Physics 9702 · Paper 2

Decay Chains & Counting

A nucleus often decays many times in a row. You can find the final nucleus by counting how much each step changes $A$ and $Z$ .

What each step does

Alpha: $A$ down by 4, $Z$ down by 2 (2 protons and 2 neutrons leave).
Beta-minus: $A$ stays the same, $Z$ up by 1 (a neutron becomes a proton).
Gamma: no change to $A$ or $Z$ .
To find the number of neutrons at any point, use $N = A - Z$ .

Counting through a chain

Worked example

A chain of alpha and beta decays

A nucleus ${}^{228}_{88}\text{Ra}$ decays through a chain of 5 alpha and 4 beta-minus steps. How many protons and neutrons does the final nucleus have?

Nucleon number: only alpha changes it. $A = 228 - 5 \times 4 = 208$ .
Proton number: alpha gives $-2$ each, beta-minus gives $+1$ each. $Z = 88 - 5 \times 2 + 4 \times 1 = 82$ .
Neutrons: $N = A - Z = 208 - 82 = 126$ .

Answer

82\ \text{protons},\ 126\ \text{neutrons}\ \ ({}^{208}_{82}\text{Pb})

Your turn— tap to reveal the worked answer (9702/12/M/J/24 Q39)

A nucleus ${}^{218}_{84}\text{X}$ decays to ${}^{210}_{83}\text{Y}$ . How many alpha and beta-minus particles are emitted? (9702/12/M/J/24 Q39)

Answer: $A$ falls by 8, and only alpha changes $A$ , so there are $8 \div 4 = 2$ alpha decays. After 2 alpha, $Z$ = $84 - 4 = 80$ . The final $Z$ is 83, so $Z$ rose by 3, which needs 3 beta-minus decays. So 2 alpha and 3 beta-minus.

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