Two radioactive nuclide $P$ and $Q$ have their decay constant in the ratio 3: 2 . One mole of each is taken separately and allowed to decay for a time interval of three times of half-life of $A$ . If 0.2 mole of $P$ are left, what mole of $Q$ will be left? (multiply $X \times 100$ )

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Solution

Let decay constant $λ_{P}$ and $λ_{Q}$ be $3 a$ and $2 a$ respectively.

$t_{1 / 2} o f P = \frac{0.693}{3 a}; t_{1 / 2} o f Q = \frac{0.693}{2 a}$
$T = 3 \times t_{1 / 2} o f P = \frac{3 \times 0.693}{3 a} = \frac{0.693}{a}$
$∵ T = \frac{2.303}{λ} log \frac{a}{a - x}$
$F o r P : \frac{0.693}{a} = \frac{2.303}{3 a} log \frac{1}{n_{P}}$
$F o r Q : \frac{0.693}{a} = \frac{2.303}{2 a} log \frac{1}{n_{Q}}$
$= \frac{log \frac{1}{n_{P}}}{log \frac{1}{n_{Q}}} = \frac{3}{2}$
or, $log \frac{1}{n_{P}} = \frac{3}{2} log \frac{1}{n_{Q}}$
$log \frac{1}{n_{P}} = log {(\frac{1}{n_{Q}})}^{3 / 2}$
or, $n_{P} = (n_{Q})^{2 / 3}$
if $n_{P} = 0.2, t h e n n_{Q} = 0.09$ mole of $Q$ will be left.
$0.09 \times 100 = 9$ .

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