A Level H1 Physics Modern Physics Quiz

A-Level Physics H1 Quiz - Modern Physics

Answer Key and Teaching Notes

Teaching Note: This answer key provides full step-by-step solutions, marking notes, and explanations suitable for students new to each concept. Common errors and marking descriptors are included where relevant.

Section A: Multiple Choice

1. C — $\frac{1}{8}$ [2 marks]

Explanation:
The half-life is the time taken for half of the radioactive nuclei to decay. After each half-life, the remaining fraction is halved:

• After 1 half-life (8 h): $\frac{1}{2}$ remains
• After 2 half-lives (16 h): $\frac{1}{4}$ remains
• After 3 half-lives (24 h): $\frac{1}{8}$ remains

Number of half-lives: $n = \frac{24}{8} = 3$.
Fraction remaining = $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$.

Common mistake: Students sometimes divide $\frac{1}{24/8} = \frac{1}{3}$ instead of computing $\left(\frac{1}{2}\right)^3$.

2. B — 1.1 eV [2 marks]

Explanation:
Using the photoelectric equation: $E_{\text{photon}} = hf = \frac{hc}{\lambda}$

$E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \text{ J}$

Converting to eV: $E = \frac{4.97 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.11 \text{ eV}$

Maximum kinetic energy: $K_{\text{max}} = E - \phi = 3.11 - 2.0 = 1.1 \text{ eV}$

Common mistake: Forgetting to convert from joules to eV, or subtracting the work function the wrong way.

3. B — The greater the binding energy per nucleon, the more stable the nucleus. [2 marks]

Explanation:
Binding energy per nucleon is a measure of nuclear stability. A higher binding energy per nucleon means more energy is required to remove a nucleon from the nucleus, indicating greater stability.

• Option A is incorrect: binding energy refers to the total energy needed to completely separate all nucleons, not just one proton.
• Option C is incorrect: the binding energy per nucleon curve rises to a peak near iron-56 and then decreases — it does not decrease uniformly.
• Option D is incorrect: fission releases energy because the products have a higher binding energy per nucleon than the original heavy nucleus.

4. A — $^{234}_{90}\text{Th}$ [2 marks]

Explanation:
Alpha decay emits a $^4_2\text{He}$ nucleus. Conservation of mass number and atomic number gives:

$^{238}_{92}\text{U} \rightarrow ^{A}_{Z}\text{X} + ^{4}_{2}\text{He}$

Mass number: $238 = A + 4$, so $A = 234$
Atomic number: $92 = Z + 2$, so $Z = 90$

Element with $Z = 90$ is thorium (Th). The product is $^{234}_{90}\text{Th}$.

Common mistake: Forgetting to subtract the alpha particle's atomic number from the parent's atomic number.

5. B [2 marks]

Explanation:
From Einstein's photoelectric equation: $K_{\text{max}} = hf - \phi$

This is a straight line with:

• Gradient = $h$ (positive)
• y-intercept = $-\phi$ (negative, since work function $\phi > 0$)
• x-intercept = threshold frequency $f_0 = \phi/h$ (where $K_{\text{max}} = 0$)

Graph B correctly shows a straight line with positive gradient, negative y-intercept, and positive x-intercept at the threshold frequency.

Note on image: The graph should show a straight line with positive gradient crossing the frequency axis at the threshold frequency $f_0$ and the $K_{\text{max}}$ axis at $-\phi$.

Section B: Short Answer and Structured Questions

6. [3 marks]

Answer:
[B1] Most alpha particles passed through the gold foil undeflected (or with very small deflections).
[B1] A small number of alpha particles were deflected through large angles (greater than 90°).
[B1] A very small number of alpha particles were deflected backwards (nearly 180°).

Teaching Notes:
These observations led Rutherford to conclude that:

• The atom is mostly empty space (most alphas pass through).
• The nucleus is very small, dense, and positively charged (large deflections occur when alphas pass close to the nucleus).
• The nucleus contains most of the mass of the atom.

Common mistake: Students sometimes state the conclusions (nuclear model) rather than the observations from the experiment. The question asks for observations.

7. (a) [1 mark]

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

Teaching Note: The work function is a property of the material. It represents the energy barrier that binds electrons to the metal surface. Symbol: $\phi$ (phi). Units: joules (J) or electron-volts (eV).

(b) [2 marks]

Answer:
[B1] The threshold frequency is the minimum frequency of incident light that can cause photoelectron emission from a metal surface.
[B1] Below this frequency, no photoelectrons are emitted regardless of the intensity of the light.

Teaching Note: The threshold frequency $f_0$ is related to the work function by $\phi = hf_0$. This is a key concept that distinguishes the particle (photon) model from the wave model of light.

8. (a) [2 marks]

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

$E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{2.48 \times 10^{-12}}$

$E = \frac{1.989 \times 10^{-25}}{2.48 \times 10^{-12}} = 8.02 \times 10^{-14} \text{ J}$

[M1] for correct substitution; [A1] for correct answer.

(b) [1 mark]

Answer:
$E = \frac{8.02 \times 10^{-14}}{1.60 \times 10^{-19}} = 5.01 \times 10^5 \text{ eV} = 0.501 \text{ MeV}$

Or using $E = \frac{hc}{\lambda}$ with $hc = 1240 \text{ eV·nm}$ and $\lambda = 2.48 \times 10^{-3} \text{ nm}$:

$E = \frac{1240}{2.48 \times 10^{-3}} = 5.0 \times 10^5 \text{ eV} \approx 0.50 \text{ MeV}$

[A1] for correct answer (accept 0.50 MeV or 0.501 MeV).

Teaching Note: This photon is a gamma ray, consistent with the very short wavelength ($2.48 \times 10^{-12}$ m is in the gamma-ray range).

9. (a) [2 marks]

Answer:
$^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}e + \bar{\nu}_e$

[M1] for correct mass number and atomic number conservation; [A1] for complete correct equation including antineutrino (or electron antineutrino).

Teaching Note: In beta-minus decay, a neutron converts to a proton, an electron, and an electron antineutrino. The mass number stays the same (14), and the atomic number increases by 1 (6 → 7), producing nitrogen-14.

(b) [2 marks]

Answer: Any TWO of the following [1 mark each]:

• Beta particles are high-speed electrons.
• They have a charge of $-1.60 \times 10^{-19}$ C (or $-e$).
• They have a range of penetration in matter (can penetrate a few mm of aluminium).
• They are deflected by electric and magnetic fields.
• They have a continuous energy spectrum (unlike alpha particles which are monoenergetic).

10. (a) [2 marks]

Answer:
The decay constant $\lambda$ is related to half-life by:

$\lambda = \frac{\ln 2}{t_{1/2}}$

Converting half-life to seconds: $t_{1/2} = 5.3 \times 365.25 \times 24 \times 3600 = 1.67 \times 10^8 \text{ s}$

$\lambda = \frac{0.693}{1.67 \times 10^8} = 4.15 \times 10^{-9} \text{ s}^{-1}$

[M1] for correct formula and conversion; [A1] for correct answer.

(b) [2 marks]

Answer:
Number of half-lives: $n = \frac{15.9}{5.3} = 3$

Activity after 3 half-lives: $A = A_0 \times \left(\frac{1}{2}\right)^3 = 3.2 \times 10^{10} \times \frac{1}{8} = 4.0 \times 10^9 \text{ Bq}$

[M1] for correct method; [A1] for correct answer.

Alternative method using $A = A_0 e^{-\lambda t}$ also accepted.

11. [4 marks]

Answer:
[B1] Wave theory predicts that light of any frequency should eventually cause photoelectron emission if the intensity is high enough, because energy accumulates over time. However, the photoelectric effect shows that no electrons are emitted below the threshold frequency, regardless of intensity.

[B1] Wave theory predicts that increasing the intensity of light should increase the kinetic energy of emitted electrons (more energy delivered). However, experiments show that increasing intensity only increases the number of photoelectrons (photocurrent), not their maximum kinetic energy.

[B1] Wave theory predicts a time delay between illumination and electron emission at low intensities, as energy needs to accumulate. However, photoelectron emission is instantaneous (within $\sim 10^{-9}$ s).

[B1] The maximum kinetic energy of photoelectrons depends on the frequency of the light, not the intensity, which is consistent with the photon model $K_{\text{max}} = hf - \phi$.

Teaching Note: This is a classic A-Level question testing understanding of why the wave model fails and the photon model succeeds. Students should address both intensity and frequency aspects.

12. (a) [2 marks]

Answer:
$\Delta E = E_3 - E_2 = (-1.51) - (-3.40) = 1.89 \text{ eV}$

The energy of the emitted photon is 1.89 eV.

[M1] for correct energy difference; [A1] for correct answer.

(b) [2 marks]

Answer:
$\Delta E = E_4 - E_2 = (-0.85) - (-3.40) = 2.55 \text{ eV}$

Converting to joules: $E = 2.55 \times 1.60 \times 10^{-19} = 4.08 \times 10^{-19} \text{ J}$

$\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{4.08 \times 10^{-19}} = 4.87 \times 10^{-7} \text{ m} = 487 \text{ nm}$

[M1] for correct energy and wavelength calculation; [A1] for correct answer (accept 486–487 nm).

Teaching Note: This wavelength (487 nm) is in the visible spectrum (blue-green), part of the Balmer series (transitions to $n = 2$).

(c) [2 marks]

Answer:
[B1] Transition C ($n = 5 \rightarrow n = 2$) produces the photon with the highest frequency.
[B1] This is because the energy difference is greatest for transition C ($E_5 - E_2 = 2.86$ eV, compared to 1.89 eV for A and 2.55 eV for B), and since $E = hf$, the largest energy corresponds to the highest frequency.

Teaching Note: Students should calculate or compare the energy differences to justify their answer.

13. (a) [3 marks]

Answer:
Using $A = \lambda N$, so $N = \frac{A}{\lambda}$.

First, find the decay constant:

$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{8.0 \times 24 \times 3600} = \frac{0.693}{6.912 \times 10^5} = 1.00 \times 10^{-6} \text{ s}^{-1}$

Now calculate $N$:

$N = \frac{4.0 \times 10^9}{1.00 \times 10^{-6}} = 4.0 \times 10^{15} \text{ nuclei}$

[M1] for correct decay constant; [M1] for using $N = A/\lambda$; [A1] for correct answer.

(b) [3 marks]

Answer:
Using $A = A_0 \left(\frac{1}{2}\right)^n$ where $A = 2.5 \times 10^7$ Bq and $A_0 = 4.0 \times 10^9$ Bq:

$\frac{A}{A_0} = \frac{2.5 \times 10^7}{4.0 \times 10^9} = 6.25 \times 10^{-3}$

$\left(\frac{1}{2}\right)^n = 6.25 \times 10^{-3}$

Taking logarithms: $n \ln(0.5) = \ln(6.25 \times 10^{-3})$

$n = \frac{\ln(6.25 \times 10^{-3})}{\ln(0.5)} = \frac{-5.075}{-0.693} = 7.32$

Time: $t = n \times t_{1/2} = 7.32 \times 8.0 = 58.6$ days $\approx 59$ days

[M1] for correct ratio and logarithmic method; [M1] for correct calculation of $n$; [A1] for correct answer (accept 58–59 days).

Alternative: Using $A = A_0 e^{-\lambda t}$ and solving for $t$ is also acceptable.

14. (a) [1 mark]

Answer:
From the graph, the binding energy per nucleon for uranium-235 ($A = 235$) is approximately 7.6 MeV.

[A1] for value in range 7.5–7.7 MeV.

(b) [3 marks]

Answer:
[B1] When uranium-235 undergoes fission, it splits into two (or more) lighter nuclei (fission fragments) with mass numbers typically in the range 80–160.
[B1] From the graph, these lighter fragments have a higher binding energy per nucleon (approximately 8.5 MeV) than the original uranium-235 nucleus (approximately 7.6 MeV).
[B1] The increase in binding energy per nucleon means that energy is released (the products are more tightly bound), in accordance with the mass-energy equivalence $E = \Delta m c^2$.

Teaching Note: The energy released per nucleon is approximately $8.5 - 7.6 = 0.9$ MeV, and for 235 nucleons, the total energy released is approximately $0.9 \times 235 \approx 210$ MeV per fission event.

(c) [2 marks]

Answer:
[B1] When two very light nuclei (e.g., hydrogen isotopes) undergo fusion, they combine to form a heavier nucleus.
[B1] From the graph, the product nucleus has a higher binding energy per nucleon than the original light nuclei (the curve rises steeply from low mass numbers toward the peak near $A = 56$), so energy is released.

Teaching Note: Fusion of light nuclei releases energy because the products are more tightly bound per nucleon. This is the energy source of stars, including our Sun.

15. (a) [2 marks]

Answer:
$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{250 \times 10^{-9}} = 7.956 \times 10^{-19} \text{ J}$

$E = \frac{7.956 \times 10^{-19}}{1.60 \times 10^{-19}} = 4.97 \text{ eV} \approx 5.0 \text{ eV}$

[M1] for correct calculation; [A1] for correct answer (accept 4.97 eV or 5.0 eV).

(b) [2 marks]

Answer:
Using $K_{\text{max}} = eV_s$ where $V_s$ is the stopping potential:

$K_{\text{max}} = 2.75 \text{ eV}$

$\phi = E - K_{\text{max}} = 4.97 - 2.75 = 2.22 \text{ eV} \approx 2.2 \text{ eV}$

[M1] for using photoelectric equation; [A1] for correct answer (accept 2.2 eV or 2.22 eV).

(c) [2 marks]

Answer:
[B1] The stopping potential will increase.
[B1] Decreasing the wavelength increases the frequency and hence the photon energy ($E = hc/\lambda$). From $K_{\text{max}} = hf - \phi$, the maximum kinetic energy increases, so a larger stopping potential is required to halt the fastest photoelectrons.

Teaching Note: The stopping potential is directly proportional to the maximum kinetic energy: $eV_s = K_{\text{max}}$.

Section C: Free Response

16. (a) [2 marks]

Answer:
[B1] From $K_{\text{max}} = hf - \phi$, photoelectrons are only emitted when $K_{\text{max}} > 0$, i.e., when $hf > \phi$.
[B1] This gives a minimum (threshold) frequency $f_0 = \phi/h$. Below this frequency, $hf < \phi$ and $K_{\text{max}}$ would be negative, which is physically impossible, so no photoelectrons are emitted regardless of intensity.

(b) [2 marks]

Answer:
[B1] The student's claim is incorrect because the photoelectric effect is a single-photon process — each electron absorbs one photon.
[B1] If the frequency is below threshold, each individual photon has insufficient energy ($hf < \phi$) to liberate an electron, so no amount of intensity (number of photons) can cause emission. Increasing intensity only increases the number of photons, not the energy of each photon.

(c) [3 marks]

Answer:
Photon energy: $E = hf = (6.63 \times 10^{-34})(8.0 \times 10^{14}) = 5.304 \times 10^{-19} \text{ J}$

Converting to eV: $E = \frac{5.304 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.315 \text{ eV}$

Maximum kinetic energy: $K_{\text{max}} = 3.315 - 2.5 = 0.815 \text{ eV} = 1.304 \times 10^{-19} \text{ J}$

Using $K_{\text{max}} = \frac{1}{2}m_e v^2$:

$v = \sqrt{\frac{2K_{\text{max}}}{m_e}} = \sqrt{\frac{2 \times 1.304 \times 10^{-19}}{9.11 \times 10^{-31}}} = \sqrt{2.86 \times 10^{11}} = 5.35 \times 10^5 \text{ m s}^{-1}$

[M1] for correct photon energy and kinetic energy; [M1] for using kinetic energy formula; [A1] for correct answer (accept $5.3 \times 10^5$ to $5.4 \times 10^5$ m/s).

17. (a) [2 marks]

Answer:
$^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + ^{4}_{2}\text{He}$

[M1] for correct mass and atomic number conservation; [A1] for complete correct equation.

(b) (i) [1 mark]

Answer:
$\Delta m = m_{\text{Ra}} - (m_{\text{Rn}} + m_{\alpha}) = 226.0254 - (222.0176 + 4.0026) = 226.0254 - 226.0202 = 0.0052 \text{ u}$

[A1] for correct answer: $\Delta m = 0.0052$ u.

(ii) [2 marks]

Answer:
Using $1 \text{ u} = 931.5 \text{ MeV}/c^2$:

$E = 0.0052 \times 931.5 = 4.84 \text{ MeV}$

[M1] for correct conversion; [A1] for correct answer (accept 4.8–4.9 MeV).

(iii) [2 marks]

Answer:
[B1] By conservation of momentum, the total momentum before decay is zero (the radium nucleus is at rest), so the momenta of the alpha particle and the radon nucleus must be equal in magnitude and opposite in direction: $m_\alpha v_\alpha = m_{\text{Rn}} v_{\text{Rn}}$.
[B1] Since kinetic energy $K = \frac{p^2}{2m}$, for equal momentum, the lighter particle (alpha particle, mass ~4 u) has much more kinetic energy than the heavier particle (radon, mass ~222 u). Specifically, $\frac{K_\alpha}{K_{\text{Rn}}} = \frac{m_{\text{Rn}}}{m_\alpha} = \frac{222}{4} \approx 55.5$, so the alpha particle carries about 98% of the kinetic energy.

Teaching Note: This is an important concept in nuclear physics — the lighter decay product carries most of the kinetic energy.

18. (a) [2 marks]

Answer:
[M1] From the graph, the activity decreases from 800 Bq to 400 Bq in 10 days.
[A1] Therefore, the half-life is 10 days.

Method: The half-life is the time for the activity to fall to half its initial value. From the graph, at $t = 0$, $A = 800$ Bq; at $t = 10$ days, $A = 400$ Bq = $\frac{1}{2} \times 800$ Bq. Hence $t_{1/2} = 10$ days.

(b) [2 marks]

Answer:
$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{10 \times 24 \times 3600} = \frac{0.693}{8.64 \times 10^5} = 8.02 \times 10^{-7} \text{ s}^{-1}$

Or in day$^{-1}$: $\lambda = \frac{0.693}{10} = 0.0693 \text{ day}^{-1}$

[M1] for correct formula; [A1] for correct answer.

(c) [2 marks]

Answer:
Number of half-lives in 35 days: $n = \frac{35}{10} = 3.5$

$A = A_0 \left(\frac{1}{2}\right)^{3.5} = 800 \times \left(\frac{1}{2}\right)^{3.5} = 800 \times 0.0884 = 70.7 \text{ Bq}$

Or using $A = 800 \times e^{-0.0693 \times 35} = 800 \times e^{-2.426} = 800 \times 0.0884 = 70.7$ Bq

[M1] for correct method; [A1] for correct answer (accept 70–71 Bq).

19. (a) [3 marks]

Answer:
[B1] The variable power supply is connected with the polarity reversed (anode negative with respect to cathode) so that it creates a retarding potential that opposes the motion of photoelectrons.
[B1] The voltage is gradually increased until the photocurrent falls to zero. This voltage is the stopping potential $V_s$.
[B1] At this point, the maximum kinetic energy of the photoelectrons equals the work done by the retarding field: $K_{\text{max}} = eV_s$, so $K_{\text{max}}$ can be determined.

Teaching Note: The stopping potential is the minimum potential needed to stop the fastest photoelectrons from reaching the anode. It is independent of the intensity of the light.

(b) (i) [1 mark]

Answer:
The maximum kinetic energy increases. From $K_{\text{max}} = hf - \phi$, increasing the frequency $f$ increases $K_{\text{max}}$.

(ii) [1 mark]

Answer:
The photocurrent remains the same. The intensity (number of photons per second) is constant, so the number of photoelectrons emitted per second is unchanged, and hence the current is unchanged.

Teaching Note: This distinction — frequency affects kinetic energy, intensity affects photocurrent — is a key concept in the photoelectric effect.

20. (a) [2 marks]

Answer:
[B1] Extremely high temperatures (on the order of $10^7$ to $10^8$ K) are required so that the nuclei have sufficient kinetic energy to overcome the electrostatic (Coulomb) repulsion between them.
[B1] A high density (or high pressure) of the reacting nuclei is needed to ensure a sufficient collision rate for the fusion reactions to be sustained.

Teaching Note: These conditions are found in the cores of stars. On Earth, achieving these conditions in a controlled manner is a major engineering challenge (e.g., tokamak magnetic confinement or inertial confinement).

(b) [2 marks]

Answer:
$^2_1\text{H} + ^2_1\text{H} \rightarrow ^3_2\text{He} + ^1_0\text{n}$

[M1] for correct mass and atomic number conservation; [A1] for identifying the other particle as a neutron ($^1_0\text{n}$).

Teaching Note: This is one of the deuterium-deuterium (D-D) fusion reactions. The other possible D-D reaction produces tritium and a proton: $^2_1\text{H} + ^2_1\text{H} \rightarrow ^3_1\text{H} + ^1_1\text{p}$.

(c) [2 marks]

Answer:
[B1] From the binding energy per nucleon curve, light nuclei (such as deuterium, with $A = 2$) have a low binding energy per nucleon (approximately 1.1 MeV), while the product helium-3 ($A = 3$) has a higher binding energy per nucleon (approximately 2.6 MeV).
[B1] The increase in binding energy per nucleon means the product nucleus is more tightly bound, and the mass defect is converted to energy according to $E = \Delta m c^2$. Hence, energy is released in the fusion reaction.

Teaching Note: Fusion releases energy for nuclei lighter than iron-56 (the peak of the binding energy curve). This is why fusion of light elements and fission of heavy elements both release energy — they both move toward more tightly bound nuclei.

END OF ANSWER KEY