Question

The half life of a radioactive substance is $20\text{minutes}$. The approximate time interval $(t_2-t_1)$ between the time $t_2$, when $\frac{2}{3}$ of it has decayed and time $t_1$, when $1/3$ of it has decayed is

• A. $7\text{min}$
• B. $14\text{min}$
• C. $20\text{min}$
• D. $28\text{min}$

Answer 1

The correct option is C $20\text{min}$

After $t_1$, the fraction remaining

$\frac{N_1}{N_0}=1-\frac{1}{3}=\frac{2}{3}=\frac{1}{2^{n_1}}$

$\frac{2}{3}=\frac{1}{2^{t_1/20}}...\left(1\right)$

After time $t_2$, the fraction remaining is

$\frac{N_2}{N_0}=1-\frac{2}{3}=\frac{1}{3}=\frac{1}{2^{n_2}}$

$\frac{1}{3}=\frac{1}{2^{t_2/20}}...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$2\left(2^{t_1/20}\right)=2^{t_2/20}$

$\Rightarrow 1+t_1/20=\frac{t_2}{20}$

$\Rightarrow 1=\frac{t_2-t_1}{20}$

$\therefore t_2-t_1=20\text{min}$

Hence, $\left(\text{C}\right)$ is the correct answer.