Question

To determine the half-life of a radioactive element, a student plots graph of $\ln \left|\frac{d N(t)}{d t}\right|$ versus $t$. Here $\frac{d N(t)}{d t}$ 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 $4.16\, y r$, the value of $p$ is $\left(e^{-2.08}=0.125\right)$

Answer 1

$\left|\frac{d N}{d t}\right|=\mid $ Activity of radioactive substance $| $
$=\frac{d N}{d t}=\frac{d}{d t} N_{0} e^{-\lambda t} \left(\because N=N_{0} e^{-\lambda t}\right)$
Taking log both sides
$\ln \left|\frac{d N}{d t}\right|=\ln \left(-(\lambda N)_{0}\right) e^{-\lambda t}$
Hence, In $\left|\frac{d N}{d t}\right|$ versus $t$ graph is a straight line with slope- $\lambda$.
From the graph, we can see that,
$\lambda=\frac{1}{2}=0.5 yr ^{-1}$
Now applying the equation
$N=N_{0} e^{-\lambda t}=N_{0} e^{-0.5 \times 4.16}$
$=N_{0} e^{-2.08}=N_{0} e^{-2 \times(0.693)}=0.125 N_{0}$
$=\frac{N_{0}}{8}$
i.e. nuclei decrease by a factor of $8$ .
Hence the answer is $8$ .