A radioactive sample emit `n beta` - particles is `2` sec , In next 2 sec it emits` 0.75 n beta`- particle , what is the mean life of the sample?

Answer

`1.75n=N_(0)(1-e^(-4 lambda)),6.95 sec, (2)/(ln((4)/(3)))`

Explanation

Let initial number of nuclei `=N_(0)`
`N_(0)overset(2 "sec")rarr (N_(0)-n) overset(2 "sec")rarr(N_(0)-n-0.75 n)`
(No. of `beta` particles emitted=No. of nuclei disintegrated)
No. of nuclei disintegrated in time is given by `=N_(0)(1-e^(-lambdat t)`
`:.` for initial 2 seconds
`n=N_(0)(1-e^(-2lambda)) " " `...........(i)
for next 2 seconds
`0.75 n=(N_(0)-n)(1-e^(-2 lambda)) " " ` ..........(ii)
Subtracting (ii) from (i) ((i)-(ii))
`.25 n=n (1-e^(-2lambda))`
`e^(-2lambda)=0.75`
`-2lambda=ln. (3)/(4) " " rArr 2lambda =ln .(4)/(3)`
`t_("avg")=(1)/(lambda)=(2)/(ln((4)/(3)))=6.95 "sec"`