Nuclear activity of a radioactive element becomes left(frac{1}{8}right)^{text {th }} of its initial

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13.Nuclei

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The nuclear activity of a radioactive element becomes $\left(\frac{1}{8}\right)^{\text {th }}$ of its initial value in $30\, years.$ The half-life of radioactive element is $....\,years.$

A

$15$

B

$10$

C

$20$

D

$25$

(JEE MAIN-2021)

Solution

$A = A _{0} e ^{-\lambda t }$

$\frac{ A _{0}}{8}= A _{0} e ^{-\lambda t } \Rightarrow \lambda t =\ln 8$

$\lambda t =3 \ln 2$

$\frac{\ln 2}{\lambda}=\frac{ t }{3}=\frac{30}{3}$

$10\;years$

Standard 12

Physics

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