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For a substance the average life for $\alpha $ -emission is $1620\ years$ and for $\beta $ emission is $405\ years$ . After how much time the $\frac {1}{4}$ of the material remains by simultaneous emission ............ $years$
$648$
$324$
$449$
$810$
Solution
$\lambda_{\alpha}=\frac{1}{1620}$ per year and $\lambda_{\beta}=\frac{1}{405}$ per year and it is given that the fraction of the remained activity $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\frac{1}{4}$
Total decay constant
$\lambda=\lambda_{\alpha}+\lambda_{\beta}=\frac{1}{1620}+\frac{1}{405}=\frac{1}{324}$ per year
We know that $\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{-\lambda t} \Rightarrow \mathrm{t}=\frac{1}{\lambda} \log _{\mathrm{e}} \frac{\mathrm{A}_{0}}{\mathrm{A}}$
$ \Rightarrow \mathrm{t} =\frac{1}{\lambda} \log _{\mathrm{e}} 4=\frac{2}{\lambda} \log _{\mathrm{e}} 2 $
$=324 \times 2 \times 0.693=449 \mathrm{\,years} $
