Question

To determine the half life of a radioactive element, a student plots a graph of $\ln \frac{dN(t)}{dt}$ versus $t.$ Here, $\frac{dN(t)}{dt}$ is the rate of radioactive decay at time $t.$ If the number of radioactive nuclei of this element decreases by a factor of $p$ after $4.16$ years, the value of $p$ is:

Answer 1

Answer: 8

According to radioactive decay law,
$N\left(t\right)=N_0{e}^{-\lambda t}\Rightarrow \left|\frac{dN}{dt}\right|=N_0\lambda e^{-\lambda t}$
Taking $log$ both sides,
$\ln \left|\frac{dN}{dt}\right|=\left[\ln \left(N_0\lambda \right)\right]-\lambda t$
This is a straight line equation with slope $m=-\lambda$
From the given graph,
$m=\frac{4-3}{4-6}=-0.5$
$\Rightarrow \lambda =0.5y{r}^{-1}$
$t_{1/2}=\frac{0.693}{\lambda }=\frac{0.693}{0.5}=1.386yr$
Let the number of half life $=n$
$4.16=n{t}_{1/2}\Rightarrow n=3$
So, $p=2^n=2^3=8$