At a given instant there are 25% undecayed radioactive nuclei in a sample . After 10 sec the number of undecayed nuclei remains `12.5%` Calculate :
(i) mean - life of the nuclei and
(ii) The time in which the number of undecayed nuclears will further reduce to `6.25 %` of the reduced number.

Answer

`t_(avg)=14.43 s` (ii) 40 second

Explanation

`(i) t_(75%)=`
`t_(1//2)=(1)/(lambda) " " `25% undecayed
`t_("avg")=(t_(1//2))/(ln 2) " " darr 10 "sec" =t_(1//2)`
`" " 12.5%` unchecayed
`=(10 "sec")/(ln 2) =14.426 "sec"`
(ii) `6.25% "of" 12.5 "is" 12.5 xx (6.25)/(100)=0.78125 ` left
`t=(1)/(lambda)ln ((12.5)/(12.5 xx(12.5)/(100))) rArr t=(t_(1//2))/(ln2) ln ((100)/(6.25)) rArr t=40 "sec"`