A radioactive element decays by `beta-`emission. A detector racords `n`-beta particles in `2` sec and in next `2` sec it records `0.65n`-beta particles. Find mean life.

Explanation

Let `N_(0)=` initial number of nuclei at time `t =0`
where `N:` number of undecayed nuclei in time `t`.
`N =N_(0)e^(-lambdat)`
The number of nuclei decayed in time `t`
`N' = N_(0) - N = N_(0)(1-e^(-lambdat)) = n …..(i)`
The number of undecayed nuclei in next time `t`
`N'_(0) = Ne^(-lambdat) = N_(0)e^(-lambdat).e^(-lambdat) = N_(0)e^(-2lambdat)`
Number of decayed nuclei in next time `t`
`N'' = N-N'_(0) = N_(0)e^(-lambdat) - N_(0)e^(-2lambdat) = N_(0)e^(-lambdat) (1-e^(-lambdat))`
`= 0.75n ....(ii)`
`(ii)//(i)`
`e^(-lambdat) = 0.75 = (3)/(4)`
`e^(lambdat) = (4)/(3)`
`lambdat = ln((4)/(3))`
`lambda xx2 = ln((4)/(3))`
`lambda = (1)/(2)ln((4)/(3))`
Mean life `bar(T) = (1)/(lambda) = (1)/(ln(4//3))`