A Level H2 Physics Modern Physics Quiz

A-Level Physics H2 Quiz - Modern Physics

Answer Key and Marking Scheme

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Section A: Multiple Choice [20 marks]

1. Answer: C [2]

Explanation: Einstein's photoelectric equation is $K_{\max} = hf - \phi$. Initially, $K_{\max} = hf - \phi$. When the frequency is doubled: $K'_{\max} = h(2f) - \phi = 2hf - \phi = (hf - \phi) + hf = K_{\max} + hf$. So the new maximum kinetic energy is $K_{\max} + hf$.

Common mistake: Students may choose B ($2K_{\max}$), forgetting that the work function $\phi$ does not change. The increase in $K_{\max}$ is $hf$, not a doubling of the original $K_{\max}$.

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2. Answer: A [2]

Explanation: Using $E = \frac{hc}{\lambda}$:

$E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{450 \times 10^{-9}} = \frac{1.989 \times 10^{-25}}{4.50 \times 10^{-7}} = 4.42 \times 10^{-19} \text{ J}$

Common mistake: Forgetting to convert nm to m (i.e., not multiplying by $10^{-9}$).

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3. Answer: B [2]

Explanation: The de Broglie wavelength is given by $\lambda = \frac{h}{p}$, where $p$ is the momentum of the particle. Therefore, the wavelength is inversely proportional to the momentum. This is a fundamental concept in quantum mechanics — all matter exhibits wave-like behaviour, with the wavelength decreasing as momentum increases.

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4. Answer: B [2]

Explanation: When an electron transitions from a higher energy level ($n = 4$) to a lower energy level ($n = 2$), it emits a photon. The energy of the photon equals the difference between the two energy levels: $E_{\text{photon}} = E_4 - E_2$. Since $E_4 > E_2$ (less negative), the photon energy is positive. Alpha particles and neutrons are not emitted in electronic transitions.

Common mistake: Choosing C ($E_2 - E_4$), which would be negative. Photon energy must be positive.

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5. Answer: C [2]

Explanation: The threshold frequency is found from $\phi = hf_0$:

$f_0 = \frac{\phi}{h} = \frac{2.3 \times 1.60 \times 10^{-19}}{6.63 \times 10^{-34}} = \frac{3.68 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.55 \times 10^{14} \text{ Hz}$

Photoemission only occurs when $f > f_0$. Option C ($5.6 \times 10^{14}$ Hz) is very close to the threshold. Given the precision of the data, $5.6 \times 10^{14}$ Hz is approximately equal to $f_0$ and at this level of precision, it would not cause photoemission (or barely would). More precisely, $5.0 \times 10^{14}$ Hz is clearly below threshold. Rechecking: $f_0 = 5.55 \times 10^{14}$ Hz, so $5.0 \times 10^{14}$ Hz is below threshold.

Corrected Answer: A ($5.0 \times 10^{14}$ Hz is below $5.55 \times 10^{14}$ Hz)

Revised Explanation: The threshold frequency is $f_0 = 5.55 \times 10^{14}$ Hz. Any photon with frequency below this will NOT cause photoemission. Option A ($5.0 \times 10^{14}$ Hz) is below the threshold frequency, so it will not cause photoemission.

Marking note: The answer is A. Option C ($5.6 \times 10^{14}$ Hz) is marginally above threshold and would cause photoemission. Option A is clearly below.

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6. Answer: B [2]

Explanation: After each half-life, the activity is halved. After 3 half-lives:

$A = A_0 \times \left(\frac{1}{2}\right)^3 = 800 \times \frac{1}{8} = 100 \text{ Bq}$

Common mistake: Dividing by 3 instead of halving three times (which would give ~267 Bq, option C).

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7. Answer: A [2]

Explanation: In alpha decay, the parent nucleus emits an alpha particle ($^4_2$He). The nucleon number decreases by 4 and the proton number decreases by 2:

$^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\text{He}$

Conservation check: $238 = 234 + 4$ ✓ and $92 = 90 + 2$ ✓.

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8. Answer: B [2]

Explanation: Total binding energy = binding energy per nucleon × number of nucleons:

$E_B = 8.8 \times 56 = 492.8 \text{ MeV}$

Common mistake: Confusing binding energy per nucleon with total binding energy (choosing A or C).

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9. Answer: B [2]

Explanation: Control rods (made of neutron-absorbing materials such as cadmium or boron) absorb excess neutrons to control the rate of fission. By inserting or withdrawing the rods, the number of neutrons available to cause further fission events is regulated, thus controlling the chain reaction.

• Option A describes the role of a neutron source, not control rods.
• Option C describes the moderator.
• Option D describes the coolant.

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10. Answer: A [2]

Explanation: The relationship between half-life and decay constant is:

$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{1.2 \times 10^{-4}} = 5.78 \times 10^{3} \text{ s}$

Common mistake: Using $t_{1/2} = \frac{1}{\lambda}$ instead of $\frac{\ln 2}{\lambda}$, which would give option B.

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Section B: Structured Questions [30 marks]

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11. (a) [2]

Answer: Einstein's photoelectric equation states that the maximum kinetic energy of a photoelectron is equal to the energy of the incident photon minus the work function of the metal:

$K_{\max} = hf - \phi$

or equivalently:

$hf = \phi + K_{\max}$

Marking: [1] for correct equation, [1] for defining terms or stating it as a sentence.

(b)(i) [2]

Answer:

$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{200 \times 10^{-9}}$

$E = \frac{1.989 \times 10^{-25}}{2.00 \times 10^{-7}} = 9.95 \times 10^{-19} \text{ J}$

Marking: [1] for correct substitution, [1] for correct answer.

(b)(ii) [2]

Answer:

$\phi = 2.14 \text{ eV} = 2.14 \times 1.60 \times 10^{-19} = 3.424 \times 10^{-19} \text{ J}$

$K_{\max} = E - \phi = 9.95 \times 10^{-19} - 3.424 \times 10^{-19} = 6.52 \times 10^{-19} \text{ J}$

Converting to eV:

$K_{\max} = \frac{6.52 \times 10^{-19}}{1.60 \times 10^{-19}} = 4.08 \text{ eV}$

Marking: [1] for correct conversion/subtraction, [1] for answer in eV.

(b)(iii) [2]

Answer:

$K_{\max} = \frac{1}{2}m_ev^2$

$v = \sqrt{\frac{2K_{\max}}{m_e}} = \sqrt{\frac{2 \times 6.52 \times 10^{-19}}{9.11 \times 10^{-31}}}$

$v = \sqrt{1.431 \times 10^{12}} = 1.20 \times 10^{6} \text{ m/s}$

Marking: [1] for correct substitution, [1] for correct answer with unit.

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12. (a) [2]

Answer: The de Broglie wavelength of a particle is the wavelength associated with a moving particle, given by $\lambda = \frac{h}{p} = \frac{h}{mv}$, where $h$ is Planck's constant and $p$ is the momentum of the particle. It describes the wave-like behaviour of matter.

Marking: [1] for defining the concept (wave nature of matter), [1] for the correct formula.

(b) [4]

Answer:

First, find the kinetic energy of the electron after being accelerated through 150 V:

$K = eV = (1.60 \times 10^{-19})(150) = 2.40 \times 10^{-17} \text{ J}$

Find the momentum:

$K = \frac{p^2}{2m_e} \implies p = \sqrt{2m_e K}$

$p = \sqrt{2 \times 9.11 \times 10^{-31} \times 2.40 \times 10^{-17}}$

$p = \sqrt{4.373 \times 10^{-47}} = 6.61 \times 10^{-24} \text{ kg m/s}$

De Broglie wavelength:

$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} = 1.00 \times 10^{-10} \text{ m} = 0.100 \text{ nm}$

Alternative (faster) method:

$\lambda = \frac{h}{\sqrt{2m_e eV}} = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 150}}$

$= \frac{6.63 \times 10^{-34}}{\sqrt{4.373 \times 10^{-47}}} = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} = 1.00 \times 10^{-10} \text{ m}$

Marking: [1] for kinetic energy, [1] for momentum, [1] for de Broglie formula, [1] for correct answer with unit.

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13. (a) [2]

Answer:

$E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.511 \text{ eV}$

$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.40 \text{ eV}$

$E_{\text{photon}} = E_3 - E_2 = -1.511 - (-3.40) = 1.89 \text{ eV}$

Marking: [1] for correct energy levels, [1] for correct photon energy.

(b) [2]

Answer:

$E = 1.89 \text{ eV} = 1.89 \times 1.60 \times 10^{-19} = 3.024 \times 10^{-19} \text{ J}$

$\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{3.024 \times 10^{-19}} = 6.58 \times 10^{-7} \text{ m} = 658 \text{ nm}$

Marking: [1] for correct conversion and formula, [1] for correct answer.

(c) [1]

Answer: Visible (red light) region.

Explanation: A wavelength of 658 nm falls within the visible spectrum (approximately 400–700 nm), specifically in the red part of the spectrum. This is the well-known H-alpha line in the Balmer series.

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14. (a) [2]

Answer: The binding energy of a nucleus is the minimum energy required to completely separate all the nucleons (protons and neutrons) in the nucleus. Equivalently, it is the energy equivalent of the mass defect — the difference between the total mass of the separate nucleons and the actual mass of the nucleus.

Marking: [1] for "energy to separate nucleons" or equivalent, [1] for reference to mass defect or energy equivalence.

(b) [3]

Answer:

The $^{12}_{6}\text{C}$ nucleus has 6 protons and 6 neutrons.

Total mass of separate nucleons:

$m_{\text{total}} = 6 \times 1.0073 + 6 \times 1.0087 = 6.0438 + 6.0522 = 12.0960 \text{ u}$

Mass defect:

$\Delta m = 12.0960 - 11.9967 = 0.0993 \text{ u}$

Marking: [1] for correct number of protons and neutrons, [1] for correct total mass calculation, [1] for correct mass defect.

(c) [2]

Answer:

$E_B = 0.0993 \times 931.5 = 92.50 \text{ MeV}$

Marking: [1] for correct substitution, [1] for correct answer with unit.

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15. (a) [2]

Answer:

$t_{1/2} = 8.0 \text{ days} = 8.0 \times 24 \times 3600 = 6.912 \times 10^{5} \text{ s}$

$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{6.912 \times 10^{5}} = 1.00 \times 10^{-6} \text{ s}^{-1}$

Marking: [1] for correct conversion to seconds, [1] for correct decay constant.

(b) [2]

Answer:

24 days = 3 half-lives (since $t_{1/2} = 8.0$ days)

$A = A_0 \times \left(\frac{1}{2}\right)^3 = 4.0 \times 10^{5} \times \frac{1}{8} = 5.0 \times 10^{4} \text{ Bq}$

Marking: [1] for identifying 3 half-lives, [1] for correct answer.

(c) [2]

Answer: Iodine-131 is suitable for medical tracer applications because:

• It has a half-life of 8.0 days, which is long enough to allow for medical procedures and imaging but short enough to minimise prolonged radiation exposure to the patient.
• It is a gamma (and beta) emitter, and gamma rays can be detected externally, allowing non-invasive imaging.
• Iodine is biologically absorbed by the thyroid gland, making it particularly useful for thyroid imaging and treatment.

Marking: [1] for half-life reasoning, [1] for emission type/biological uptake reasoning.

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Section C: Data Interpretation Question [10 marks]

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16. (a) [3]

Answer: The graph should show:

• Points plotted correctly: (5.0, 0.20), (6.0, 0.62), (7.0, 1.03), (8.0, 1.45), (9.0, 1.86), (10.0, 2.28)
• A straight line of best fit through the points
• Axes correctly labelled with units

Marking: [1] for correct scales and labelled axes, [1] for correct plotting of points, [1] for straight line of best fit.

Expected graph features: The graph is a straight line with a positive gradient. The line does NOT pass through the origin — it intersects the f-axis at the threshold frequency.

(b)(i) [2]

Answer: The threshold frequency is found by extrapolating the line to where $V_s = 0$. From the data, using the two-point form or extrapolation:

Taking the first two points: gradient = $\frac{0.62 - 0.20}{(6.0 - 5.0) \times 10^{14}} = \frac{0.42}{1.0 \times 10^{14}} = 4.2 \times 10^{-15}$ V/Hz

Using $V_s = 0$ at $f = f_0$: $0 = 0.20 + 4.2 \times 10^{-15}(f_0 - 5.0 \times 10^{14})$

$f_0 = 5.0 \times 10^{14} - \frac{0.20}{4.2 \times 10^{-14}} = 5.0 \times 10^{14} - 0.476 \times 10^{14} = 4.52 \times 10^{14}$ Hz

Threshold frequency: $f_0 \approx 4.5 \times 10^{14}$ Hz

Marking: [1] for correct extrapolation method, [1] for answer in range $4.4 - 4.6 \times 10^{14}$ Hz.

(b)(ii) [2]

Answer:

$\phi = hf_0 = (6.63 \times 10^{-34})(4.52 \times 10^{14}) = 2.997 \times 10^{-19} \text{ J}$

$\phi = \frac{2.997 \times 10^{-19}}{1.60 \times 10^{-19}} = 1.87 \text{ eV}$

Work function: $\phi \approx 1.9$ eV

Marking: [1] for correct formula, [1] for answer in range 1.8–2.0 eV.

(c) [3]

Answer: From Einstein's photoelectric equation: $eV_s = hf - \phi$, so $V_s = \frac{h}{e}f - \frac{\phi}{e}$.

The gradient of the $V_s$ vs $f$ graph is $\frac{h}{e}$.

Using two widely separated points, e.g., (5.0, 0.20) and (10.0, 2.28):

$\text{gradient} = \frac{2.28 - 0.20}{(10.0 - 5.0) \times 10^{14}} = \frac{2.08}{5.0 \times 10^{14}} = 4.16 \times 10^{-15} \text{ V/Hz}$

$h = \text{gradient} \times e = 4.16 \times 10^{-15} \times 1.60 \times 10^{-19} = 6.66 \times 10^{-34} \text{ J s}$

Planck's constant: $h \approx 6.6 \times 10^{-34}$ J s

Marking: [1] for identifying gradient = $h/e$, [1] for correct gradient calculation, [1] for correct value of $h$.

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17. (a) [3]

Answer:

Kinetic energy of electron: $K = eV = (1.60 \times 10^{-19})(54) = 8.64 \times 10^{-18}$ J

Momentum: $p = \sqrt{2m_e K} = \sqrt{2 \times 9.11 \times 10^{-31} \times 8.64 \times 10^{-18}}$

$p = \sqrt{1.574 \times 10^{-47}} = 3.97 \times 10^{-24} \text{ kg m/s}$

De Broglie wavelength:

$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{3.97 \times 10^{-24}} = 1.67 \times 10^{-10} \text{ m} = 0.167 \text{ nm}$

Marking: [1] for kinetic energy, [1] for momentum, [1] for wavelength.

(b) [2]

Answer: Using Bragg's law with $n = 1$:

$\lambda = 2d\sin\theta$

$\sin\theta = \frac{\lambda}{2d} = \frac{1.67 \times 10^{-10}}{2 \times 0.215 \times 10^{-9}} = \frac{1.67 \times 10^{-10}}{4.30 \times 10^{-10}} = 0.388$

$\theta = \arcsin(0.388) = 22.9°$

Marking: [1] for correct substitution, [1] for correct angle.

(c) [2]

Answer: Diffraction is a wave phenomenon — it occurs when a wave encounters an obstacle or aperture comparable in size to its wavelength. The observation of a diffraction pattern when electrons are fired at a crystal demonstrates that electrons exhibit wave-like behaviour. The regular spacing of atoms in the crystal acts as a diffraction grating for the electron waves, producing constructive and destructive interference at specific angles. This is direct experimental evidence supporting de Broglie's hypothesis that particles possess wave properties.

Marking: [1] for identifying diffraction as a wave phenomenon, [1] for linking to de Broglie hypothesis / wave-particle duality.

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18. (a) [2]

Answer:

Nucleon number (mass number) conservation: Left side: $235 + 1 = 236$ Right side: $141 + 92 + 3(1) = 141 + 92 + 3 = 236$ ✓

Proton number (atomic number) conservation: Left side: $92 + 0 = 92$ Right side: $56 + 36 + 3(0) = 56 + 36 = 92$ ✓

Both nucleon number and proton number are conserved, so the equation is balanced.

Marking: [1] for nucleon number check, [1] for proton number check.

(b) [3]

Answer: Energy is released in this fission reaction because the binding energy per nucleon of the products (barium-141 and krypton-92) is greater than that of the reactant (uranium-235). On the binding energy per nucleon curve, uranium-235 is a heavy nucleus with a binding energy per nucleon of approximately 7.6 MeV, while the fission products lie closer to the peak of the curve (near iron-56) with binding energy per nucleon of approximately 8.3–8.5 MeV. The increase in total binding energy means that the products are more tightly bound, and the difference in binding energy is released (primarily as kinetic energy of the fission fragments and neutrons). This can also be understood in terms of mass defect: the total mass of the products is less than the total mass of the reactants, and this mass difference is converted to energy via $E = \Delta m c^2$.

Marking: [1] for stating that products have higher binding energy per nucleon, [1] for reference to the binding energy curve / position relative to iron peak, [1] for explaining energy release via mass defect or $E = \Delta mc^2$.

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19. (a) [3]

Answer:

• Continuous spectrum: Produced by hot, dense objects (e.g., incandescent filament, hot metal). It contains all wavelengths across a continuous range with no gaps — the colours merge smoothly from red to violet.

• Emission line spectrum: Produced by excited low-density gases (e.g., gas discharge tubes). It consists of discrete, bright lines at specific wavelengths against a dark background. Each element has a unique set of emission lines corresponding to electronic transitions between energy levels.

• Absorption line spectrum: Produced when white light passes through a cooler low-density gas. It appears as a continuous dark background with dark lines at specific wavelengths. The dark lines correspond to wavelengths absorbed by the gas atoms, and they occur at exactly the same wavelengths as the emission lines of that element.

Marking: [1] for each correct description.

(b) [2]

Answer: When light from a star's hot interior passes through the cooler outer atmosphere, atoms in the atmosphere absorb photons at specific wavelengths corresponding to their electronic transitions. This produces an absorption line spectrum. By comparing the wavelengths of the dark absorption lines with known emission spectra of elements (obtained in laboratory experiments), the chemical composition of the star's atmosphere can be identified. Each element produces a unique pattern of absorption lines, acting as a "fingerprint" for that element.

Marking: [1] for explaining absorption by atmospheric atoms, [1] for comparing with known spectra to identify elements.

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20. (a) [4]

Answer:

Let the number of alpha decays be $x$ and the number of beta-minus decays be $y$.

Each alpha decay: $A$ decreases by 4, $Z$ decreases by 2. Each beta-minus decay: $A$ unchanged, $Z$ increases by 1.

Conservation of nucleon number (A):

$232 - 4x = 208$

$4x = 24$

$x = 6$

Conservation of proton number (Z):

$90 - 2x + y = 82$

$90 - 2(6) + y = 82$

$90 - 12 + y = 82$

$y = 82 - 78 = 4$

Therefore: 6 alpha decays and 4 beta-minus decays.

Verification:

• Nucleon number: $232 - 6(4) = 232 - 24 = 208$ ✓
• Proton number: $90 - 6(2) + 4(1) = 90 - 12 + 4 = 82$ ✓

Marking: [1] for setting up nucleon number equation, [1] for $x = 6$, [1] for setting up proton number equation, [1] for $y = 4$.

(b) [2]

Answer: Alpha decay is more likely than proton emission for heavy nuclei because:

• The alpha particle ($^4_2$He) is an exceptionally stable nucleus with a very high binding energy per nucleon for its size (7.1 MeV/nucleon). This means it can easily form inside the nucleus as a pre-formed cluster.
• The Coulomb barrier for alpha emission, while significant, is lower than for proton emission in terms of the energy balance (Q-value) because the alpha particle carries away more binding energy, making the decay energetically more favourable.
• Proton emission requires overcoming the Coulomb barrier for a single proton, and for most heavy nuclei, the separation energy for a single proton is positive (i.e., it costs energy), making proton emission energetically forbidden. Alpha decay, by contrast, releases energy (positive Q-value) for most heavy nuclei.

Marking: [1] for alpha particle stability/pre-formation argument, [1] for energy/Coulomb barrier reasoning.

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Total: 60 marks