At a given instant there are `25%` undecayed radioactive nuclei in a sample. After `10 s` the number of undecayed nuclei reduces to `12.5%`. Calculate
(a) mean life of the nuclei,
(b) the time in which the number of undecayed nuclei will further reduce to` 6.25%` of the reduced number.

Explanation

Half-life of radioactive sample, i.e., the time in which number of undecayed nuclei becomes half(T) is 10 s.
(a) Mena life `tau=(T)/(log_(e) 2) =(10)/(0.693 s) =1.443 xx 10 =14.43 s`
(b) The reduced number further reduces to `6.25%` in n half -lives given by
`(N)/(100) =((1)/(2))^(n) rArr (6.25)/(100)=((1)/(2))^(n)`
`rArr (1)/(16)=((1)/(2))^(n) rArr ((1)/(2))^(4) =((1)/(2))^(n) rArr n=4`
Time, `t=4T = 4xx 10 =40 s`.