Question

To determine the half life of a radioactive element, student plots a graph of $\ln \left|\frac{dN\left(t\right)}{dt}\right|$ versus $t$. Here $dN\left(t\right)/dt$ 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 $4.16\text{years}$, the value of $p$ is ____. (Round off the nearest integer) $(e^{4.16}=64.07)$

Answer 1

From the law of radioactivity, first order decay is

$N\left(t\right)=N_0{e}^{-\lambda t}$

$\frac{dN}{dt}=N_0\lambda e^{-\lambda t}$

Taking natural $\log$ both side,

$\ln \left(\frac{dN}{dt}\right)=\ln \left(N_0\lambda \right)-\lambda t=-\lambda t+\ln \left(N_0\lambda \right)$

After comparing with eq. $y=mx+c$, we get,

Slope, $m=-\lambda$

From graph, slope will be

$m=\frac{3-6}{7-1}=-0.5$

$\therefore \lambda =0.5\text{per year}$

So, from given in question,

$\frac{N}{N_0}=\frac{1}{p}=e^{-\lambda t}=e^{-0.5\times 4.16}=\frac{1}{8}$

$\therefore p=8$

Hence, $8$ will be the right answer.

Why this question: To understand the way to analyse half- life period graphically.