A Level H2 Physics Modern Physics Quiz

Answer Key - A-Level Physics H2 Quiz (Modern Physics)

1. Photon Hypothesis [2]

• Light consists of discrete packets of energy called photons [1].
• The energy of a photon is proportional to its frequency ($E = hf$) [1].

2. Photon Energy [2]

• $E = hc / \lambda = (6.63 \times 10^{-34} \times 3.00 \times 10^8) / (320 \times 10^{-9})$
• $E = 6.21 \times 10^{-19} \text{ J}$
• $E = (6.21 \times 10^{-19}) / (1.60 \times 10^{-19}) = 3.88 \text{ eV}$ [2]

3. Threshold Frequency [2]

• $\Phi = h f_0 \implies f_0 = \Phi / h$
• $\Phi = 2.3 \times 1.60 \times 10^{-19} = 3.68 \times 10^{-19} \text{ J}$
• $f_0 = (3.68 \times 10^{-19}) / (6.63 \times 10^{-34}) = 5.55 \times 10^{14} \text{ Hz}$ [2]

4. Work Function [3]

• $E_{photon} = hc / \lambda = (6.63 \times 10^{-34} \times 3.00 \times 10^8) / (200 \times 10^{-9}) = 9.945 \times 10^{-19} \text{ J} = 6.21 \text{ eV}$ [1]
• $K_{\max} = eV_s = 3.9 \text{ eV}$ [1]
• $\Phi = E_{photon} - K_{\max} = 6.21 - 3.9 = 2.31 \text{ eV}$ [1]

5. Frequency and KE [3]

• According to Einstein's equation $hf = \Phi + K_{\max}$ [1].
• Work function $\Phi$ is a constant property of the metal [1].
• Therefore, any increase in photon energy ($hf$) results in a direct increase in the maximum kinetic energy of the emitted electrons [1].

6. Intensity Effects [3]

• Photoelectric current: Increases. Higher intensity means more photons per second, leading to more photoelectrons emitted per second [2].
• Stopping potential: Remains constant. Stopping potential depends only on the maximum KE of electrons, which is determined by frequency, not intensity [1].

7. de Broglie Wavelength [3]

• $K = eV = 1.5 \times 10^3 \times 1.60 \times 10^{-19} = 2.40 \times 10^{-16} \text{ J}$ [1]
• $p = \sqrt{2mK} = \sqrt{2 \times 9.11 \times 10^{-31} \times 2.40 \times 10^{-16}} = 2.09 \times 10^{-23} \text{ kg m/s}$ [1]
• $\lambda = h/p = (6.63 \times 10^{-34}) / (2.09 \times 10^{-23}) = 3.17 \times 10^{-11} \text{ m}$ [1]

8. Binding Energy per Nucleon [2]

• The total binding energy of a nucleus divided by the number of nucleons (protons + neutrons) it contains [2].

9. Binding Energy Calculation [4]

• $\Delta m = [ (2 \times 1.007276) + (2 \times 1.008665) ] - 4.002603$ [1]
• $\Delta m = 4.023882 - 4.002603 = 0.021279 \text{ u}$ [1]
• $BE = 0.021279 \times 931.5 \text{ MeV} = 19.82 \text{ MeV}$ [2]

10. Stability [2]

• A higher BE per nucleon means more energy is required to remove a single nucleon from the nucleus [1].
• This indicates a stronger net attractive force relative to the electrostatic repulsion, making the nucleus more stable [1].

11. Half-Life [3]

• $A/A_0 = (1.50 \times 10^2) / (1.20 \times 10^3) = 0.125 = 1/8$ [1]
• $1/8 = (1/2)^3$, so 3 half-lives have passed [1]
• $T_{1/2} = 12.0 / 3 = 4.0 \text{ hours}$ [1]

12. Beta Plus Decay [3]

• A proton in the nucleus converts into a neutron [1].
• A positron ($e^+$) and a neutrino ($\nu_e$) are emitted [1].
• The atomic number decreases by 1, while the mass number remains constant [1].

13. Alpha Decay [2]

• Atomic number decreases by 2 [1].
• Mass number decreases by 4 [1].

14. Fusion Energy [3]

• The mass of the resulting nucleus is less than the sum of the masses of the original light nuclei [1].
• This mass defect ($\Delta m$) is converted into energy according to $E = \Delta mc^2$ [1].
• Because the binding energy per nucleon increases (the product is more stable), energy is released [1].

15. Population Inversion [3]

• A state where there are more atoms in a higher energy excited state than in a lower energy state [2].
• Necessary because stimulated emission can only dominate over absorption if the higher state is more populated [1].

16. Spontaneous vs Stimulated [3]

• Spontaneous: An atom in an excited state decays naturally, emitting a photon in a random direction and phase [1.5].
• Stimulated: An incident photon triggers an excited atom to decay, emitting a second photon with the same frequency, phase, and direction [1.5].

17. Forward Bias [3]

• The positive terminal repels holes toward the junction and the negative terminal repels electrons toward the junction [1].
• This narrows the depletion region [1].
• This lowers the potential barrier, allowing majority carriers to cross the junction and increasing current flow [1].

18. Reverse Bias [3]

• The positive terminal attracts electrons and the negative terminal attracts holes away from the junction [1].
• This widens the depletion region [1].
• The potential barrier increases, preventing majority carriers from crossing; only a negligible minority carrier current flows [1].

19. Zener Regulation [3]

• The Zener diode is operated in the breakdown region (reverse bias) [1].
• In this region, a small increase in voltage leads to a large increase in current [1].
• This allows the diode to "shunt" excess current to ground, keeping the voltage across the Zener diode (and thus the load) constant [1].

20. Conductivity [4]

• n-type has higher conductivity [1].
• In intrinsic semiconductors, only a few electrons are thermally excited into the conduction band [1].
• In n-type, pentavalent impurities provide many additional free electrons (donor levels) [1].
• This significantly increases the charge carrier concentration, thus increasing conductivity [1].