Question

The half-life of a radioactive substance is $30$ minutes. The time interval (in minutes) between $40$% decay and $85$% decay, of the same radioactive substance, is,

• A. $60$ min
• B. $45$ min
• C. $30$ min
• D. $15$ min

Answer 1

The correct option is A $60$ min

(i) After $40$% decay, remaining is $60$%.
$N_1=60$%, $N_0=\frac{60}{100}{N}_0$

$\frac{N_1}{N_0}=\frac{60}{100}=\frac{3}{5}=\frac{1}{2^{n_1}}.......\left(i\right)$

Now, $t_1=n_{1}T\Rightarrow n_1=\frac{t_1}{T}$

So, equation $\left(i\right)$ becomes,

$\frac{3}{5}=\frac{1}{2^{\left(\frac{t_1}{T}\right)}}.......\left(ii\right)$

(ii) After $85$% decay, remaining is $15$%.

$\frac{N_2}{N_0}=\frac{15}{100}=\frac{3}{20}=\frac{1}{2^{n_2}}$

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

Dividing $\left(ii\right)$ by $\left(iii\right)$, we get,
$4=\frac{2^{\left(\frac{t_2}{T}\right)}}{2^{\left(\frac{t_1}{T}\right)}}$

$2^2=2^{\left(\frac{t_2-t_1}{T}\right)}$

Comparing powers on both sides, we get,

$2=\frac{t_2-t_1}{T}$
$t_2-t_1=2\times T=2\times 30=60\text{min.}$

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