13.Nuclei
medium

एक रेडियोधर्मी तत्व की औसत-आयु के दौरान, विघटित भाग है

A

$e$

B

$\frac{1}{e}$

C

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

D

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

Solution

$N = {N_0}{e^{ – \lambda t}}$ एवं औसत आयु $t = \frac{1}{\lambda }$

इसलिए, $N = {N_0}{e^{ – \lambda  \times 1/\lambda }} = {N_0}{e^{ – 1}} $

$\Rightarrow \frac{N}{{{N_0}}} = {e^{ – 1}} = \frac{1}{e}$

अब विघटित भाग $ = 1 – \frac{N}{{{N_0}}} = 1 – \frac{1}{e} = \frac{{e – 1}}{e}$

Standard 12
Physics

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