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13.Nuclei
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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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