Certain radio-active substance reduces to 25% of its value in 16 days. Its half-life is ........ day

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13.Nuclei

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Certain radio-active substance reduces to $25\%$ of its value in $16$ days. Its half-life is ........ $days$

A

$32$

B

$8$

C

$64$

D

$28$

Solution

(d) $N = {N_0}{\left( {\frac{1}{2}} \right)^{\frac{t}{T}}} $

$\Rightarrow \frac{{{N_0}}}{4} = {N_0}{\left( {\frac{1}{2}} \right)^{\frac{{16}}{T}}}$

$ \Rightarrow T = 8\,\, days.$

Standard 12

Physics

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