excess holes). [1 mark]
(b) Explain, with reference to charge carriers, how doping silicon with phosphorus produces an n-type semiconductor.
[2 marks]
Answer: Phosphorus is a Group V element with five valence electrons. When it replaces a silicon atom in the crystal lattice, four electrons form covalent bonds with neighbouring silicon atoms, leaving one electron free. [1 mark] This extra electron becomes a mobile negative charge carrier, and the phosphorus atom becomes a fixed positive ion. The majority charge carriers are electrons (negative), hence it is called n-type. [1 mark]
Section E: Integrated Modern Physics (Questions 17–20)
Total: 12 marks
17. (a) Calculate the minimum wavelength of the X-rays produced.
[2 marks]
Answer: The minimum wavelength corresponds to electrons giving all their kinetic energy to a single photon.
E = eV = 1.60 × 10⁻¹⁹ × 50 × 10³ = 8.00 × 10⁻¹⁵ J [1 mark]
λ_min = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (8.00 × 10⁻¹⁵) = 2.49 × 10⁻¹¹ m [1 mark for correct answer with unit]
(b) Explain why the X-ray spectrum from such a tube consists of a continuous background with sharp peaks superimposed on it.
[3 marks]
Answer: The continuous background (Bremsstrahlung) is produced when high-speed electrons are decelerated by the target atoms. [1 mark] Different electrons lose different amounts of kinetic energy in these interactions, producing photons with a continuous range of energies up to the maximum (corresponding to complete stopping of the electron). [1 mark] The sharp peaks (characteristic X-rays) occur when incident electrons eject inner-shell electrons from the target atoms. When outer electrons fall into these vacancies, they emit photons of specific energies equal to the energy differences between the atomic energy levels, producing sharp peaks characteristic of the target material. [1 mark]
18. (a) Write a balanced nuclear equation for this decay.
[2 marks]
Answer: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He [1 mark for correct products, 1 mark for balanced mass numbers and atomic numbers]
(b) Calculate the energy released in this alpha decay, expressing your answer in MeV.
[3 marks]
Answer: Mass of products = 234.04363 + 4.00260 = 238.04623 u [1 mark]
Mass defect = 238.05079 - 238.04623 = 0.00456 u [1 mark]
Energy released = 0.00456 × 931.5 = 4.25 MeV [1 mark for correct answer with unit]
19. Determine whether photoelectrons will be emitted. Show your reasoning.
[2 marks]
Answer: Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (200 × 10⁻⁹) = 9.95 × 10⁻¹⁹ J = 6.22 eV [1 mark for correct photon energy calculation]
Since E = 6.22 eV > Φ = 4.5 eV, the photon energy exceeds the work function, so photoelectrons will be emitted. [1 mark for correct conclusion with reasoning]
20. Explain why this statement is incorrect, making reference to the photon model of light.
[2 marks]
Answer: In the photon model, each photon interacts with a single electron and transfers its entire energy hf to that electron. [1 mark] The maximum kinetic energy of emitted electrons depends only on the photon energy (frequency) and the work function: K_max = hf - Φ. Increasing intensity increases the number of photons arriving per second, which increases the number of emitted electrons (and hence the photocurrent), but does not change the energy of individual photons or the maximum kinetic energy of the electrons. [1 mark]
END OF ANSWER KEY
Total Marks: 60