$32 m g$ of pure $_{94}^{238} P u O_{2}$ has an activity of $6.4 \times 10^{7} {s e c}^{- 1}$ .
(i) What will be the half life of $_{94}^{238} P u$ in years?
(ii) What amount of $P u O_{2}$ will remain if $100 m g$ $P u O_{2}$ is kept for $5000$ years?

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Solution

(i) Mass of $^{238} P u = \frac{238}{270} \times 32 = 28.207 m g$
Rate $= \frac{0.693}{t_{1 / 2}} \times \frac{w}{A t . w t} \times N$
$6.4 \times 10^{7} = \frac{0.693}{t_{1 / 2}} \times \frac{28.207 \times 10^{- 3}}{238} \times 6.023 \times 10^{23}$
$t_{1 / 2} = 7.729 \times 10^{11} s e c = 2.45 \times 10^{4}$ years

(ii) $\frac{0.693}{t_{1 / 2}} = \frac{2.303}{t} {log}_{10} (\frac{N_{0}}{N}$
$\frac{0.693}{2.45 \times 10^{4}} = \frac{2.303}{5000} {log}_{10} (\frac{100}{N})$
$N = 86.7 m g$

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