During mean life of a radioactive element, th | vedclass

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Question

(c) By using $N = {N_0}{e^{ - \lambda t}}$ and average life time $t = \frac{1}{\lambda }$

So $N = {N_0}{e^{ - \lambda 	imes 1/\lambda }} = {N_0}{e

Vedclass · Accepted Answer

$\frac{{e - 1}}{e}$

Vedclass · Answer

(c) By using $N = {N_0}{e^{ - \lambda t}}$ and average life time $t = \frac{1}{\lambda }$

So $N = {N_0}{e^{ - \lambda 	imes 1/\lambda }} = {N_0}{e^{ - 1}}$

$\Rightarrow \frac{N}{{{N_0}}} = {e^{ - 1}} = \frac{1}{e}$

Now disintegrated fraction $ = 1 - \frac{N}{{{N_0}}} = 1 - \frac{1}{e} = \frac{{e - 1}}{e}$

During mean life of a radioactive element, the fraction that disintegrates is

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