Gujarati
Hindi
13.Nuclei
medium

The mean life time of a radionuclide, if its activity decrease by $4\%$ for every $1h$ , would be  .......... $h$ [product is non-radioactive i.e. stable]

A

$25$

B

$1.042$

C

$2$

D

$30$

Solution

$\frac{\mathrm{dN}}{\mathrm{dt}}=\frac{0.04 \mathrm{N}}{3600}$  (given) 

But $\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=-\lambda \mathrm{N}$

So $-\lambda N=\frac{-0.04}{3600} \mathrm{N}$ or $\lambda=\frac{0.04}{3600} \mathrm{\,s}^{-1}$

Mean life $T=\frac{1}{\lambda}=\frac{3600}{0.04} s=25\, h$

Standard 12
Physics

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