Question

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`.

Answer 1

Correct Answer - `6.954 sec`
Let `n_(0)` be the number of radioactive nuclei at time `t=0`. Number of nuclei decayed in time `t` are given `n_(0)(1-e^(-2lambda))`, which is also equal to the number of beta particles emitted 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 (ii) by (i) we get
`1.75=(1-e^(-4lambda))/(1-e^(-2lambda))` or `1.75-1.75e^(-2lambda)=1-e^(-4lambda)`
`:. 1.75e^(-2lambda)=(3)/(4)`.......(iii)
Let us take `e^(-2lambda)=x`
Then the above equation is,
`x^(2)-1.75x+0.75=0`
or `x=(1//75sqrt((1.75)^(2)-(4)(0.75)))/(2)` or `x=1` and `(3)/(4)`
`:.` From equation (iii) either `e^(-2lambda)=1 or e^(-2lambda)=(3)/(4)`
but `e^(-2lambda)=1` is not accepted because which means `lambda=0`. Hence
`e^(-2lambda)=(3)/(4) or " " -2lambda ln (e )= l n (3) - l n(4) = l n (3)-2 l n(2):. " "lambda= l n (2)-(1)/(2) l n(3)`
Substituting the given values, `lambda=0.6931-(1)/(2)xx(1.0986)=0.14395s^(-)`
`:.` Mean life `t_("means")=(1)/(lambda)=6.947 sec`