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` be the number of radioactive nuclei at time `t=0`. Number of nuclei decayed in time t are given by `n_0(1-e^(-lambdat))`, which is also equal to the number of beta particles emitted during the same interval of time. For the given condition,
`n=n_0(1-e^(-2lambda))` ...(i)
`(n+0.75n)=n_0(1-e^(-4lambda))` ...(ii)
Dividing Eq. (ii) by Eq. (i), we get
`1.75=(1-e^(-4lambda))/(1-e^(-2lambda))`
or `1.75 - 1.75 e^(-2lambda) = 1-e^(-4lambda)`
:. `1.75e^(-2lambda)-e^(-4lambda)=3/4` ...(iii)
Let us take `e^(-2lambda)=x`
Then, the above equation is
`x^2-1.75x+0.75=0`
or `x=(1.75+-sqrt((1.75)^2-(4)(0.75)))/(2)`
or `x=1 and 3/4`
:. From Eq. (iii) either
`e^(-2lambda)=1`
or `e^(-2lambda)=3/4`
but `e^(-2lambda)=1` is not acceptable because which means `lambda=0`.
Hence, `e^(-2lambda)=3/4`
or `-2lambda1n(e)=1n(3)-1n(4)=1n(3)-21n(2)`
:. `lambda=1n(2)-(1)/(2)1n(3)`
`Substituting the given values,
`lambda=0.6931-1/2xx(1.0986)=0.14395s^-1`
:. Mean life, `t_(mean)=1/lambda=6.947`s
:. The correct answer is 7.