A Level H2 Physics Modern Physics Quiz

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

1.
The minimum energy required to remove an electron from the surface of a metal. [1]

2.
$hf = \Phi + E_{k,max}$ [1]
Where:
$h$ is Planck’s constant,
$f$ is the frequency of incident radiation,
$\Phi$ is the work function of the metal,
$E_{k,max}$ is the maximum kinetic energy of the emitted photoelectron. [1]
(1 mark for correct equation, 1 mark for defining at least 2 symbols correctly)

3.
(a) $E = \frac{hc}{\lambda}$ [1]
$E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{450 \times 10^{-9}}$
$E = 4.42 \times 10^{-19} \text{ J}$ [1]

(b) $E (\text{eV}) = \frac{4.42 \times 10^{-19}}{1.60 \times 10^{-19}}$ [1]
$E = 2.76 \text{ eV}$

4.
(a) $f_0$ is the threshold frequency. It is the minimum frequency of incident radiation required to emit photoelectrons from the metal surface. [1]

(b) The graph intersects the y-axis at $-\Phi$ because energy is required to overcome the work function. If $f < f_0$, the photon energy $hf$ is less than the work function $\Phi$, so no electrons are emitted (kinetic energy cannot be negative). The equation $E_{k,max} = hf - \Phi$ shows a negative intercept. [2]

5.
(a) No change. [1]
The maximum kinetic energy depends only on the frequency of the incident photons and the work function of the metal ($E_{k,max} = hf - \Phi$). Intensity affects the number of photons, not their individual energy. [1]

(b) Current doubles (increases proportionally). [1]
Intensity is proportional to the number of photons incident per unit time. Doubling intensity doubles the number of photons, which doubles the number of photoelectrons emitted per second, hence doubling the current. [1]

6.
(a) $E_k = eV$ [1]
$E_k = 1.60 \times 10^{-19} \times 2500$
$E_k = 4.00 \times 10^{-16} \text{ J}$ [1]

(b) $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}}$ [1]
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 4.00 \times 10^{-16}}}$ [1]
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{7.288 \times 10^{-46}}}$
$\lambda = \frac{6.63 \times 10^{-34}}{2.70 \times 10^{-23}}$
$\lambda = 2.46 \times 10^{-11} \text{ m}$ [1]

7.
The photoelectric effect. [1]
Wave theory predicts that energy accumulates over time, so low-intensity light should eventually eject electrons. However, experiments show that emission is instantaneous and only occurs if frequency exceeds a threshold, supporting the particle (photon) model where energy is quantized in packets $E=hf$. [1]

8.
The energy required to completely separate a nucleus into its constituent protons and neutrons (nucleons). [1]
Alternatively: The energy released when constituent nucleons combine to form a nucleus. [1]

9.
(a) $E = mc^2$
$1 \text{ u} = 1.6605 \times 10^{-27} \text{ kg}$
$E = 1.6605 \times 10^{-27} \times (3.00 \times 10^8)^2 = 1.494 \times 10^{-10} \text{ J}$ [1]
$1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$
$E (\text{MeV}) = \frac{1.494 \times 10^{-10}}{1.60 \times 10^{-13}} \approx 934 \text{ MeV}$ (Accept 931.5 based on precise constants) [1]

(b) The mass difference (mass defect) is converted into binding energy according to $E=mc^2$. This energy is released when the nucleus forms, making the bound system more stable (lower energy state) than the separated nucleons. [2]

10.
(a) Mass of reactants $= 2.0141 + 3.0160 = 5.0301 \text{ u}$
Mass of products $= 4.0026 + 1.0087 = 5.0113 \text{ u}$
Mass defect $\Delta m = 5.0301 - 5.0113 = 0.0188 \text{ u}$ [2]

(b) Energy $= 0.0188 \times 931.5$ [1]
Energy $= 17.51 \text{ MeV}$ [1]

11.
(a) The time taken for half the number of radioactive nuclei in a sample to decay (or for the activity to fall to half its initial value). [1]

(b) Time elapsed $= 2 \text{ days} = 48 \text{ hours}$.
Number of half-lives $n = \frac{48}{12} = 4$. [1]
$A = A_0 \left(\frac{1}{2}\right)^n = 800 \times \left(\frac{1}{2}\right)^4 = 800 \times \frac{1}{16} = 50 \text{ Bq}$. [1]

12.
(a) In fusion, light nuclei combine to form a heavier nucleus with a higher binding energy per nucleon. [1]
This increase in binding energy per nucleon means the final nucleus is more tightly bound, and the difference in binding energy is released as kinetic energy/radiation. [1]

(b) In fission, a heavy nucleus splits into lighter nuclei which have a higher binding energy per nucleon (closer to the peak at Fe-56). [1]
The increase in binding energy per nucleon results in a release of energy. [1]

13.
$^{14}_6\text{C} \rightarrow ^{14}_7\text{N} + ^0_{-1}\beta + \bar{\nu}_e$ [2]
(1 mark for correct Nucleus, 1 mark for electron and antineutrino)

14.
(a) If only two particles (nucleus and electron) were involved, conservation of energy and momentum would require the electron to have a fixed discrete energy. The continuous spectrum implies a third particle (antineutrino) shares the energy and momentum variably. [2]

(b) Any two from:

1. Zero charge. [1]
2. Very small/negligible mass. [1]
3. Weakly interacting (low interaction cross-section). [1]

15.
(a) Incident electrons are decelerated by the electric fields of the target nuclei. [1]
The loss in kinetic energy is converted into a photon. Since electrons lose varying amounts of energy (from zero to maximum), a continuous range of photon energies (wavelengths) is produced. [1]

(b) Incident electrons collide with and eject inner-shell electrons from the target atoms. [1]
Outer-shell electrons drop down to fill the vacancies, emitting photons with specific energies corresponding to the difference between discrete atomic energy levels. [1]

16.
(a) Maximum photon energy corresponds to an electron losing all its kinetic energy in a single collision.
$E_{max} = hf_{max} = \frac{hc}{\lambda_{min}}$ [1]
Kinetic energy of electron $K = eV$.
Equating energies: $eV = \frac{hc}{\lambda_{min}} \Rightarrow \lambda_{min} = \frac{hc}{eV}$. [1]

(b) $\lambda_{min}$ is halved. [1]

17.
(a) Energy of one photon $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{633 \times 10^{-9}} = 3.14 \times 10^{-19} \text{ J}$. [1]
Power $P = 2.0 \text{ mW} = 2.0 \times 10^{-3} \text{ J s}^{-1}$. [1]
Number of photons $N = \frac{P}{E} = \frac{2.0 \times 10^{-3}}{3.14 \times 10^{-19}} = 6.37 \times 10^{15} \text{ s}^{-1}$. [1]

(b) Any two from:

1. Monochromatic (single wavelength/frequency). [1]
2. Coherent (constant phase difference). [1]
3. Collimated (low divergence/parallel beam). [1]
4. High intensity. [1]

18.
(a) As temperature increases, more covalent bonds are broken, releasing more electron-hole pairs. [1]
This increases the number density of charge carriers, thereby increasing conductivity. [1]

(b) Intrinsic: Pure semiconductor where charge carriers are generated only by thermal excitation across the band gap. [1]
Extrinsic: Semiconductor doped with impurities (Group 3 or 5) to increase the number of majority charge carriers (holes or electrons). [1]

19.
(a) UV photons have energy greater than the work function of zinc. [1]
Photoelectrons are emitted, removing negative charge from the plate. The electroscope discharges, causing the leaf to collapse. [1]

(b) Visible light photons have frequency lower than the threshold frequency of zinc ($f < f_0$). [1]
Individual photon energy $hf$ is less than the work function $\Phi$, so no electrons are emitted regardless of how many photons (intensity) strike the surface. [1]

20.
(a) Both involve a mass defect which is converted into energy according to $E=mc^2$. [1]
(b) Fusion requires extremely high temperatures and pressures to overcome electrostatic repulsion between nuclei; Fission can occur at lower temperatures (often initiated by neutron absorption). [1]
(c) Fusion fuel (isotopes of hydrogen) is more abundant / Fusion produces no long-lived high-level radioactive waste / Fusion has higher energy yield per unit mass. [1]