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A Level H1 Physics Modern Physics Quiz

Free A Level H1 Physics Modern Physics quiz, HY3 Exam version, with questions, answers, and A Level-style practice for Singapore students.

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A Level H1 Physics From Real Exams Generated by Tencent HY3 Free Updated 2026-08-17

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A-Level Physics H1 Quiz - Modern Physics (Answers)

Total Marks: 40


Section A

1. [1] A nuclide is a species of atom characterised by its number of protons and neutrons (nucleon number A and proton number Z).
Teaching note: Use "specific nucleus with given Z and A". Common mistake: calling it just "atom".

2. [1] Mass defect is the difference between the sum of masses of free constituent nucleons and the actual mass of the bound nucleus.
Δm=Zmp+(AZ)mnmnucleus\Delta m = Z m_p + (A-Z)m_n - m_{\text{nucleus}}

3. [1] 88226Ra 86222Rn+ 24α^{226}_{88}\text{Ra} \rightarrow\ ^{222}_{86}\text{Rn} +\ ^{4}_{2}\alpha
Check: A: 226→222+4, Z: 88→86+2.

4. [2] Radioactive decay: random and spontaneous; nucleus decays independently. Half-life: time for half the undecayed nuclei (or activity) to decay. [1+1]

5. [1] Binding energy = minimum energy required to separate nucleus into its constituent protons and neutrons.


Section B

6. [2] A=λN=(2.5×106)(4.0×1020)=1.0×1015 BqA = \lambda N = (2.5\times10^{-6})(4.0\times10^{20}) = 1.0\times10^{15}\ \text{Bq}. [M1 for formula, A1 answer]

7. [2] 36 h = 3 half-lives. N=N0(1/2)3=8.0×1018/8=1.0×1018N = N_0 (1/2)^3 = 8.0\times10^{18} / 8 = 1.0\times10^{18}. [M1: 3 half-lives, A1]

8. [3] Sum = 1.00728+1.00867=2.01595 u1.00728 + 1.00867 = 2.01595\ u. Defect = 2.015952.01355=0.00240 u2.01595 - 2.01355 = 0.00240\ u. In kg: 0.00240×1.66×1027=3.98×1030 kg0.00240 \times 1.66\times10^{-27} = 3.98\times10^{-30}\ \text{kg}. [M1 sum, M1 diff, A1 kg]

9. [2] Δm=0.00240×1.66×1027=3.98×1030 kg\Delta m = 0.00240 \times 1.66\times10^{-27} = 3.98\times10^{-30}\ \text{kg}. E=mc2=(3.98×1030)(3.00×108)2=3.59×1013 JE = mc^2 = (3.98\times10^{-30})(3.00\times10^8)^2 = 3.59\times10^{-13}\ \text{J}. [M1 mass, A1 energy]

10. [2] N=A/λ=(1.2×108)/(4.0×109)=3.0×1016N = A/\lambda = (1.2\times10^8)/(4.0\times10^{-9}) = 3.0\times10^{16}. [M1 formula, A1]

11. [3] Total BE = 56×8.79=492.2 MeV56 \times 8.79 = 492.2\ \text{MeV}. In J: 492.2×106×1.60×1019=7.88×1011 J492.2\times10^6 \times 1.60\times10^{-19} = 7.88\times10^{-11}\ \text{J}. [M1 MeV, M1 J, A1]

12. [2] 25% = (1/2)2(1/2)^2 → 2 half-lives. Age = 2×5730=11460 years2 \times 5730 = 11460\ \text{years}. [M1 factor, A1]


Section C

13. [3] (a) Alpha: helium nucleus 24He^4_2\text{He}, high ionisation, low penetration; Beta: electron, lower ionisation, higher penetration. [1+1] (b) Gamma is photon, no mass/charge change. [1]

14. [2] ZAX Z+1AY+ 10e+νˉ^A_ZX \rightarrow\ ^A_{Z+1}Y +\ ^0_{-1}e + \bar{\nu}. Z increases by 1, A unchanged. [1 eq, 1 statement]

15. [3] (a) Half-life = 2 min (800→400). [1] (b) λ=ln2/T1/2=0.693/2=0.347 min1\lambda = \ln2 / T_{1/2} = 0.693/2 = 0.347\ \text{min}^{-1}. [M1, A1]

16. [3] (a) Gradient = (1420)/(1500)=0.040 s1(14-20)/(150-0) = -0.040\ \text{s}^{-1}. [1] (b) λ=grad=0.040 s1\lambda = -\text{grad} = 0.040\ \text{s}^{-1}. [1] (c) At t=0, ln N = 20 → N0=e20=4.85×108N_0 = e^{20} = 4.85\times10^8. [1]
Graph must show straight line, negative slope.

17. [3] Energy/day = 500×106×86400=4.32×1013 J500\times10^6 \times 86400 = 4.32\times10^{13}\ \text{J}. m=E/c2=4.32×1013/9×1016=4.8×104 kgm = E/c^2 = 4.32\times10^{13} / 9\times10^{16} = 4.8\times10^{-4}\ \text{kg}. [M1 E, M1 m, A1]

18. [3] (a) Activity = number of decays per unit time, A=λNA = \lambda N. [1] (b) A0/4A_0/4 = 2 half-lives → T1/2=10T_{1/2}=10 days. [M1, A1]

19. [3] Fission: heavy nucleus splits, moves toward higher BE/nucleon (mid-mass), releases energy. Fusion: light nuclei join, also to higher BE/nucleon, releases energy. [1+1+1]

20. [4] (a) N0=m/u=0.20×106/1.66×1027=1.20×1020N_0 = m/u = 0.20\times10^{-6} / 1.66\times10^{-27} = 1.20\times10^{20} (using uu as nucleon mass approx; or molar mass method). (b) 16 d = 2 half-lives, A=A0/4A = A_0/4, A0=λN0A_0 = \lambda N_0, λ=ln2/(8×86400)=1.00×106 s1\lambda = \ln2/ (8\times86400) = 1.00\times10^{-6}\ \text{s}^{-1}, A=(1.00×106×1.20×1020)/4=3.0×1013 BqA = (1.00\times10^{-6}\times1.20\times10^{20})/4 = 3.0\times10^{13}\ \text{Bq}. [M1 N0, M1 λ, M1 A0, A1]