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Standard XII

Physics

Binding Energy

84Po218 t1/2=...

Question

$_{84} P o^{218} (t_{1 / 2} = 3.05 m i n)$ decay to $_{82} P b^{214} (t_{1 / 2} = 2.68 m i n)$ by $α$ -emission,while $P b^{214}$ is a $β$ -emitter. In an experiment starting with 1gm atom of pure $P o^{218}$ ,how much time would be required for the number of nuclei of $_{82} P b^{214}$ to reach maximum?

4.125 min

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Solution

The correct option is B 4.125 min
For 1st process rate constant $K_{1}$ = $\frac{0.693}{3.05} m i n^{- 1} = 0.227 m i n^{- 1}$
For 2nd process rate constant $K_{2}$ = $\frac{0.693}{2.68} m i n^{- 1} = 0.259 m i n^{- 1}$
$t_{m a x}$ = $\frac{1}{(K_{1} - K_{2})} l n \frac{K_{1}}{K_{2}} = \frac{1}{0.227 - 0.259)} l n \frac{2.68}{3.05} = 4.125 m i n$

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84Po218(t1/2=3.05min)decay to 82Pb214(t1/2=2.68min) by α -emission,while Pb214 is a β -emitter. In an experiment starting with 1gm atom of pure Po218,how much time would be required for the number of nuclei of 82Pb214 to reach maximum?

The correct option is B 4.125 minFor 1st process rate constant K1 = 0.6933.05 min−1=0.227min−1For 2nd process rate constant K2 = 0.6932.68 min−1=0.259min−1tmax = 1(K1−K2)lnK1K2=10.227−0.259)ln2.683.05=4.125 min