A radioactive element decays by `beta-emission`. A detector records `n` beta particles in `2 s` and in next `2 s` it records `0.75 n` beta particles. Find mean life correct to nearest whole number. Given ln `|2| = 0.6931`, ln `|3|=1.0986`.

Explanation

Let `N_(0)` initial number of nuclei at time `t=0`.
The number of undecayed nuclei in time `t` is` N =N_(0) e^(- lambda t)`
The number of nuceli decayed in time t is
`n=N_(0) - N=N_(0) -N_(0) e^(-lambda t)`
The number of undecayed nuclei in next time t,
`N'=Ne^(-lambda t)=(N_(0)e^(-lambda t)) e^(-lambdat)=n_(0)e^(-2 lamn=bdat)= N_(0)e^(-2 lambda t)`
`0.75 n=N -N' =N_(0) e^(- lambda t) e^(- lambda t) -N_(0) e^(-2 lambda t)`
`=N_(0) e^(- lambda t)(1 -e^(- lambda t))`
Dividing Eq. (ii) by Eq(i), we get
`0.75 =e^(-lambda t) rArr (4)/(3) =e^(lambda t)`
Taking natural logarithm,
In `(4)/(3) = lambda t rArr lambda = (In(4)/(3))/(t)`
Given`t=2s`.
`:. lambda = (ln 4-ln 3)/(2) =(2 ln 2 -ln 3)/(2) =(2 xx 0.6931 -1.0986)/(2)`
`=0.1438 s^(-1)`
Mean life, `tau =(1)/(lambda)=(1)/(0.1438)=7s ` (whole number).