InterviewSolution
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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» 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. |
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