Question

The half-life of a radioactive substance is $20$ day. If $2/3$ part of the substance has decayed in time $t_2$ and $1/3$ part of it has decayed in time $t_1$ then the time interval between $t_2$ and $t_1$ is $(t_2-t_1)=$ __________

• A. $10$ day
• B. $5$ day
• C. $20$ day
• D. $40$ day

Answer 1

The correct option is C $20$ day

According to the question [Given, $T=20$ days]
$\frac{2}{3}{N}_0=N_0{e}^{-\lambda t_2}$ ...(i)
$\frac{1}{3}{N}_0=N_0{e}^{-\lambda t_1}$ ...(ii)
Dividing equation (i) by equation (ii), we get
$\frac{\frac{2}{3}{N}_0}{\frac{1}{3}{N}_0}=\frac{N_0{e}^{-\lambda t_2}}{N_0{e}^{-\lambda t_1}}$
$2=e^{+\lambda (t_2-t_1)}$
$\lambda (t_2-t_1)=\log 2$
$t_2-t_1=\frac{1}{\lambda }\log 2$ $[\because T=\frac{\log 2}{\lambda }]$
$t_2-t_1=T\Rightarrow t_2-t_1=20$ day