Question

The half-life of a radioactive substance is $20\text{min}.$ The approximate time interval $(t_2-t_1)$ between the time $t_2$ when $\frac{2}{3}$ of it had decayed and time $t_1$ when $\frac{1}{3}$ of it had decayed (in $\text{min})$ is ​​​​​​(integer only)

Answer 1

Given,

Half - life, $t=\frac{\ln 2}{\lambda }=20\text{min}$

The number of undecayed nuclei is given by,

$N=N_0{e}^{-\lambda t}$

When $\frac{2}{3}$ of the substance is decayed then the undecayed substance is given by,

$\therefore (1-\frac{2}{3})N_0=N_0{e}^{-\lambda t_2}$

$\frac{N_0}{3}=N_0{e}^{-\lambda t_2}......\left(1\right)$

Again, when $\frac{1}{3}$ of substance is decayed, then the undecayed substance is given by,

$(1-\frac{1}{3})N_0=N_0{e}^{\lambda t_1}$

$\frac{2}{3}{N}_0=N_0{e}^{-\lambda t_1}......\left(2\right)$

Dividing equation (1) by(2),

$\frac{1}{2}=e^{-\lambda (t_2-t_1)}$

$\therefore t_2-t_1=\frac{ln2}{\lambda }=t=20\text{min}$