Question

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

Answer 1

$N=N_o{e}^{-\lambda t}$[Radioactive nucliei of the element which is decaying after time $t$]

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

$ln\left|\frac{dN}{dt}\right|=ln\lambda N_o-\lambda t$

By graph

$\lambda =\frac{1}{2}$

$N_{final}=\frac{N_o}{P}=N_o{e}^{-\lambda t}$

$P=e^{\lambda t}$

$=e^{\frac{1}{2}\times 4.16}$

$=e^{2.08}$

$P\approx 8$