Radioactivity
5 marks · HardA sample contains the isotope chlorine-36, which decays by beta minus emission. After 72,000 years, a scientist finds that only 25% of the original chlorine-36 remains.
Write a balanced nuclear equation for the beta minus decay of chlorine-36. State the names and atomic numbers of the products formed.
Calculate the half-life of chlorine-36. Give your answer to the nearest thousand years.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) ^{36}_{17}\text{Cl} \rightarrow ^{36}_{18}\text{Ar} + ^{0}_{-1}\text{e} + \text{energy} · must show Correct symbols for all nuclei and beta particle, Atomic/mass numbers balanced, Naming argon and beta particle (electron)
(b) 36,000 years · must show Identifying two half-lives to reach 25%, Showing calculation: 72,000 \div 2 = 36,000
Common mistakes: Incorrectly balancing the nuclear equation; Using wrong half-life formula or not recognising 25% is two half-lives
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Marking points
M1 (a) · 1 mark · Correctly writes balanced nuclear equation for beta minus decay.
A1 (a) · 1 mark · States names and atomic numbers of the daughter nucleus (argon, 18) and electron (beta particle, -1).
M2 (b) · 1 mark · Identifies that 25% remaining corresponds to two half-lives.
M3 (b) · 1 mark · Divides total time by 2. (follow-through)
A2 (b) · 1 mark · Arrives at 36,000 years (or correct to nearest 1,000 years). (follow-through)
Worked solution
a) Chlorine-36 has 17 protons and 36 nucleons. In beta minus decay, a neutron changes into a proton, emitting a beta minus particle (electron).
Products are argon-36 (atomic number 18) and a beta particle (electron, atomic number -1).
b) After one half-life: 50% remains. After two half-lives: 25% remains. Thus, two half-lives have passed in 72,000 years.