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InterviewSolution
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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`. |
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Answer» 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` |
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