A radioactive element decays by `beta`-emission. A detector records n-beta particles in 2 s and in next 2 s it records 0.45 n beta particles. Find mean life.

Explanation

Let `N_(0)` specifies initial number of nuclei at time t=0,
where, `N` = number of undecayed nuclei in time t.
According to the law of radioactivity, `N=N_(0)e^(-lamdat)`
The number of decayed nuclei in time t,
`" "N'=N_(0)-N=N_(0)(1-e^(-lamdat))=n" "`...(i)
The number of undecayed nuclei in next time t,
`" "N'_(0)=Ne^(-lamdat)=N_(0)e^(-lamdat)*e^(-lamdat)=N_(0)e^(-2lamdat)`
Number of decayed nuclei in next time t,
`" "N''=N-N'_(0)`
`" "=N_(0)e^(-lamdat)-N_(0)e^(-2lamdat)`
`" "=N_(0)e^(-lamdat)(1-e^(-lamdat))" "`...(ii)
`" "=0.45 n`
On dividing Eq. (ii) by Eq. (i), we get
`" "e^(-lamdat)=0.45=(9)/(20)`
`rArr" "e^(lamdat)=(20)/(9)rArrlamdat= In ((20)/(9))`
As, time = 2s `" "` (given)
`rArr" "2xxlamda=In ((20)/(9))rArrlamda=(1)/(2)In ((20)/(9))`
So, mean life, `tau=(1)/(lamda)=(2)/(In((20)/(9)))`