A Level H1 Physics Modern Physics Quiz

A-Level Physics H1 Quiz - Modern Physics (Answer Key)

Topic: Modern Physics (Syllabus 8867 Topic 11)
Total Marks: 40
Note: Content generated from syllabus-first LLM templates; not claimed as past-year derived.

Section A: Concepts and Recall

1. [1 mark]
A nuclide is a species of atom characterised by the number of protons and neutrons in its nucleus (i.e. a specific combination of $Z$ and $A$).
Teaching note: "Nuclide" refers to the nucleus type, not the element alone. Isotopes are nuclides with same $Z$ but different $A$.

2. [2 marks]
Mass defect is the difference between the sum of the masses of the separate nucleons (protons and neutrons) and the actual mass of the bound nucleus.
[1] for "difference in mass"
[1] for "free nucleons vs bound nucleus"
Common mistake: Confusing with binding energy (energy equivalent).

3. [2 marks]
Conservation of nucleon number: $238 = A + 4 \Rightarrow A = 234$
Conservation of charge: $92 = Z + 2 \Rightarrow Z = 90$
$A = 234$, $Z = 90$ [1 each]

4. [2 marks]
In $\beta^-$ decay, a neutron transforms into a proton and an electron (and antineutrino): $n \rightarrow p + e^- + \bar{\nu}_e$.
The proton stays in nucleus so $Z$ increases by 1; the electron is emitted, mass number (nucleon count) unchanged.
[1] for neutron→proton explanation, [1] for mass number unchanged.

5. [1 mark]
Half-life is the time taken for half of the radioactive nuclei in a sample to decay (or activity to halve).

6. [2 marks]
$^{22}_{11}\text{Na} \rightarrow ^{22}_{10}\text{Ne} + ^{0}_{+1}e$ (or $\beta^+$)
[1] for daughter Ne with Z=10, A=22; [1] for positron emitted.

7. [1 mark]
Any one: use tongs/remote handling; store in lead container; limit exposure time; wear badge dosimeter.

Section B: Calculations

8. [2 marks]
Mass defect $\Delta m = 12.0984 - 12.0000 = 0.0984\ \text{u}$
[2] for correct subtraction and unit.

9. [2 marks]
$E_b = 0.0984 \times 931.5 = 91.7\ \text{MeV}$
[1] method, [1] answer to 3 s.f.

10. [2 marks]
$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.050} = 13.9\ \text{h}$
[1] formula, [1] answer.

11. [2 marks]
$100 = 800 \times (1/2)^{12/T}$
$1/8 = (1/2)^{12/T} \Rightarrow 3 = 12/T \Rightarrow T = 4.0\ \text{days}$
[1] use of halving, [1] answer.

12. [3 marks]
$\Delta m = 0.200 \times 1.66 \times 10^{-27} = 3.32 \times 10^{-28}\ \text{kg}$
$E = \Delta m c^2 = 3.32 \times 10^{-28} \times (3.00 \times 10^8)^2 = 2.99 \times 10^{-11}\ \text{J}$
[1] mass in kg, [1] use $E=mc^2$, [1] answer.

13. [3 marks]
Equation: $^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + ^{4}_{2}\alpha$ [1]
$E = 0.0053 \times 931.5 = 4.94\ \text{MeV}$ [2]

14. [2 marks]
$A = \lambda N \Rightarrow N = A/\lambda = 2.4\times10^6 / 1.2\times10^{-6} = 2.0\times10^{12}$
[1] formula, [1] answer.

Section C: Data Interpretation

15. [2 marks]
Gradient $= (2.7 - 6.9)/(150 - 0) = -4.2/150 = -0.028\ \text{s}^{-1}$
$\lambda = -\text{gradient} = 0.028\ \text{s}^{-1}$
[1] gradient, [1] lambda positive.

16. [1 mark]
$t_{1/2} = 0.693 / 0.028 = 24.8\ \text{s}$

17. [3 marks]
Curve peaks near Fe because nuclei are most tightly bound there. [1]
Fusion of light nuclei moves toward peak, releasing energy. [1]
Fission of heavy nuclei also moves toward peak, releasing energy. [1]

18. [1 mark]
Corrected rate $= 80 - 20 = 60\ \text{s}^{-1}$

19. [2 marks]
After ~50k years, C-14 remaining is too small to measure accurately (less than ~0.1% left). [1]
Background and contamination dominate. [1]

20. [3 marks]
GM tube: ionizing radiation enters, ionizes gas, pulse counted. [2]
Limitation: cannot distinguish radiation types or measure energy. [1]