Question

To determine the half-life of a radioactive element, a student plots graph of $\ln \left|\frac{dN(t)}{dt}\right|$ 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 $4.16yr$, the value of $p$ is $\left(e^{-2.08}=0.125\right)$

Answer 1

Answer: 8

$\left|\frac{dN}{dt}\right|=\mid$ Activity of radioactive substance $|$
$=\frac{dN}{dt}=\frac{d}{dt}{N}_0{e}^{-\lambda t}\left(\because N=N_0{e}^{-\lambda t}\right)$
Taking log both sides
$\ln \left|\frac{dN}{dt}\right|=\ln \left(-(\lambda N{)}_0\right)e^{-\lambda t}$
Hence, In $\left|\frac{dN}{dt}\right|$ versus $t$ graph is a straight line with slope- $\lambda$.
From the graph, we can see that,
$\lambda =\frac{1}{2}=0.5y{r}^{-1}$
Now applying the equation
$N=N_0{e}^{-\lambda t}=N_0{e}^{-0.5\times 4.16}$
$=N_0{e}^{-2.08}=N_0{e}^{-2\times (0.693)}=0.125N_0$
$=\frac{N_0}{8}$
i.e. nuclei decrease by a factor of $8$ .
Hence the answer is $8$ .