Secondary 4 Pure Physics Modern Physics Quiz

Secondary 4 Pure 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 photoelectrons are emitted only when the incident light frequency exceeds a threshold frequency $f_0$ (dependent on the work function $\phi$). The maximum kinetic energy depends on frequency ($K_{\text{max}} = hf - \phi$), not intensity. Intensity affects the number of photoelectrons (current), not their energy. Emission is instantaneous (no time delay).

2. [1 mark] Answer: A

Working: $E = 4.0 \text{ eV} = 4.0 \times 1.60 \times 10^{-19} = 6.40 \times 10^{-19} \text{ J}$ $\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{6.40 \times 10^{-19}} = 3.10 \times 10^{-7} \text{ m} = 310 \text{ nm}$

3. [1 mark] Answer: C

Explanation: The minimum wavelength (cutoff wavelength) of bremsstrahlung X-rays is given by $\lambda_{\text{min}} = \frac{hc}{eV}$, which depends only on the accelerating voltage $V$. The target material affects characteristic X-ray lines, not the continuous spectrum's minimum wavelength.

4. [1 mark] Answer: B

Explanation: Electron diffraction (e.g., Davisson-Germer experiment, or diffraction through graphite) demonstrates the wave nature of electrons. The photoelectric effect and Compton scattering demonstrate particle nature of light. Line emission spectra show quantised energy levels in atoms.

5. [1 mark] Answer: A

Working: Photon energy: $E = \frac{hc}{\lambda} = \frac{1240 \text{ eV·nm}}{400 \text{ nm}} = 3.1 \text{ eV}$ (using $hc = 1240 \text{ eV·nm}$) $K_{\text{max}} = E - \phi = 3.1 - 2.5 = 0.6 \text{ eV}$

6. [1 mark] Answer: B

Working: Period: 4 horizontal divisions × 2 ms/div = 8 ms Peak voltage: (3 vertical divisions / 2) × 5 V/div = 1.5 × 5 = 7.5 V (Peak-to-peak is 3 divisions, so amplitude is 1.5 divisions)

7. [1 mark] Answer: C

Explanation: Characteristic X-rays are produced when inner-shell electrons are ejected by bombarding electrons, and outer-shell electrons drop down to fill the vacancies, emitting photons with specific energies (wavelengths) characteristic of the target material. They form a line spectrum, not continuous. The continuous spectrum (bremsstrahlung) is from deceleration of electrons.

8. [1 mark] Answer: A

Working: Kinetic energy: $E = eV = (1.60 \times 10^{-19})(500) = 8.00 \times 10^{-17} \text{ J}$ Momentum: $p = \sqrt{2m_e E} = \sqrt{2(9.11 \times 10^{-31})(8.00 \times 10^{-17})} = 1.21 \times 10^{-23} \text{ kg m/s}$ de Broglie wavelength: $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.21 \times 10^{-23}} = 5.48 \times 10^{-11} \text{ m} \approx 5.5 \times 10^{-11} \text{ m}$

9. [1 mark] Answer: B

Working: Half-value layer: $I = \frac{I_0}{2} = I_0 e^{-\mu x_{\text{HVL}}}$ $\frac{1}{2} = e^{-\mu \times 3.0}$ $\ln 2 = \mu \times 3.0$ $\mu = \frac{\ln 2}{3.0} = \frac{0.693}{3.0} = 0.231 \text{ mm}^{-1} \approx 0.23 \text{ mm}^{-1}$

10. [1 mark] Answer: B

Explanation: A CRO displays voltage vs time, allowing measurement of period (hence frequency), peak voltage, waveform shape, and phase difference. It cannot directly measure resistance, power, or capacitance without additional circuitry and calculations.

Section B: Structured Questions (30 marks)

11. [3 marks] Photoelectric Effect

(a) [1 mark] Answer: 4.96 eV (accept 5.0 eV) Working: $E = \frac{hc}{\lambda} = \frac{1240 \text{ eV·nm}}{250 \text{ nm}} = 4.96 \text{ eV}$ Or: $E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{250 \times 10^{-9}} = 7.956 \times 10^{-19} \text{ J} = \frac{7.956 \times 10^{-19}}{1.60 \times 10^{-19}} = 4.97 \text{ eV}$

(b) [1 mark] Answer: $1.06 \times 10^{-19} \text{ J}$ (accept $1.1 \times 10^{-19} \text{ J}$) Working: $K_{\text{max}} = E - \phi = 4.96 - 4.3 = 0.66 \text{ eV}$ $K_{\text{max}} = 0.66 \times 1.60 \times 10^{-19} = 1.056 \times 10^{-19} \text{ J}$

(c) [1 mark] Answer: The photoelectric current doubles (increases proportionally with intensity). Explanation: Intensity is proportional to the number of photons per second. Doubling intensity doubles the number of incident photons per second, hence doubles the number of photoelectrons emitted per second (current), provided frequency > threshold frequency.

Common mistake: Saying kinetic energy increases. Kinetic energy depends only on frequency, not intensity.

12. [4 marks] X-ray Production

(a) [1 mark] Answer: $1.28 \times 10^{-14} \text{ J}$ Working: $K_{\text{max}} = eV = (1.60 \times 10^{-19})(80 \times 10^3) = 1.28 \times 10^{-14} \text{ J}$

(b) [1 mark] Answer: $1.55 \times 10^{-11} \text{ m}$ Working: $\lambda_{\text{min}} = \frac{hc}{eV} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{1.28 \times 10^{-14}} = 1.55 \times 10^{-11} \text{ m}$ Or using shortcut: $\lambda_{\text{min}} (\text{nm}) = \frac{1240}{V(\text{volts})} = \frac{1240}{80000} = 0.0155 \text{ nm} = 1.55 \times 10^{-11} \text{ m}$

(c) [1 mark] Answer: 192 W Working: Electron beam power: $P_{\text{beam}} = VI = (80 \times 10^3)(15 \times 10^{-3}) = 1200 \text{ W}$ X-ray power: $P_{\text{X-ray}} = 0.01 \times 1200 = 12 \text{ W}$ Wait, let me recalculate: $80,000 \times 0.015 = 1200 \text{ W}$. 1% of 1200 = 12 W. Correction: Answer is 12 W.

(d) [1 mark] Answer: The continuous spectrum (bremsstrahlung) is produced when incident electrons are decelerated by the target nuclei, losing kinetic energy which is emitted as X-ray photons of a continuous range of energies up to a maximum (cutoff wavelength). Characteristic lines are produced when incident electrons knock out inner-shell electrons, and outer-shell electrons transition down to fill the vacancies, emitting photons with specific energies characteristic of the target material.

13. [4 marks] Wave-Particle Duality

(a) [2 marks] Answer: $1.00 \times 10^{-10} \text{ m}$ (or 0.100 nm) Working: $E = eV = (1.60 \times 10^{-19})(150) = 2.40 \times 10^{-17} \text{ J}$ $p = \sqrt{2m_e E} = \sqrt{2(9.11 \times 10^{-31})(2.40 \times 10^{-17})} = 6.62 \times 10^{-24} \text{ kg m/s}$ $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{6.62 \times 10^{-24}} = 1.00 \times 10^{-10} \text{ m}$

Mark breakdown: 1 mark for correct kinetic energy/momentum calculation, 1 mark for correct wavelength.

(b) [1 mark] Answer: The diffraction pattern of concentric rings shows that electrons interfere constructively and destructively after passing through the graphite crystal lattice. This interference is a wave phenomenon, demonstrating that electrons exhibit wave-like behaviour with wavelength given by the de Broglie relation $\lambda = h/p$.

(c) [1 mark] Answer: The diameter of the diffraction rings decreases. Explanation: Increasing accelerating voltage to 600 V increases electron kinetic energy and momentum. Since $\lambda = h/p$, the de Broglie wavelength decreases. For diffraction, $d \sin \theta = n\lambda$; smaller $\lambda$ gives smaller diffraction angle $\theta$, so rings become smaller (closer to centre).

14. [4 marks] X-ray Attenuation

(a) [1 mark] Answer: 0.368 (or 36.8%) Working: $\frac{I}{I_0} = e^{-\mu x} = e^{-0.50 \times 2.0} = e^{-1.0} = 0.3679 \approx 0.368$

(b) [1 mark] Answer: 1.39 mm (or 1.4 mm) Working: $\mu x_{\text{HVL}} = \ln 2$ $x_{\text{HVL}} = \frac{\ln 2}{\mu} = \frac{0.693}{0.50} = 1.386 \text{ mm} \approx 1.39 \text{ mm}$

(c) [1 mark] Answer: 0.223 (or 22.3%) Working: Total thickness = 3.0 mm $\frac{I}{I_0} = e^{-\mu x_{\text{total}}} = e^{-0.50 \times 3.0} = e^{-1.5} = 0.2231$ Alternatively: $0.368 \times e^{-0.50 \times 1.0} = 0.368 \times 0.607 = 0.223$

(d) [1 mark] Answer: Low-energy (soft) X-rays are more strongly absorbed by the patient's soft tissues (photoelectric effect dominates at low energies), increasing radiation dose without contributing to image contrast (they don't reach the detector). Removing them reduces patient dose and improves image quality by reducing scatter.

15. [3 marks] Cathode Ray Oscilloscope (CRO)

(a) [1 mark] Answer: 20 ms Working: 2.5 cycles occupy 10 divisions horizontally. 1 cycle = 10 / 2.5 = 4 divisions Period = 4 div × 5 ms/div = 20 ms

(b) [1 mark] Answer: 4 V Working: Peak-to-peak = 4 divisions → Amplitude (peak) = 2 divisions Peak voltage = 2 div × 2 V/div = 4 V

(c) [1 mark] Answer: Square wave with period 20 ms (4 divisions per cycle), peak voltage ±4 V (2 divisions above and below centre line), showing at least 2 complete cycles (8 divisions horizontally). Marking: Correct period (4 div/cycle), correct amplitude (2 div peak), square wave shape, at least 2 cycles shown.

16. [4 marks] Line Spectra and Energy Levels

(a) [2 marks] Answer: 486 nm (accept 486.1 nm) Working: Energy difference: $\Delta E = E_4 - E_2 = (-0.85) - (-3.4) = 2.55 \text{ eV}$ $\lambda = \frac{hc}{\Delta E} = \frac{1240 \text{ eV·nm}}{2.55 \text{ eV}} = 486.3 \text{ nm} \approx 486 \text{ nm}$

Mark breakdown: 1 mark for correct energy difference, 1 mark for correct wavelength calculation.

(b) [1 mark] Answer: Electrons in hydrogen atoms occupy discrete energy levels. Photons are emitted only when electrons transition between these specific levels. The photon energy equals the difference between the two levels ($E = hf = \frac{hc}{\lambda}$), so only discrete wavelengths corresponding to these energy differences are emitted.

(c) [1 mark] Answer: Transitions from n=1 to higher levels (n=2, 3, 4, ...), i.e., the Lyman series. Explanation: At very low temperature, virtually all atoms are in the ground state (n=1). Absorption can only occur from n=1 to excited states, producing the Lyman series in the UV region.

17. [4 marks] Photoelectric Effect Experiment

(a) [1 mark] Answer: $5.0 \times 10^{14} \text{ Hz}$ Explanation: Threshold frequency is the x-intercept where $V_s = 0$. From graph: $f_0 = 5.0 \times 10^{14} \text{ Hz}$.

(b) [2 marks] Answer: $6.4 \times 10^{-34} \text{ J s}$ (accept $6.6 \times 10^{-34} \text{ J s}$) Working: Gradient = $\frac{\Delta V_s}{\Delta f} = \frac{h}{e}$ From graph: gradient = $\frac{1.5 - 0.5}{(8.0 - 6.0) \times 10^{14}} = \frac{1.0}{2.0 \times 10^{14}} = 5.0 \times 10^{-15} \text{ V/Hz}$ $h = e \times \text{gradient} = (1.60 \times 10^{-19})(5.0 \times 10^{-15}) = 8.0 \times 10^{-34} \text{ J s}$ Wait, let me check the graph values again. Points: (6.0, 0.5), (8.0, 1.5), (10.0, 2.5) in units of $10^{14}$ Hz and V. Gradient = $\frac{2.5 - 0.5}{(10.0 - 6.0) \times 10^{14}} = \frac{2.0}{4.0 \times 10^{14}} = 5.0 \times 10^{-15} \text{ V/Hz}$ $h = (1.60 \times 10^{-19})(5.0 \times 10^{-15}) = 8.0 \times 10^{-34} \text{ J s}$ But actual $h = 6.63 \times 10^{-34}$. The graph data gives a higher value. This is acceptable as experimental data. Answer: $8.0 \times 10^{-34} \text{ J s}$ (based on given graph data)

Mark breakdown: 1 mark for correct gradient calculation, 1 mark for $h = e \times \text{gradient}$.

(c) [1 mark] Answer: Yes, the graph is consistent. Working: Work function $\phi = h f_0 = (6.63 \times 10^{-34})(5.0 \times 10^{14}) = 3.315 \times 10^{-19} \text{ J} = \frac{3.315 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.07 \text{ eV} \approx 2.1 \text{ eV}$. Or using graph gradient: $\phi = e \times |y\text{-intercept}| = (1.60 \times 10^{-19})(2.0) = 3.2 \times 10^{-19} \text{ J} = 2.0 \text{ eV}$. Both close to 2.1 eV.

18. [3 marks] X-ray Imaging

(a) [1 mark] Answer: "Harden the beam" means removing low-energy (soft) X-rays from the beam using a filter (e.g., aluminium), so that the remaining beam has a higher average energy (harder X-rays) which penetrates tissue better and reduces patient dose.

(b) [1 mark] Answer: 36.8% Working: $\frac{I}{I_0} = e^{-\mu x} = e^{-0.40 \times 2.5} = e^{-1.0} = 0.3679 = 36.8\%$

(c) [1 mark] Answer: To minimise the risk of stochastic effects (cancer induction, genetic mutations) and deterministic effects (tissue damage) from ionising radiation, following the ALARA (As Low As Reasonably Achievable) principle.

19. [4 marks] Electron Microscope

(a) [2 marks] Answer: $3.70 \times 10^{-12} \text{ m}$ (or 3.70 pm) Working: $E = eV = (1.60 \times 10^{-19})(100 \times 10^3) = 1.60 \times 10^{-14} \text{ J}$ $m_e c^2 = (9.11 \times 10^{-31})(3.00 \times 10^8)^2 = 8.20 \times 10^{-14} \text{ J}$ $p = \frac{1}{c} \sqrt{E^2 + 2E m_e c^2} = \frac{1}{3.00 \times 10^8} \sqrt{(1.60 \times 10^{-14})^2 + 2(1.60 \times 10^{-14})(8.20 \times 10^{-14})}$ $= \frac{1}{3.00 \times 10^8} \sqrt{2.56 \times 10^{-28} + 2.62 \times 10^{-27}} = \frac{1}{3.00 \times 10^8} \sqrt{2.88 \times 10^{-27}}$ $= \frac{1}{3.00 \times 10^8} (5.37 \times 10^{-14}) = 1.79 \times 10^{-22} \text{ kg m/s}$ $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.79 \times 10^{-22}} = 3.70 \times 10^{-12} \text{ m}$

Mark breakdown: 1 mark for correct relativistic momentum calculation, 1 mark for correct wavelength.

(b) [1 mark] Answer: The de Broglie wavelength of 100 keV electrons (~3.7 pm) is much smaller than the wavelength of visible light (~400-700 nm). Resolution is

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Secondary 4 Pure 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 photoelectrons are emitted only when the incident light frequency exceeds a threshold frequency $f_0$ (dependent on the work function $\phi$). The maximum kinetic energy depends on frequency ($K_{\text{max}} = hf - \phi$), not intensity. Intensity affects the number of photoelectrons (current), not their energy. Emission is instantaneous (no time delay).

2. [1 mark] Answer: A

Working: $E = 4.0 \text{ eV} = 4.0 \times 1.60 \times 10^{-19} = 6.40 \times 10^{-19} \text{ J}$ $\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{6.40 \times 10^{-19}} = 3.10 \times 10^{-7} \text{ m} = 310 \text{ nm}$

3. [1 mark] Answer: C

Explanation: The minimum wavelength (cutoff wavelength) of X-rays produced in an X-ray tube is given by $\lambda_{\text{min}} = \frac{hc}{eV}$, which depends only on the accelerating voltage $V$. The target material affects the characteristic X-ray lines, not the continuous spectrum cutoff.

4. [1 mark] Answer: B

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

5. [1 mark] Answer: A

Working: Photon energy: $E = \frac{hc}{\lambda} = \frac{1240 \text{ eV·nm}}{400 \text{ nm}} = 3.1 \text{ eV}$ $K_{\text{max}} = E - \phi = 3.1 \text{ eV} - 2.5 \text{ eV} = 0.6 \text{ eV}$

6. [1 mark] Answer: B

Working: Period = horizontal divisions × time-base = 4 div × 2 ms/div = 8 ms Peak voltage = (vertical divisions / 2) × Y-gain = (3 div / 2) × 5 V/div = 1.5 × 5 = 7.5 V

7. [1 mark] Answer: C

Explanation: Characteristic X-rays are produced when inner-shell electrons are ejected by bombarding electrons, and outer-shell electrons fall to fill the vacancies, emitting photons with specific energies (wavelengths) characteristic of the target material. They form a line spectrum, not continuous. Their wavelengths depend on the target material, not the accelerating voltage (above the threshold).

8. [1 mark] Answer: A

Working: $K = eV = (1.60 \times 10^{-19})(500) = 8.00 \times 10^{-17} \text{ J}$ $p = \sqrt{2m_e K} = \sqrt{2(9.11 \times 10^{-31})(8.00 \times 10^{-17})} = 1.21 \times 10^{-23} \text{ kg m/s}$ $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.21 \times 10^{-23}} = 5.48 \times 10^{-11} \text{ m} \approx 5.5 \times 10^{-11} \text{ m}$

9. [1 mark] Answer: B

Working: HVL = $\frac{\ln 2}{\mu} \Rightarrow \mu = \frac{\ln 2}{\text{HVL}} = \frac{0.693}{3.0 \text{ mm}} = 0.231 \text{ mm}^{-1} \approx 0.23 \text{ mm}^{-1}$

10. [1 mark] Answer: B

Explanation: A CRO displays voltage vs time waveforms. It can measure frequency (from period), peak voltage, phase difference, etc. It cannot directly measure resistance, power, or capacitance without additional circuits/calculations.

Section B: Structured Questions (30 marks)

11. [3 marks] Photoelectric Effect

(a) [1 mark] Answer: 4.96 eV (accept 5.0 eV) Working: $E = \frac{hc}{\lambda} = \frac{1240 \text{ eV·nm}}{250 \text{ nm}} = 4.96 \text{ eV}$

(b) [1 mark] Answer: $1.06 \times 10^{-19} \text{ J}$ (accept $1.1 \times 10^{-19} \text{ J}$) Working: $K_{\text{max}} = 4.96 \text{ eV} - 4.3 \text{ eV} = 0.66 \text{ eV}$ $K_{\text{max}} = 0.66 \times 1.60 \times 10^{-19} = 1.056 \times 10^{-19} \text{ J}$

(c) [1 mark] Answer: The photoelectric current doubles (or increases proportionally). Explanation: Doubling intensity doubles the number of photons per second, hence doubles the number of photoelectrons emitted per second (current), provided frequency > threshold.

12. [4 marks] X-ray Production

(a) [1 mark] Answer: $1.28 \times 10^{-14} \text{ J}$ Working: $K_{\text{max}} = eV = (1.60 \times 10^{-19})(80 \times 10^3) = 1.28 \times 10^{-14} \text{ J}$

(b) [1 mark] Answer: $1.55 \times 10^{-11} \text{ m}$ Working: $\lambda_{\text{min}} = \frac{hc}{eV} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{1.28 \times 10^{-14}} = 1.55 \times 10^{-11} \text{ m}$

(c) [1 mark] Answer: 19.2 W Working: Electron beam power = $VI = (80 \times 10^3)(15 \times 10^{-3}) = 1200 \text{ W}$ X-ray power = 1% × 1200 W = 12 W Correction: 1% of 1200 W = 12 W, not 19.2 W. Let me recalculate. $P_{\text{electron}} = IV = 0.015 \times 80000 = 1200 \text{ W}$ $P_{\text{X-ray}} = 0.01 \times 1200 = 12 \text{ W}$ Correct Answer: 12 W

(d) [1 mark] Answer: The continuous spectrum (bremsstrahlung) is produced when incident electrons are decelerated by target nuclei, losing varying amounts of kinetic energy. Characteristic lines are produced when inner-shell electrons are ejected and outer-shell electrons fill the vacancies, emitting photons of specific energies unique to the target material.

13. [4 marks] Wave-Particle Duality

(a) [2 marks] Answer: $1.00 \times 10^{-10} \text{ m}$ (or 0.100 nm) Working: $K = eV = (1.60 \times 10^{-19})(150) = 2.40 \times 10^{-17} \text{ J}$ $p = \sqrt{2m_e K} = \sqrt{2(9.11 \times 10^{-31})(2.40 \times 10^{-17})} = 6.61 \times 10^{-24} \text{ kg m/s}$ $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} = 1.00 \times 10^{-10} \text{ m}$

(b) [1 mark] Answer: Electrons behave as waves with de Broglie wavelength $\lambda = h/p$. When the electron beam passes through the graphite crystal, the atomic planes act as a diffraction grating. Constructive interference occurs at specific angles satisfying Bragg's law $2d\sin\theta = n\lambda$, producing concentric diffraction rings on the screen.

(c) [1 mark] Answer: The diameter of the diffraction rings decreases. Explanation: Increasing accelerating voltage to 600 V increases electron kinetic energy and momentum, decreasing de Broglie wavelength ($\lambda \propto 1/\sqrt{V}$). Smaller $\lambda$ means smaller diffraction angles (Bragg's law), so rings shrink.

14. [4 marks] X-ray Attenuation

(a) [1 mark] Answer: 0.368 (or 36.8%) Working: $\frac{I}{I_0} = e^{-\mu x} = e^{-(0.50)(2.0)} = e^{-1.0} = 0.3679$

(b) [1 mark] Answer: 1.39 mm (or 1.4 mm) Working: $\text{HVL} = \frac{\ln 2}{\mu} = \frac{0.693}{0.50} = 1.386 \text{ mm}$

(c) [1 mark] Answer: 0.223 (or 22.3%) Working: Total thickness = 3.0 mm $\frac{I}{I_0} = e^{-(0.50)(3.0)} = e^{-1.5} = 0.2231$ Alternative: $0.3679 \times e^{-0.50} = 0.3679 \times 0.6065 = 0.2231$

(d) [1 mark] Answer: Low-energy (soft) X-rays are more strongly absorbed by the patient's soft tissues, increasing radiation dose without contributing to image contrast (they don't reach the detector). Filters (e.g., aluminium) preferentially absorb low-energy photons, "hardening" the beam, reducing patient dose and improving image quality.

15. [3 marks] Cathode Ray Oscilloscope (CRO)

(a) [1 mark] Answer: 20 ms Working: 2.5 cycles occupy 10 horizontal divisions. 1 cycle = 10 div / 2.5 = 4 div Period = 4 div × 5 ms/div = 20 ms

(b) [1 mark] Answer: 4 V Working: Peak-to-peak = 4 divisions Amplitude (peak) = 2 divisions Peak voltage = 2 div × 2 V/div = 4 V

(c) [1 mark] Answer: Sketch showing a square wave with:

• Period = 4 divisions (2.5 cycles across 10 divisions)
• Peak-to-peak = 4 divisions (amplitude = 2 divisions above and below centre)
• Vertical transitions at correct time intervals
• At least 2 complete cycles shown

16. [4 marks] Line Spectra and Energy Levels

(a) [2 marks] Answer: 486 nm (accept 486.1 nm) Working: $\Delta E = E_4 - E_2 = (-0.85) - (-3.4) = 2.55 \text{ eV}$ $\lambda = \frac{hc}{\Delta E} = \frac{1240 \text{ eV·nm}}{2.55 \text{ eV}} = 486.3 \text{ nm}$

(b) [1 mark] Answer: Electrons in hydrogen atoms occupy discrete energy levels. Photons are emitted only when electrons transition between these specific levels. The photon energy equals the energy difference between levels ($hf = E_i - E_f$), so only specific wavelengths (corresponding to allowed transitions) are emitted.

(c) [1 mark] Answer: Only transitions from n=1 to higher levels (n=2, 3, 4, ...) — the Lyman series in UV. Explanation: At very low temperature, virtually all atoms are in the ground state (n=1). Absorption can only occur from n=1 to excited states.

17. [4 marks] Photoelectric Effect Experiment

(a) [1 mark] Answer: $5.0 \times 10^{14} \text{ Hz}$ Explanation: Threshold frequency is the x-intercept where $V_s = 0$.

(b) [2 marks] Answer: $6.6 \times 10^{-34} \text{ J s}$ (accept $6.63 \times 10^{-34} \text{ J s}$) Working: Gradient = $\frac{\Delta V_s}{\Delta f} = \frac{2.5 - 0.5}{(10.0 - 6.0) \times 10^{14}} = \frac{2.0}{4.0 \times 10^{14}} = 5.0 \times 10^{-15} \text{ V/Hz}$ $e \times \text{gradient} = h$ $h = (1.60 \times 10^{-19})(5.0 \times 10^{-15}) = 8.0 \times 10^{-34} \text{ J s}$ Wait, let me check the graph values again. Graph shows: points at (6.0, 0.5), (8.0, 1.5), (10.0, 2.5) in units of $10^{14}$ Hz and V. Gradient = $\frac{2.5 - 0.5}{(10.0 - 6.0) \times 10^{14}} = \frac{2.0}{4.0 \times 10^{14}} = 0.5 \times 10^{-14} = 5.0 \times 10^{-15} \text{ V/Hz}$ $h = e \times \text{gradient} = 1.60 \times 10^{-19} \times 5.0 \times 10^{-15} = 8.0 \times 10^{-34} \text{ J s}$ But the description says "gradient = 0.5 V per 10^14 Hz" which is $0.5 / 10^{14} = 5.0 \times 10^{-15} \text{ V/Hz}$. Same result. However, the accepted value is $6.63 \times 10^{-34}$. The graph data gives $8.0 \times 10^{-34}$. Students should calculate from the given graph data. Answer: $8.0 \times 10^{-34} \text{ J s}$ (based on graph data provided)

(c) [1 mark] Answer: The graph gives a work function of 2.0 eV, not 2.1 eV. The y-intercept is -2.0 V, so $\phi = e \times |y\text{-intercept}| = 2.0 \text{ eV}$. This is close to but not exactly 2.1 eV; the discrepancy could be due to experimental error or contact potential difference.

18. [3 marks] X-ray Imaging

(a) [1 mark] Answer: "Harden the beam" means removing low-energy (soft) X-ray photons from the beam using a filter (e.g., aluminium), increasing the average energy (penetrating power) of the beam.

(b) [1 mark] Answer: 36.8% (accept 37%) Working: $\frac{I}{I_0} = e^{-\mu x} = e^{-(0.40)(2.5)} = e^{-1.0} = 0.3679 = 36.8\%$

(c) [1 mark] Answer: To minimise the risk of radiation-induced cancer and genetic damage to the patient (stochastic effects), and to avoid deterministic effects (e.g., skin burns) at high doses.

19. [4 marks] Electron Microscope

(a) [2 marks] Answer: $3.70 \times 10^{-12} \text{ m}$ (or 3.70 pm) Working: $E = eV = (1.60 \times 10^{-19})(100 \times 10^3) = 1.60 \times 10^{-14} \text{ J}$ $m_e c^2 = (9.11 \times 10^{-31})(3.00 \times 10^8)^2 = 8.20 \times 10^{-14} \text{ J}$ $p = \frac{1}{c} \sqrt{E^2 + 2E m_e c^2} = \frac{1}{3.00 \times 10^8} \sqrt{(1.60 \times 10^{-14})^2 + 2(1.60 \times 10^{-14})(8.20 \times 10^{-14})}$ $p = \frac{1}{3.00 \times 10^8} \sqrt{2.56 \times 10^{-28} + 2.62 \times 10^{-27}} = \frac{1}{3.00 \times 10^8} \sqrt{2.88 \times 10^{-27}} = \frac{5.37 \times 10^{-14}}{3.00 \times 10^8} = 1.79 \times 10^{-22} \text{ kg m/s}$ $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.79 \times 10^{-22}} = 3.70 \times 10^{-12} \text{ m}$

(b) [1 mark] Answer: The de Broglie wavelength of 100 keV electrons (~3.7 pm) is much smaller than the wavelength of visible light (~400-700 nm). Resolution is limited by diffraction to approximately the wavelength used, so electrons achieve ~10^5 times better resolution.

(c) [1 mark] Answer: Biological samples must be placed in high vacuum, which causes dehydration and structural damage. Also, the high-energy electron beam causes radiation damage to delicate biological structures. Samples require complex preparation (fixation, staining, sectioning) which may introduce artifacts.

20. [4 marks] Applications of Modern Physics

(a) [1 mark] Answer: Conservation of momentum. The positron-electron system has near-zero total momentum before annihilation (thermal motion negligible). To conserve momentum, the two gamma photons must have equal and opposite momenta, so they travel in opposite directions (180° apart).

(b) [1 mark] Answer: $2.43 \times 10^{-12} \text{ m}$ (or 2.43 pm) Working: $E = 511 \text{ keV} = 511 \times 10^3 \times 1.60 \times 10^{-19} = 8.18 \times 10^{-14} \text{ J}$ $\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{8.18 \times 10^{-14}} = 2.43 \times 10^{-12} \text{ m}$

(c) [1 mark] Answer: Cooling reduces thermal noise (dark current) caused by thermally generated electron-hole pairs in the semiconductor. This improves energy resolution and signal-to-noise ratio, allowing accurate measurement of gamma photon energies.

(d) [1 mark] Answer: PET scans show metabolic/physiological function (e.g., glucose uptake, blood flow) rather than just anatomical structure. They can detect disease at cellular level before structural changes appear, and provide 3D functional images with quantitative data.

End of Answer Key