Question

The activity of a radioactive element drops to 1/16th of its initial value in 32 years find the mean life of the sample
(1 mark type question)

Answer 1

$\large{\sf\underline{ \red{\underline{Question }}}}:- \:$

→ The activity of a radioactive elements drops to $\dfrac{1}{16}th$ of its initial value in 32 years .find the mean life of the sample .

$\large\sf\underline{\underline{Answer }}:- \: \\ \implies \sf mean \: life (\tau) = 11.544 \: \\ \\ \large\sf\underline{ \green{\underline{To \: Find }}}:- \\ \to \sf mean \: life \: of \: the \: sample \\ \\ \large\sf\underline{ \pink{\underline{Explanation }}}:- \:$

Given that :

• Activity of element $\dfrac{1}{16}$

• time (t) = 32 years

We know that ,

→ Activity of a radioactive substances at any time "t" is -

$\implies \sf \red{ N = N_0 \: }{e}^{ - \lambda \:t} \: \: \: \\ \\ \implies \sf \frac{N}{N_0} = {e}^{ - \lambda \: t} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \because \boxed{ \sf\frac{N}{N_0} = \frac{1}{16} } \\ \\ \implies \sf \: \frac{1}{16} = {e}^{ - \lambda \: t} \\ \\ \implies \sf \: {e}^{ 32\lambda \: } = 16 \\ \\ \sf \: taking \: \log() \: on \: both \: sides \: \\ \\ \implies \sf \log( {e}^{32 \lambda} ) = \log(16) \\ \\ \: \: \: \: \: \because \boxed{ \sf \: \log( {m}^{n} ) = n \log m} \\ \\ \implies \sf \: 32 \lambda \log e = \log( {2}^{4} ) \\ \\ \because \boxed{ \sf \log e = 1} \\ \\ \implies \: \: 32 \lambda = 4 \log 2 \\ \\ \implies \boxed{ \sf \: \lambda = \frac{ \log 2}{8} \: }$

We know that ,

$\to \sf \boxed{ \sf mean \: life \:(\tau) = \frac{ 1}{ \lambda} } \\ \\ \to \sf \tau= \frac{ 1}{ \frac{ \log 2}{8} } \\ \\ \to { \sf \tau = \frac{8}{ \log 2} \: } \: \: \because \boxed{ \sf \log 2 = 0.693} \\ \\ \to \sf \: \tau= \frac{8}{0.693} \\ \\ \to \boxed{ \sf \tau= 11.544 \: }$