If a radioactive substance reduces to `(1)/(16)` of its original mass in `40` days, what is its half-life ?

Options

A. 10
B. 5
C. 2.5
D. 20

Answer

A

Explanation

(a) From Rutherford and Soddy law for radioactive decay `N = N_0 e^(-lamda t)`
where `N_0` and `N` are number of atom in a radioactive substance at time` t = 0` and `t = t` and `lamda` is decay constant.
Also, half-life `T_(1//2) = (0.693)/(lamda)`
`:. (N)/(N_0) = e^((0.683)/(-T^(1//2)) t)`
Given, `t = 40 days, (N)/(N_0) = (1)/(16)`
`:. (1)/(16) = e^((-(0.693 xx 40))/(T^(1//2)))`
`rArr (0.693 xx 40)/(T^(1//2)) = log 16`
`T_(1//2) = (0.693 xx 40)/(log 16)`
=`9.99 ~~ 10 days`.