Secondary 3 Physics Modern Physics Quiz

Secondary 3 Physics Quiz - Modern Physics (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions (10 marks)

1. [1 mark] Answer: B

Explanation: The photoelectric effect shows that electrons are emitted only when the incident light frequency exceeds a threshold frequency $f_0$, regardless of intensity. This contradicts classical wave theory which predicts energy depends on intensity. The kinetic energy of emitted electrons depends on frequency (not intensity), emission is nearly instantaneous (no time delay), and the number of electrons per second (photocurrent) is proportional to intensity.

Marking note: 1 mark for correct choice.

2. [1 mark] Answer: B

Working: $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{500 \times 10^{-9}} = 3.978 \times 10^{-19} \text{ J}$ Convert to eV: $\frac{3.978 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.486 \text{ eV} \approx 2.48 \text{ eV}$

Alternative: Use $hc = 1240 \text{ eV·nm}$, so $E = \frac{1240}{500} = 2.48 \text{ eV}$.

Marking note: 1 mark for correct choice.

3. [1 mark] Answer: C

Explanation: The stopping potential $V_s$ is related to maximum kinetic energy by $K_{\text{max}} = eV_s$. If $V_s = 2.5 \text{ V}$, then $K_{\text{max}} = 2.5 \text{ eV}$. In joules: $2.5 \times 1.60 \times 10^{-19} = 4.0 \times 10^{-19} \text{ J}$. Option C gives the answer directly in eV, which is the conventional unit for photoelectron energies.

Marking note: 1 mark for correct choice. Common mistake: confusing eV and J values.

4. [1 mark] Answer: B

Working: Photon energy: $E = hf = (6.63 \times 10^{-34})(8.0 \times 10^{14}) = 5.304 \times 10^{-19} \text{ J}$ In eV: $\frac{5.304 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.315 \text{ eV}$ $K_{\text{max}} = E - \phi = 3.315 - 2.0 = 1.315 \text{ eV} \approx 1.3 \text{ eV}$

Marking note: 1 mark for correct choice.

5. [1 mark] Answer: C

Explanation: Einstein's photoelectric equation: $K_{\text{max}} = hf - \phi = hf - hf_0$. This is a linear equation $y = mx + c$ with slope $h$ and x-intercept at threshold frequency $f_0$. Graph C shows a straight line with positive slope crossing the frequency axis at $f_0$ (negative intercept on $K_{\text{max}}$ axis).

Marking note: 1 mark for correct choice. Graph B is incorrect because it passes through origin (would imply zero work function).

6. [1 mark] Answer: A

Working: Electron kinetic energy: $K = eV = (1.60 \times 10^{-19})(100) = 1.60 \times 10^{-17} \text{ J}$ Momentum: $p = \sqrt{2m_eK} = \sqrt{2(9.11 \times 10^{-31})(1.60 \times 10^{-17})} = 5.40 \times 10^{-24} \text{ kg m/s}$ de Broglie wavelength: $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.40 \times 10^{-24}} = 1.23 \times 10^{-10} \text{ m}$

Shortcut formula: $\lambda = \frac{h}{\sqrt{2m_e eV}} \approx \frac{1.23}{\sqrt{V}} \text{ nm} = \frac{1.23}{\sqrt{100}} = 0.123 \text{ nm} = 1.23 \times 10^{-10} \text{ m}$

Marking note: 1 mark for correct choice.

7. [1 mark] Answer: C

Explanation: Electron diffraction (e.g., Davisson-Germer experiment) demonstrates the wave nature of electrons by showing interference patterns. The photoelectric effect and Compton scattering demonstrate particle nature of light. Line emission spectra demonstrate quantised atomic energy levels.

Marking note: 1 mark for correct choice.

8. [1 mark] Answer: A

Working: $E_3 = -\frac{13.6}{3^2} = -1.51 \text{ eV}$ $E_2 = -\frac{13.6}{2^2} = -3.40 \text{ eV}$ Photon energy: $\Delta E = E_3 - E_2 = (-1.51) - (-3.40) = 1.89 \text{ eV}$

Marking note: 1 mark for correct choice.

9. [1 mark] Answer: C

Explanation:

• A is incorrect: Characteristic X-ray lines depend on target material (atomic number), not accelerating voltage.
• B is incorrect: Minimum wavelength $\lambda_{\text{min}} = \frac{hc}{eV}$ depends only on accelerating voltage, not target material.
• C is correct: Higher accelerating voltage → more energetic electrons → more intense bremsstrahlung (continuous spectrum).
• D is incorrect: Characteristic lines appear at specific wavelengths; they can be shorter or longer than $\lambda_{\text{min}}$ depending on voltage.

Marking note: 1 mark for correct choice.

10. [1 mark] Answer: C

Working: 36 hours = 3 half-lives (36/12 = 3). Fraction remaining = $(\frac{1}{2})^3 = \frac{1}{8}$.

Marking note: 1 mark for correct choice.

Section B: Structured Questions (30 marks)

11. [3 marks]

(a) [2 marks] Any two of:

• Electrons are emitted only if the frequency of incident light exceeds a threshold frequency $f_0$, independent of intensity. (Classical theory predicts emission at any frequency given sufficient intensity.)
• Maximum kinetic energy of emitted electrons depends on frequency, not intensity. (Classical theory predicts energy depends on intensity.)
• Emission is nearly instantaneous (no time lag) even at very low intensities. (Classical theory predicts a time delay for energy accumulation.)
• The photocurrent (number of electrons per second) is directly proportional to light intensity. (Classical theory can explain this, but not the other observations.)

Marking: 1 mark per valid observation, max 2 marks.

(b) [1 mark] Einstein proposed light consists of photons with energy $E = hf$. An electron absorbs one photon. If $hf > \phi$ (work function), the electron is emitted with $K_{\text{max}} = hf - \phi$. This explains:

• Threshold frequency: need $hf > \phi$ → $f > f_0 = \phi/h$
• $K_{\text{max}}$ depends on $f$, not intensity
• Instantaneous emission: one photon → one electron interaction

Marking: 1 mark for correct explanation linking photon model to one observation.

12. [4 marks]

(a) [2 marks] $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{400 \times 10^{-9}} = 4.9725 \times 10^{-19} \text{ J}$ Answer: $4.97 \times 10^{-19} \text{ J}$ (or $4.97 \times 10^{-19} \text{ J}$)

Marking: 1 mark for correct substitution, 1 mark for correct answer with unit.

(b) [2 marks] Photon energy in eV: $\frac{4.9725 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.108 \text{ eV}$ (or use $hc = 1240 \text{ eV·nm}$: $\frac{1240}{400} = 3.10 \text{ eV}$) $K_{\text{max}} = E - \phi = 3.108 - 2.1 = 1.008 \text{ eV} \approx 1.01 \text{ eV}$

Answer: $1.01 \text{ eV}$ (accept $1.0 \text{ eV}$ or $1.01 \text{ eV}$)

Marking: 1 mark for converting photon energy to eV or using $hc = 1240 \text{ eV·nm}$, 1 mark for correct subtraction and answer with unit.

13. [3 marks]

(a) [2 marks] Energy difference: $\Delta E = E_4 - E_2 = (-0.85) - (-3.40) = 2.55 \text{ eV}$ Convert to joules: $2.55 \times 1.60 \times 10^{-19} = 4.08 \times 10^{-19} \text{ J}$ Wavelength: $\lambda = \frac{hc}{\Delta E} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{4.08 \times 10^{-19}} = 4.87 \times 10^{-7} \text{ m} = 487 \text{ nm}$

Alternative using $hc = 1240 \text{ eV·nm}$: $\lambda = \frac{1240}{2.55} = 486 \text{ nm}$

Answer: $487 \text{ nm}$ (accept $486 \text{ nm}$ or $4.87 \times 10^{-7} \text{ m}$)

Marking: 1 mark for correct energy difference, 1 mark for correct wavelength calculation with unit.

(b) [1 mark] Balmer series (transitions to $n=2$).

Marking: 1 mark for correct series name.

14. [4 marks]

(a) [2 marks] Minimum wavelength occurs when all electron kinetic energy converts to one photon: $eV = \frac{hc}{\lambda_{\text{min}}}$ $\lambda_{\text{min}} = \frac{hc}{eV} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{(1.60 \times 10^{-19})(50 \times 10^3)} = 2.486 \times 10^{-11} \text{ m}$

Alternative: $\lambda_{\text{min}} (\text{nm}) = \frac{1240}{V(\text{V})} = \frac{1240}{50000} = 0.0248 \text{ nm} = 2.48 \times 10^{-11} \text{ m}$

Answer: $2.49 \times 10^{-11} \text{ m}$ (or $0.0249 \text{ nm}$)

Marking: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

(b) [2 marks] The minimum wavelength corresponds to the maximum photon energy, which occurs when a single electron loses all its kinetic energy in one collision (bremsstrahlung). The electron's kinetic energy is $eV$, so the maximum photon energy is $eV$, giving a minimum wavelength $\lambda_{\text{min}} = hc/eV$. No photon can have energy greater than the electron's initial kinetic energy, hence the sharp cut-off.

Marking: 1 mark for stating maximum photon energy equals electron kinetic energy, 1 mark for explaining the cut-off.

15. [3 marks]

(a) [2 marks] Momentum: $p = m_e v = (9.11 \times 10^{-31})(2.0 \times 10^6) = 1.822 \times 10^{-24} \text{ kg m/s}$ $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.822 \times 10^{-24}} = 3.64 \times 10^{-10} \text{ m}$

Answer: $3.64 \times 10^{-10} \text{ m}$ (or $0.364 \text{ nm}$)

Marking: 1 mark for correct momentum calculation, 1 mark for correct wavelength with unit.

(b) [1 mark] Macroscopic objects have large mass, so their momentum $p = mv$ is very large even at ordinary speeds. Their de Broglie wavelength $\lambda = h/p$ is extremely small (typically $< 10^{-30} \text{ m}$), far too small to produce observable diffraction or interference effects.

Marking: 1 mark for correct explanation (large mass → large momentum → negligible wavelength).

16. [4 marks]

(a) [1 mark] From the table, activity halves every 10 days (8000 → 4000 → 2000 → 1000 → 500). Half-life = 10 days

Marking: 1 mark for correct value with unit.

(b) [2 marks] $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{10} = 0.0693 \text{ day}^{-1}$

Answer: $0.0693 \text{ day}^{-1}$ (accept $6.93 \times 10^{-2} \text{ day}^{-1}$)

Marking: 1 mark for correct formula, 1 mark for correct calculation with unit.

(c) [1 mark] After 55 days = 5.5 half-lives. $A = A_0 (\frac{1}{2})^{t/t_{1/2}} = 8000 \times (\frac{1}{2})^{5.5} = 8000 \times 0.0221 = 177 \text{ Bq}$

Alternative using decay constant: $A = A_0 e^{-\lambda t} = 8000 e^{-0.0693 \times 55} = 8000 e^{-3.8115} = 8000 \times 0.0221 = 177 \text{ Bq}$

Answer: $177 \text{ Bq}$ (accept $180 \text{ Bq}$ or $176 \text{ Bq}$)

Marking: 1 mark for correct answer with unit.

17. [3 marks]

(a) [1 mark] Corrected count rate = measured count rate - background = $1200 - 30 = 1170 \text{ counts/min}$

Answer: $1170 \text{ counts/min}$

Marking: 1 mark for correct subtraction.

(b) [1 mark] Corrected count rate with absorber = $150 - 30 = 120 \text{ counts/min}$ Fraction transmitted = $\frac{120}{1170} = 0.1026 \approx 0.103$ (or $10.3\%$)

Answer: $0.103$ (or $10.3\%$)

Marking: 1 mark for correct calculation (must subtract background from both readings).

(c) [1 mark] $\gamma$-radiation (gamma rays). Reasoning: $\alpha$-particles are stopped by a few cm of air or paper; $\beta$-particles are stopped by a few mm of aluminium. A 2 cm thick lead sheet transmits about 10% of the radiation, which is characteristic of penetrating $\gamma$-radiation.

Marking: 1 mark for correct radiation type with valid reasoning.

18. [3 marks]

(a) [1 mark] Electrons are produced by thermionic emission: the cathode filament is heated by a low-voltage current, giving electrons enough thermal energy to overcome the work function of the metal and escape.

Marking: 1 mark for "thermionic emission" or description of heating filament to release electrons.

(b) [1 mark] Kinetic energy of electrons → X-ray photons (electromagnetic radiation) + heat. (Accept: Electrical potential energy → kinetic energy of electrons → X-rays + heat)

Marking: 1 mark for correct energy conversion.

(c) [1 mark] Most of the electron kinetic energy (over 99%) is converted to heat at the anode target. Tungsten is used for its high melting point (3422°C), and cooling fins increase surface area for heat dissipation to prevent the anode from melting.

Marking: 1 mark for explaining heat production and need for high melting point/cooling.

19. [3 marks]

(a) [2 marks] The photoelectric effect requires photons with energy $E = hf$ greater than the work function $\phi$ of zinc. UV light has higher frequency (shorter wavelength) than red light, so UV photons have sufficient energy to eject electrons ($hf_{\text{UV}} > \phi$), while red photons do not ($hf_{\text{red}} < \phi$). Intensity only affects the number of photons per second, not the energy per photon.

Marking: 1 mark for linking UV frequency to work function, 1 mark for explaining why red light fails (photon energy too low) and intensity is irrelevant.

(b) [1 mark] Moving the UV source further away reduces the intensity (photon flux) at the zinc plate. Fewer photons per second strike the surface, so fewer electrons are emitted per second, reducing the discharge current and causing the leaf to fall more slowly.

Marking: 1 mark for linking distance to intensity to emission rate.

20. [3 marks]

(a) [2 marks] Activity ratio: $\frac{A}{A_0} = 0.25 = (\frac{1}{2})^{t/t_{1/2}}$ $(\frac{1}{2})^{t/5730} = (\frac{1}{2})^2$ $\frac{t}{5730} = 2$ $t = 2 \times 5730 = 11460 \text{ years}$

Alternative using decay constant: $\lambda = \frac{\ln 2}{5730} = 1.209 \times 10^{-4} \text{ yr}^{-1}$ $0.25 = e^{-\lambda t} \Rightarrow \ln 0.25 = -\lambda t \Rightarrow t = \frac{\ln 4}{\lambda} = \frac{1.386}{1.209 \times 10^{-4}} = 11460 \text{ years}$

Answer: $11460 \text{ years}$ (accept $11500 \text{ years}$ or $1.15 \times 10^4 \text{ years}$)

Marking: 1 mark for correct method (half-life or decay constant), 1 mark for correct answer with unit.

(b) [1 mark] Any one of:

• The initial carbon-14 activity in living organisms has been constant over time (constant atmospheric $^{14}\text{C}/^{12}\text{C}$ ratio).
• The sample has not been contaminated by carbon from other sources.
• The decay constant/half-life of carbon-14 has not changed.
• The sample was in equilibrium with the atmosphere when the organism died.

Marking: 1 mark for any valid assumption.

End of Answer Key