Question

The half life of a radioactive nucleus is $50$ days. The time interval (t$_2-t_1$) between the time t$_2$ when $\frac{2}{3}$ ot it has decayed and the time $t_1$, when $\frac{1}{3}$ of it had decayed is

• A. 30 days
• B. 50 days
• C. 60 days
• D. 15 days

Answer 1

Answer: (B)

According to radioactive decay law
$N=N_0e^{-\lambda t}$
where N$_0$ = Number of radioactive nuclei at
time t= 0
N = Number of radioactive nuclei left
undecayed at any time t
$\lambda$= decay constant
At time $t_2,\frac{2}{3}$ of the sample had decayed
$\therefore N=\frac{1}{3}N_0$
$\quad\therefore \frac{1}{3} N_{0}=N_{0}e^{-\lambda t_{2}} \quad\ldots\left(i\right)$
At time $t_1, \frac{1}{3}$ of the sample had decayed,
$\therefore N=\frac{2}{3}N_0$
$\quad\therefore \frac{2}{3}N_{0}=N_{0}e^{-\lambda t_{1}} \quad\ldots\left(ii\right)$
Divide (i) by (ii), we get
$\frac{1}{2}=\frac{e^{-\lambda t_2}}{e ^{-\lambda t_1}}$
$\frac{1}{2}=e^{-\lambda t_2}(t_2 -t_1)$
$\lambda(t_2-t_1)=In 2$
$ t_{2}-t_{1}=\frac{In \,2}{\lambda}=\frac{In 2}{\left(\frac{In\, 2}{T_{1 /2}}\right)} \left(\because\lambda=\frac{In\, 2}{T _{1/ 2}}\right)$
$=T_{1 2}=50$ days