CIE iGCSE Co-Ordinated Science P5.2.1 Detection of radioactivity Exam Style Questions Paper 4
Question
Fig. 6.2 shows how the number of undecayed nuclei in a sample changes over time.
Use the correct nuclide notation to complete the decay equation.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C1.1 — Solids, liquids and gases (Part (a))
• Topic C2.3 — Isotopes (Part (b)(i))
• Topic P5.2.1 — Detection of radioactivity (Part (b)(ii)–(iii))
• Topic P1.4 — Density (Part (c))
▶️ Answer/Explanation
(a)(i) Liquid: random arrangement, molecules touching/close together. Gas: random arrangement, molecules far apart
In both liquids and gases the particles are arranged randomly (no regular pattern), unlike in a solid.
However, liquid particles remain close together and touching, while gas particles are spread far apart with large spaces between them.
(a)(ii) Liquid molecules move around/slide past each other (slower); gas molecules move freely and rapidly in all directions
Liquid particles can flow over one another but stay relatively close, moving with less kinetic energy than gas particles.
Gas particles move completely freely at high speed, only interacting when they collide.
(b)(i) Similarity: same number of protons (17). Difference: different number of neutrons
Both isotopes have the same proton (atomic) number, 17, since they are both chlorine atoms.
Chlorine-35 has 18 neutrons while chlorine-37 has 20 neutrons — they differ in nucleon number/mass.
(b)(ii) 300 thousand years
From the decay graph, the number of undecayed nuclei falls from about 700 to 350 (half) over roughly 300 thousand years.
This time taken for the activity (or undecayed nuclei) to halve is, by definition, the half-life.
(b)(iii) \({}^{36}_{17}Cl \rightarrow {}^{36}_{18}Ar + {}^{\;\;0}_{-1}\beta\)
Mass (nucleon) number is conserved: \(36 = 36 + 0\).
Charge (proton number) is conserved: \(17 = 18 + (-1)\), confirming the emitted particle is a beta particle.
(c) 1.6 kg
Mass of liquid chlorine \(= \rho V = 570 \times 0.020 = 11.4\) kg.
Mass of empty canister \(= 13 – 11.4 = 1.6\) kg.
Question
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P5.2.1 — Detection of Radioactivity (Parts a(i), a(ii) & a(iii))
• Topic P6.2.1 — The Sun as a star / Orbital motion (Part b)
▶️ Answer/Explanation
(a)(i) Complete the nuclear decay equation:
\(^{90}_{38}\text{Sr} \rightarrow ^{90}_{39}\text{Y} + ^{0}_{-1}\beta\)
During beta decay, a neutron in the nucleus is converted into a proton and an electron (beta particle). The electron is emitted from the nucleus, increasing the atomic number by 1 (from 38 to 39) while the mass number remains the same (90). The new element formed is Yttrium-90.
(a)(ii) Mass remaining after 58 years:
Half-life = 29 years.
58 years = 2 × 29 years, so 2 half-lives have passed.
After each half-life, the mass reduces by half:
• After 1st half-life (29 years): 1.6 ÷ 2 = 0.8 mg
• After 2nd half-life (58 years): 0.8 ÷ 2 = 0.4 mg
Mass remaining = 0.4 mg
(a)(iii) Three ways to keep workers safe from radiation:
1. Minimise exposure time — workers should spend as little time as possible near the radioactive source to reduce the total radiation dose received.
2. Maximise distance from the source — radiation intensity decreases rapidly with distance (inverse square law). Workers should stand as far away as possible from the source.
3. Use shielding — thick lead or concrete barriers should be placed between the workers and the source to absorb or block the radiation.
4. Wear radiation detection badges — to monitor and record the cumulative radiation dose received by each worker, ensuring it stays within safe limits.
(b) Calculate the mean orbital speed of the Moon:
The Moon travels in a circular orbit around the Earth. The orbital speed is calculated using:
\(v = \frac{2\pi r}{T}\)
Where:
\(r = 3.84 \times 10^8 \, \text{m}\)
\(T = 27.3 \, \text{days} = 27.3 \times 24 \times 3600 = 2,358,720 \, \text{s}\)
\(v = \frac{2 \times \pi \times 3.84 \times 10^8}{2,358,720}\)
\(v = \frac{2.412 \times 10^9}{2.35872 \times 10^6}\)
\(v \approx 1023 \, \text{m/s}\)
Mean orbital speed ≈ 1020 m/s