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.
Energy of incident photon $E = \frac{hc}{\lambda}$
$E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{250 \times 10^{-9}} = 7.956 \times 10^{-19} \text{ J}$ [1]
Work function $\Phi = 4.3 \text{ eV} = 4.3 \times 1.60 \times 10^{-19} = 6.88 \times 10^{-19} \text{ J}$ [1]
Max KE $= E - \Phi = 7.956 \times 10^{-19} - 6.88 \times 10^{-19} = 1.076 \times 10^{-19} \text{ J}$
Answer: $1.1 \times 10^{-19} \text{ J}$ [1]

3.
(a) No change. [1]
Max KE depends only on the frequency (energy) of individual photons and the work function ($KE_{max} = hf - \Phi$). Intensity does not change photon energy. [1]
(b) Increases. [1]
Intensity is proportional to the number of photons incident per unit time. More photons cause more photoelectrons to be emitted per second, increasing the current. [1]

4.
Wave theory predicts that energy accumulates over time, so emission should occur at any frequency given enough intensity/time. [1]
Particle theory states energy is quantized in packets ($E=hf$). [1]
If $hf < \Phi$, no single photon has enough energy to eject an electron, regardless of intensity. This explains the threshold frequency. [1]

5.
(a) Planck constant $h$ (or $h/e$ if plotting $V_s$ vs $f$, since $eV_s = hf - \Phi \Rightarrow V_s = \frac{h}{e}f - \frac{\Phi}{e}$). Note: Standard syllabus accepts $h/e$ for gradient of $V_s$ vs $f$. [1]
(b) Threshold frequency $f_0$. [1]

6. The minimum energy required to remove an electron from the ground state of an atom to infinity (where it is free from the nucleus). [1]

7.
Energy difference $\Delta E = E_3 - E_2 = -1.51 - (-3.40) = 1.89 \text{ eV}$ [1]
Convert to Joules: $1.89 \times 1.60 \times 10^{-19} = 3.024 \times 10^{-19} \text{ J}$
$\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.024 \times 10^{-19}}$ [1]
$\lambda = 6.59 \times 10^{-7} \text{ m}$ (or 659 nm) [1]

8.
Electrons in atoms occupy discrete/quantized energy levels. [1]
Transitions occur only between these specific levels, emitting photons of specific energies (and thus specific wavelengths/frequencies). [1]

9.
(a) Energy required for transition $n=1 \to n=2$:
$\Delta E = E_2 - E_1 = -3.40 - (-13.6) = 10.2 \text{ eV}$ [1]
Since incident electron KE (12.0 eV) > 10.2 eV, excitation can occur. [1]
(b) The remaining kinetic energy ($12.0 - 10.2 = 1.8 \text{ eV}$) is retained by the incident electron as kinetic energy. [1]

10.
High-speed electrons from the cathode collide with target atoms. [1]
They knock out inner-shell (e.g., K-shell) electrons, creating a vacancy. [1]
Outer-shell electrons drop down to fill the vacancy, emitting X-ray photons with specific energies corresponding to the difference in binding energies of the shells. [1]

11.
The energy required to completely separate a nucleus into its constituent protons and neutrons. [1]
(Or: The energy released when protons and neutrons combine to form a nucleus). [1]

12.
Mass of constituents: $2(1.00728) + 2(1.00867) = 2.01456 + 2.01734 = 4.03190 \text{ u}$ [1]
Mass defect $\Delta m = 4.03190 - 4.00151 = 0.03039 \text{ u}$
Binding Energy $BE = 0.03039 \times 931.5 = 28.308 \text{ MeV}$ [1]
BE per nucleon $= 28.308 / 4 = 7.08 \text{ MeV}$ [1]

13.

1. $\beta^-$ involves emission of an electron (and antineutrino); $\gamma$ is electromagnetic radiation (photon). [1]
2. $\beta^-$ changes the proton/neutron number (transmutation); $\gamma$ does not change the composition of the nucleus. [1]
   (Other valid answers: Charge, Mass, Penetrating power)

14.
Number of half-lives $n = 60 / 15 = 4$ [1]
Activity $A = A_0 (1/2)^n = 8.0 \times 10^4 \times (1/2)^4$
$A = 8.0 \times 10^4 / 16 = 5.0 \times 10^3 \text{ Bq}$ [1]

15.
When nucleons combine, energy is released (binding energy). [1]
By mass-energy equivalence ($E=mc^2$), this loss of energy corresponds to a loss of mass. [1]

16.
(a) $A \approx 56$ (Iron/Fe). [1]
(b) Light nuclei have lower BE per nucleon than the product nucleus formed after fusion. [1]
The product nucleus is more tightly bound (higher BE per nucleon). [1]
The increase in total binding energy corresponds to energy released to the surroundings. [1]

17.
Kinetic Energy $KE = eV = 1.60 \times 10^{-19} \times 150 = 2.40 \times 10^{-17} \text{ J}$
Momentum $p = \sqrt{2m_e KE} = \sqrt{2 \times 9.11 \times 10^{-31} \times 2.40 \times 10^{-17}}$ [1]
$p = \sqrt{4.3728 \times 10^{-47}} = 6.61 \times 10^{-24} \text{ kg m s}^{-1}$
$\lambda = h/p = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}}$ [1]
$\lambda = 1.00 \times 10^{-10} \text{ m}$ [1]

18.
(a) Wave nature (or wave-particle duality). [1]
(b) Diameter decreases. [1]
Higher voltage $\rightarrow$ higher momentum $\rightarrow$ shorter de Broglie wavelength. Shorter wavelength diffracts less, resulting in smaller ring diameters. [1]

19.
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 $P = 2.0 \text{ mW} = 2.0 \times 10^{-3} \text{ J s}^{-1}$
Number of photons $N = \frac{P}{E} = \frac{2.0 \times 10^{-3}}{3.14 \times 10^{-19}}$ [1]
$N = 6.37 \times 10^{15} \text{ s}^{-1}$ [1]

20.
An incident photon of specific energy interacts with an excited atom, causing it to drop to a lower energy level and emit a second photon. [1]
The emitted photon is identical to the incident photon in frequency, phase, direction, and polarization. [1]
Condition: Population Inversion (more atoms in excited state than ground state). [1]