Question

To determine the half life of a radioactive element, student plots a graph of $ln|dN\left(t\right)/dt|$ 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 $4.16$ years, the value of $p$ is ......

Answer 1

First order decay:
$N\left(t\right)=N_o{e}^{-\lambda t}$
$\frac{dN}{dt}=N_o\lambda e^{-\lambda t}$

Take log on both sides:
$ln\left(\frac{dN}{dt}\right)=ln(N_o\lambda$ - $\lambda t)$
So, the slope of the curve, $m=-\lambda$
From the graph:
$m=\Delta y/\Delta x=1/2=0.5$
$\lambda =0.5$
$\frac{ln2}{t_{1/2}}=0.5$
$t_{1/2}=0.3465years$

Now, $N=N_o{e}^{-0.5t}$
$\frac{1}{p}=e^{-0.5\times 4.16}$
$p=8$