Correct Answer - `6.947 s`
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 equation (ii) by (i), we get
`1.75=(1-e^(-4lambda))/(1-e^(-2lambda))implies 1.75-1.75 e^(-2lambda)=1-e^(-4lambda)`
`:. 1.75 e^(-2lambda)-e^(-4lambda)=3/4` ...(iii)
Let us take `e^(-2x)=x`
Then the above equation is `x^(2)-1.75x+0.75=0`
`implies x=(1.75 pm sqrt((1.75)^(2)-(4)(0.75)))/(2) implies x=1` and `3/4`
`:.` From equation (iii) either `e^(-2lambda)=1, e^(-2lambda)=3/4`
but `e^(-2lambda)=1` is not accelted because which means `lambda=0` Hence, `e^(-2lambda)=3/4`
`implies -2lambda ln(e)=ln (3)-ln(4)=ln(3)-2ln(2)`
`:. lambda=ln (2)-1/2 ln(3)`
Substituting the given values,
`l=0.6931-1/2 xx(1.0986)=0.14395 s^(-1)`
`:.` Mean-life `t_("mean")=1/lambda=6.947 s`
