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Question

(d) $\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n},\;n = 2$

$\Rightarrow \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}$

Vedclass · Accepted Answer

Three-fourth

Vedclass · Answer

(d) $\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n},\;n = 2$

$\Rightarrow \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}$

So disintegrated part $ = 1 - \frac{N}{{{N_0}}} = 1 - \frac{1}{4} = \frac{3}{4}$

What fraction of a radioactive material will get disintegrated in a period of two half-lives

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