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The half life of a radioactive nucleus is 50 days. The time interval $(t_2 - t_1)$ between the time $t_2$ when $\frac{2}{3}$ of it has decayed and the time $t_1$ when $\frac{1}{3}$ of it had decayed is:
A
15 days
B
30 days
C
50 days
D
60 days
Detailed Solution
At $t_1$, $\frac{1}{3}$ has decayed, so $\frac{2}{3}$ remains: $\frac{2}{3}N_0 = N_0e^{-\lambda t_1}$ ...(i)
At $t_2$, $\frac{2}{3}$ has decayed, so $\frac{1}{3}$ remains: $\frac{1}{3}N_0 = N_0e^{-\lambda t_2}$ ...(ii)
Dividing (i) by (ii): $2 = e^{\lambda(t_2 - t_1)}$
$\ln2 = \lambda(t_2 - t_1) \Rightarrow t_2 - t_1 = \frac{\ln2}{\lambda} = t_{1/2}$
$t_2 - t_1 = 50$ days
At $t_2$, $\frac{2}{3}$ has decayed, so $\frac{1}{3}$ remains: $\frac{1}{3}N_0 = N_0e^{-\lambda t_2}$ ...(ii)
Dividing (i) by (ii): $2 = e^{\lambda(t_2 - t_1)}$
$\ln2 = \lambda(t_2 - t_1) \Rightarrow t_2 - t_1 = \frac{\ln2}{\lambda} = t_{1/2}$
$t_2 - t_1 = 50$ days