The nuclear activity of a radioactive element | vedclass

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Question

$A = A _{0} e ^{-\lambda t }$

$\frac{ A _{0}}{8}= A _{0} e ^{-\lambda t } \Rightarrow \lambda t =\ln 8$

$\lambda t =3 \ln 2$

$\frac{\ln 2}{\l

Vedclass · Accepted Answer

$10$

Vedclass · Answer

$A = A _{0} e ^{-\lambda t }$

$\frac{ A _{0}}{8}= A _{0} e ^{-\lambda t } \Rightarrow \lambda t =\ln 8$

$\lambda t =3 \ln 2$

$\frac{\ln 2}{\lambda}=\frac{ t }{3}=\frac{30}{3}$

$10\;years$

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 $....\,years.$

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