Question

The count rate observed from a radioactive source at $t$ second was $N_0$ and at $4t$ second it was $\frac{N_0}{16}$. The count rate observed at $\left(\frac{11}{2}\right)t$ second will be

• A. $\frac{N_0}{128}$
• B. $\frac{N_0}{64}$
• C. $\frac{N_0}{32}$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{N_0}{64}$

after n lives,

$N=N_0\left(\frac{1}{2}\right)$

initially,

$t=t$

$t=4t-t=3t$

$\frac{N_0}{16}=N_0{\left(\frac{1}{2}\right)}^n$

$n=4$

$4t_{\frac{1}{2}}=3t$

$\Rightarrow t_{\frac{1}{2}}=\frac{3}{4}t$

$\lambda =\frac{\ln 2}{t_{\frac{1}{2}}}=\frac{4\ln 2}{3t}$

at $t=\frac{11}{2}ts$

$t=\frac{11}{2}t-t=\frac{9}{2}t$

$N=N_0{e}^{-\lambda \frac{9}{2}t}$

$N=N_0{e}^{-\lambda \frac{4\ln 2}{3t}\times \frac{9}{2}t}$

$N=N_0{e}^{-6\ln 2}$

$N=N_0{e}^{\ln \frac{1}{2^6}}$

$N=\frac{N_0}{2^6}=\frac{N_0}{64}$