The half-life of a radioactive nucleus is `50` days. The time interval `(t_2 -t_1)` between the time `t_2` when `(2)/(3)` of it has decayed and the time `t_1` when `(1)/(3)` of it had decayed is

Options

A. 30 dyas
B. 50 days
C. 60 days
D. 15 days

Explanation

According to radioactive decay law, `N=N_(0)e^(-lambdatt)`
where `N_(0)=` Number of radioactive nuclei at time t = 0
N = Number of radioactive nuclei left undecayed at any time t
`lambda=` decay constant
At time `t_(2),(2)/(3)` of the sample had decayed `therefore N=(1)/(3)N_(0)`
`therefore (1)/(3)N_(0)=N_(0)e^(-lambdat_(2))`
At time `t_(1),(1)/(3)` of the sample had decayed, `therefore N=(2)/(3)N_(0)`
`therefore" "(2)/(3)N_(0)=N_(0)e^(-lambdat_(1))`
Divide (i) by (ii), we get
`(1)/(2)=(e^(-lambdat_(2)))/(e^(-lambdat_(1)))=e^(-lambda(t_(2)-t_(1)))`
`t_(2)-t_(1)=(ln2)/(lambda)=(ln2)/(((ln2)/(T_(1//2))))" "because lambda=(ln2)/(T_(1//2))`
`=T_(1//2)="50 days"`