A Level H1 Physics Modern Physics Quiz

A-Level Physics H1 Quiz - Modern Physics — Answer Key and Marking Scheme

Total Marks: 50

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Section A: Nuclear Physics (Questions 1–7)

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Question 1 [2 marks]

Answer: Isotopes are atoms (or nuclides) of the same element that have the same number of protons (same atomic number) but different numbers of neutrons (different mass numbers).

Marking:

• [B1] Same number of protons / same atomic number
• [B1] Different number of neutrons / different mass number

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Question 2 [2 marks]

Answer: $^{14}_{7}\text{N}$

Marking:

• [B1] Mass number = 14
• [B1] Atomic number = 7 (nitrogen)

Explanation: In beta-minus decay, a neutron converts to a proton, so atomic number increases by 1 while mass number stays the same.

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Question 3 [3 marks]

Answer: 25% activity means ¼ of original remains, which corresponds to 2 half-lives (since ½ × ½ = ¼).

Age = 2 × 5730 = 11,460 years (≈ 11,500 years)

Marking:

• [M1] Recognition that 25% = (½)² or that 2 half-lives have elapsed
• [M1] Use of N = N₀(½)^n or A = A₀e^(−λt) with correct n or λ
• [A1] Correct answer: 11,460 years (accept 11,500 years)

Alternative method: λ = ln 2 / 5730 = 1.21 × 10⁻⁴ yr⁻¹ A/A₀ = 0.25 = e^(−λt) → t = −ln(0.25)/λ = 11,460 years

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Question 4 [3 marks]

Answer: The mass of a helium-4 nucleus is less than the sum of the masses of its constituent nucleons (2 protons and 2 neutrons) because some mass is converted into energy when the nucleons bind together. This energy is the binding energy of the nucleus (E = mc²). The mass difference is called the mass defect.

Marking:

• [B1] Mass is converted to energy / E = mc²
• [B1] This energy is the binding energy that holds the nucleus together
• [B1] The mass difference is called the mass defect

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Question 5 [3 marks]

Answer: Binding energy per nucleon is the average energy required to remove one nucleon from a nucleus (or the binding energy of the nucleus divided by the number of nucleons). Iron-56 is particularly stable because it has the highest binding energy per nucleon of all nuclei, meaning the most energy per nucleon would be needed to break it apart.

Marking:

• [B1] Correct definition: binding energy divided by number of nucleons / energy to remove one nucleon
• [B1] Iron-56 has the highest/maximum binding energy per nucleon
• [B1] Link to stability: most energy needed to break apart / most tightly bound

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Question 6 [4 marks]

(a) [2 marks]

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

Marking:

• [B1] Correct reactants: U-235 + neutron
• [B1] Correct products with balanced mass numbers (236 = 141 + 92 + 3) and atomic numbers (92 = 56 + 36)

(b) [2 marks]

Answer: The neutrons released in fission are fast neutrons. Uranium-235 has a much higher probability of capturing (absorbing) slow (thermal) neutrons than fast neutrons. Therefore, the neutrons must be slowed down by a moderator so they are more likely to cause further fission reactions and sustain the chain reaction.

Marking:

• [B1] Fast neutrons have low probability of causing further fission in U-235
• [B1] Slower/thermal neutrons are more likely to be captured / have higher fission cross-section

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Question 7 [3 marks]

Answer: An alpha particle consists of 2 protons and 2 neutrons (a helium-4 nucleus). Alpha radiation has a very short range in air because alpha particles are relatively large, doubly charged (+2e), and slow-moving. They cause intense ionisation along their path, losing energy rapidly through many collisions with air molecules, so they are stopped within a few centimetres.

Marking:

• [B1] Alpha particle is a helium nucleus / 2 protons + 2 neutrons
• [B1] Alpha particles are highly ionising
• [B1] Rapid energy loss through ionisation leads to short range

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Section B: Quantum Physics – Photoelectric Effect (Questions 8–13)

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Question 8 [2 marks]

Answer (any two from):

1. There is a threshold frequency below which no electrons are emitted, regardless of light intensity.
2. Electrons are emitted instantaneously (with no time delay) when light above the threshold frequency shines on the metal.
3. The maximum kinetic energy of emitted electrons depends on the frequency of light, not its intensity.
4. Increasing intensity increases the number of emitted electrons but not their maximum kinetic energy.

Marking:

• [B1] One correct observation
• [B1] Second correct observation

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Question 9 [2 marks]

Answer: The work function (Φ) of a metal is the minimum energy required to remove an electron from the surface of the metal.

Marking:

• [B1] Minimum energy required
• [B1] To remove an electron from the metal surface

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Question 10 [6 marks]

(a) [2 marks]

Answer: E = hf = hc/λ E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (200 × 10⁻⁹) E = 1.989 × 10⁻²⁵ / 2.00 × 10⁻⁷ E = 9.95 × 10⁻¹⁹ J

Marking:

• [M1] Correct formula E = hc/λ
• [A1] Correct answer with unit

(b) [2 marks]

Answer: Φ = 2.28 eV = 2.28 × 1.60 × 10⁻¹⁹ = 3.648 × 10⁻¹⁹ J K.E._max = hf − Φ = 9.95 × 10⁻¹⁹ − 3.648 × 10⁻¹⁹ = 6.30 × 10⁻¹⁹ J

Marking:

• [M1] Correct use of Einstein's equation K.E._max = hf − Φ (with Φ in joules)
• [A1] Correct answer with unit

(c) [2 marks]

Answer: eV_s = K.E._max V_s = K.E._max / e = 6.30 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3.94 V

Marking:

• [M1] Correct relationship eV_s = K.E._max
• [A1] Correct answer with unit

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Question 11 [2 marks]

(a) [1 mark]

Answer: The gradient represents Planck's constant (h).

(b) [1 mark]

Answer: The intercept on the frequency axis represents the threshold frequency (f₀) of the metal.

Marking:

• [B1] (a) Planck's constant / h
• [B1] (b) Threshold frequency / minimum frequency for photoemission

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Question 12 [4 marks]

(a) [2 marks]

Answer: The maximum kinetic energy of the photoelectrons remains unchanged. This is because the maximum kinetic energy depends only on the frequency of the incident light (K.E._max = hf − Φ), not on its intensity. Since the frequency is unchanged, the photon energy is unchanged, so the maximum kinetic energy is unchanged.

Marking:

• [B1] Maximum kinetic energy is unchanged
• [B1] Correct explanation: K.E._max depends on frequency, not intensity

(b) [2 marks]

Answer: The number of photoelectrons emitted per second doubles. Doubling the intensity means twice as many photons strike the surface per second. Each photon above the threshold frequency can liberate one electron, so twice as many photons produce twice as many photoelectrons per second.

Marking:

• [B1] Number of photoelectrons doubles
• [B1] Correct explanation: more photons per second → more electrons emitted

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Question 13 [2 marks]

Answer: Φ = hf₀ Φ = 6.63 × 10⁻³⁴ × 5.5 × 10¹⁴ = 3.6465 × 10⁻¹⁹ J Φ (in eV) = 3.6465 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.28 eV

Marking:

• [M1] Correct formula Φ = hf₀
• [A1] Correct answer in eV (2.28 eV)

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Section C: Quantum Physics – Wave-Particle Duality & Spectra (Questions 14–20)

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Question 14 [2 marks]

Answer: The de Broglie wavelength is the wavelength associated with a moving particle, given by λ = h/p where h is Planck's constant and p is the momentum of the particle. It represents the wave-like nature of matter.

Marking:

• [B1] Wavelength associated with a moving particle / matter wave
• [B1] λ = h/p or λ = h/mv

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Question 15 [4 marks]

(a) [1 mark]

Answer: K.E. = eV = 1.60 × 10⁻¹⁹ × 150 = 2.40 × 10⁻¹⁷ J

Marking:

• [B1] Correct answer with unit

(b) [3 marks]

Answer: λ = h/p = h/√(2mK.E.) λ = 6.63 × 10⁻³⁴ / √(2 × 9.11 × 10⁻³¹ × 2.40 × 10⁻¹⁷) λ = 6.63 × 10⁻³⁴ / √(4.373 × 10⁻⁴⁷) λ = 6.63 × 10⁻³⁴ / 6.613 × 10⁻²⁴ λ = 1.00 × 10⁻¹⁰ m (0.100 nm)

Marking:

• [M1] Correct formula λ = h/p or λ = h/√(2mE)
• [M1] Correct substitution with electron mass
• [A1] Correct answer with unit

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Question 16 [2 marks]

Answer: The de Broglie wavelength λ = h/mv. At the same speed, the wavelength is inversely proportional to mass. Since a proton has a much larger mass than an electron (m_p ≈ 1836 m_e), the de Broglie wavelength of the proton is much smaller (by a factor of ~1836) than that of the electron.

Marking:

• [B1] λ ∝ 1/m at constant speed
• [B1] Proton has larger mass, therefore smaller wavelength

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Question 17 [2 marks]

Answer: An emission line spectrum consists of bright lines of specific wavelengths on a dark background, produced when excited atoms emit photons as electrons transition from higher to lower energy levels. The discrete (specific) wavelengths of the lines show that only certain energy transitions are possible, which provides evidence that electrons in atoms can only occupy discrete (quantised) energy levels.

Marking:

• [B1] Emission line spectrum: bright lines at specific wavelengths / discrete lines
• [B1] Evidence for discrete energy levels: only specific energy transitions possible / quantised energy levels

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Question 18 [3 marks]

Answer: Energy difference: ΔE = E₄ − E₂ = (−0.85) − (−3.40) = 2.55 eV ΔE in joules = 2.55 × 1.60 × 10⁻¹⁹ = 4.08 × 10⁻¹⁹ J

λ = hc/ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (4.08 × 10⁻¹⁹) λ = 1.989 × 10⁻²⁵ / 4.08 × 10⁻¹⁹ = 4.88 × 10⁻⁷ m (488 nm)

Marking:

• [M1] Correct energy difference calculation (2.55 eV)
• [M1] Correct use of E = hc/λ
• [A1] Correct answer with unit (accept 4.87–4.88 × 10⁻⁷ m or 487–488 nm)

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Question 19 [2 marks]

Answer: When white light passes through a cool gas, atoms in the gas absorb photons of specific energies corresponding to transitions from lower to higher energy levels. These absorbed wavelengths are missing from the transmitted light, producing dark lines on a continuous spectrum. The dark lines in the absorption spectrum occur at exactly the same wavelengths as the bright lines in the emission spectrum of the same gas, because both involve the same energy level transitions.

Marking:

• [B1] Dark lines due to absorption of specific wavelengths by gas atoms
• [B1] Absorption lines correspond to same wavelengths as emission lines (same energy transitions)

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Question 20 [3 marks]

Answer: The minimum wavelength corresponds to the maximum photon energy, which occurs when all the kinetic energy of an electron is converted to a single X-ray photon.

E_max = hc/λ_min = eV

V = hc/(eλ_min) V = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (1.60 × 10⁻¹⁹ × 3.1 × 10⁻¹¹) V = 1.989 × 10⁻²⁵ / 4.96 × 10⁻³⁰ V = 40,100 V = 40.1 kV

Marking:

• [M1] Correct relationship: eV = hc/λ_min
• [M1] Correct substitution
• [A1] Correct answer with unit (40.1 kV or 4.01 × 10⁴ V)

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END OF ANSWER KEY