Plutonium decays with a half-life of 24000 ,years . If plutonium is stored for 72000 ,years , then f

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13.Nuclei

easy

Plutonium decays with a half-life of $24000 \,years$. If the plutonium is stored for $72000 \,years$, then the fraction of plutonium that remains is

A

$\frac{1}{2}$

B

$\frac{1}{3}$

C

$\frac{1}{4}$

D

$\frac{1}{8}$

Solution

(d) $n = \frac{{72000}}{{24000}} = 3;$

Now $\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n} = \frac{1}{8}$

Standard 12

Physics

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