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A Level H2 Physics Modern Physics Quiz

Free A Level H2 Physics Modern Physics quiz, HY3 Exam version, with questions, answers, and A Level-style practice for Singapore students.

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A Level H2 Physics From Real Exams Generated by Tencent HY3 Free Updated 2026-08-17

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Answers

A-Level Physics H2 Quiz - Modern Physics (Answers)

Total Marks: 40
Topic: Modern Physics


Section A Answers (1–10)

1. [2 marks]
Photoelectric effect: emission of electrons from a metal surface when electromagnetic radiation of sufficient frequency shines on it.
Condition: frequency of incident light must exceed threshold frequency f0f_0 (or photon energy hf>ϕhf > \phi).
Marking: 1 mark for effect, 1 mark for condition.

2. [1 mark]
E=hfE = hf
Marking: correct equation with symbols.

3. [2 marks]
ϕ=2.0×1.60×1019=3.20×1019 J\phi = 2.0 \times 1.60 \times 10^{-19} = 3.20 \times 10^{-19}\ \text{J}
f0=ϕ/h=3.20×1019/6.63×1034=4.83×1014 Hzf_0 = \phi / h = 3.20 \times 10^{-19} / 6.63 \times 10^{-34} = 4.83 \times 10^{14}\ \text{Hz}
Marking: 1 mark conversion, 1 mark calculation.

4. [1 mark]
λ=h/p\lambda = h / p
Marking: correct formula.

5. [2 marks]
K=eV=1.60×1019×100=1.60×1017 JK = eV = 1.60 \times 10^{-19} \times 100 = 1.60 \times 10^{-17}\ \text{J}
p=2mK=2×9.11×1031×1.60×1017=1.71×1023 kg m s1p = \sqrt{2mK} = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-17}} = 1.71 \times 10^{-23}\ \text{kg m s}^{-1}
λ=h/p=6.63×1034/1.71×1023=3.88×1011 m\lambda = h/p = 6.63 \times 10^{-34} / 1.71 \times 10^{-23} = 3.88 \times 10^{-11}\ \text{m}
Marking: 1 mark momentum, 1 mark wavelength.

6. [1 mark]
24He{}_{2}^{4}\text{He} (alpha particle)
Marking: correct particle.

7. [2 marks]
n=15/5=3n = 15/5 = 3 half-lives
N=N0/23=8000/8=1000N = N_0 / 2^3 = 8000 / 8 = 1000
Marking: 1 mark for n, 1 mark for N.

8. [1 mark]
Energy required to separate nucleus into its constituent protons and neutrons.
Marking: clear definition.

9. [2 marks]
Mass of parts = 2(1.00728)+2(1.00867)=4.03190 u2(1.00728) + 2(1.00867) = 4.03190\ \text{u}
Defect = 4.031904.00260=0.02930 u4.03190 - 4.00260 = 0.02930\ \text{u}
=0.02930×1.66×1027=4.86×1029 kg= 0.02930 \times 1.66 \times 10^{-27} = 4.86 \times 10^{-29}\ \text{kg}
Marking: 1 mark mass diff, 1 mark kg conversion.

10. [1 mark]
931.5 MeV931.5\ \text{MeV} (from E=mc2E = mc^2 with 1 u1\ \text{u})
Marking: correct value.


Section B Answers (11–15)

11. [3 marks]
Ephoton=hc/λ=(6.63×1034×3.00×108)/(400×109)=4.97×1019 J=3.11 eVE_{\text{photon}} = hc/\lambda = (6.63\times10^{-34} \times 3.00\times10^8)/(400\times10^{-9}) = 4.97\times10^{-19}\ \text{J} = 3.11\ \text{eV}
Kmax=3.111.5=1.61 eVK_{\max} = 3.11 - 1.5 = 1.61\ \text{eV}
Marking: 1 photon energy, 1 conversion, 1 KE.

12. [3 marks]
N=N0eλt150=1200e24λe24λ=0.125N = N_0 e^{-\lambda t} \Rightarrow 150 = 1200 e^{-24\lambda} \Rightarrow e^{-24\lambda} = 0.125
24λ=ln0.125=2.079λ=0.0866 h1-24\lambda = \ln 0.125 = -2.079 \Rightarrow \lambda = 0.0866\ \text{h}^{-1}
T1/2=ln2/λ=0.693/0.0866=8.0 hT_{1/2} = \ln 2 / \lambda = 0.693 / 0.0866 = 8.0\ \text{h}
Marking: 1 eqn, 1 lambda, 1 half-life.

13. [3 marks]
Δm=56.4493255.92066=0.52866 u\Delta m = 56.44932 - 55.92066 = 0.52866\ \text{u}
B=0.52866×931.5=492.4 MeVB = 0.52866 \times 931.5 = 492.4\ \text{MeV}
Per nucleon = 492.4/56=8.79 MeV492.4 / 56 = 8.79\ \text{MeV}
Marking: 1 defect, 1 total, 1 per nucleon.

14. [3 marks]
K=2.0×106×1.60×1019=3.20×1013 JK = 2.0\times10^6 \times 1.60\times10^{-19} = 3.20\times10^{-13}\ \text{J}
p=2mK=2×1.67×1027×3.20×1013=3.27×1020p = \sqrt{2mK} = \sqrt{2 \times 1.67\times10^{-27} \times 3.20\times10^{-13}} = 3.27\times10^{-20}
λ=h/p=6.63×1034/3.27×1020=2.03×1014 m\lambda = h/p = 6.63\times10^{-34} / 3.27\times10^{-20} = 2.03\times10^{-14}\ \text{m}
Marking: 1 K, 1 p, 1 lambda.

15. [3 marks]
E=0.20×931.5=186.3 MeVE = 0.20 \times 931.5 = 186.3\ \text{MeV}
In J: 0.20×1.66×1027×(3.00×108)2=2.98×1011 J0.20 \times 1.66\times10^{-27} \times (3.00\times10^8)^2 = 2.98\times10^{-11}\ \text{J}
Marking: 1 MeV, 1 J formula, 1 value.


Section C Answers (16–20)

16. [4 marks]

  • Evidence: electron diffraction (Davisson-Germer) shows interference pattern (2 marks).
  • de Broglie: λ=h/p\lambda = h/p gives particles wave-like property; explains diffraction (2 marks).
    Teaching: connect wavelength to momentum.

17. [4 marks]
Mass defect: difference between mass of separated nucleons and nucleus (2). Binding energy: energy equivalent via E=Δmc2E = \Delta m c^2 (1). Relation: B=(MsepM)c2B = (M_{\text{sep}} - M)c^2 (1).

18. [4 marks]
Graph: straight line, gradient hh, intercept ϕ- \phi on K axis, x-intercept f0f_0 (2). Explanation: Kmax=hfϕK_{\max} = hf - \phi (2).
Image must show labelled axes and intercepts.

19. [4 marks]
Law: rate of decay proportional to N: dN/dt=λNdN/dt = -\lambda N (2). Integrate: dN/N=λdtln(N/N0)=λtN=N0eλt\int dN/N = -\lambda \int dt \Rightarrow \ln(N/N_0) = -\lambda t \Rightarrow N = N_0 e^{-\lambda t} (2).

20. [4 marks]
Alpha: He nucleus, low penetration, high ionization (1). Beta: electron, medium penetration, medium ionization (1). Gamma: photon, high penetration, low ionization (2).