Question

The half life of a radioactive substance is $20$ minutes. At the time instant $t_1$, $\frac{2}{3}$ of the substance has decayed while at the instant $t_2$, $\frac{1}{3}$ of it has decayed. The approximate time interval $(t_2-t_1)$ is equal to,

• A. $7$ min
• B. $14$ min
• C. $20$ min
• D. $28$ min

Answer 1

The correct option is C $20$ min

After time $t_1$, the fraction remaining is,
$\frac{N_1}{N_0}=\frac{2}{3}=\frac{1}{2^{n_1}}$

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

After time $t_2$, the fraction remaining is
$\frac{N_Z}{N_0}=\frac{1}{3}=\frac{1}{2^{\left(\frac{t_2}{T}\right)}}$
$\frac{1}{3}=\frac{1}{2^{\left(\frac{t_2}{20}\right)}}......\left(ii\right)$

Dividing $\left(i\right)$ by $\left(ii\right)$ we get,

$2\times 2^{\left(\frac{t_1}{20}\right)}=2^{\left(\frac{t_2}{20}\right)}$
$2^{(1+\frac{t_1}{20})}=2^{\frac{t_2}{20}}$

Comparing the powers on both sides,

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

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

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

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