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  • A radioactive element decays by -emission. A detector records n beta particles in 2 seconds and in next 2-seconds it

A radioactive element decays by \beta-emission. A detector records n beta particles in 2 seconds and in next 2-seconds it records 0.75 n beta particles. Then the mean life correct to nearest whole number. Given ln |2| = 0.6931, ln |3| =1.0986.

Option: 1

6.45 sec


Option: 2

6.95 sec


Option: 3

7.25 sec


Option: 4

7.85


Answers (1)

best_answer

Let \mathrm{n_0} be the number of radioactive nuclei at time t = 0. Number of nuclei decayed in time t are given by\mathrm{\mathrm{n}_0\left(1-\mathrm{e}^{-\lambda t}\right)}, which is also equal to the number of beta particles emitted during the same interval of time. For the given condition,

\mathrm{\begin{aligned} n & =n_0\left(1-e^{-2 \lambda}\right) \\ (n+0.75 n) & =n_0\left(1-e^{-4 \lambda}\right) \end{aligned}}

Dividing Eq. (2) by (1), we get

\mathrm{\begin{aligned} & \qquad 1.75=\frac{1-e^{-4 \lambda}}{1-e^{-2 \lambda}} \\ & \text { or } 1.75-1.75 \mathrm{e}^{-2 \lambda} \\ & \therefore 1.75 \mathrm{e}^{-2 \lambda}-\mathrm{e}=1-\mathrm{e}^{-4 \lambda} \\ & \therefore 1.75 \mathrm{e}^{-2 \lambda}-\mathrm{e}^{-4 \lambda} \quad=1-\mathrm{e}^{-4 \lambda} \end{aligned}}            [3]

\mathrm{\text { Let us take } \mathrm{e}^{-2 \lambda}=\mathrm{x}}

Then the above equation is,

\mathrm{\begin{aligned} & x^2-1.75 x+0.75 \\ & \text { or } x=\frac{1.75 \pm \sqrt{(1.75)^2-(4)(0.75)}}{2} \\ & \text { or } \quad x=1 \text { or } e^{-2 \lambda}=\frac{3}{4} \end{aligned}}

\mathrm{\text { but } \mathrm{e}^{-2 \lambda}=1 \text { is not accepted because which means } \lambda=0 \text {. }}

\mathrm{\text { Hence, } \mathrm{e}^{-2 \lambda}=\frac{3}{4}}

\mathrm{\begin{aligned} & \text { or }-2 \lambda \ln (e)=\ln (3)-\ln (4) \\ & =\ln (3)-2 \ln (2) \\ & \therefore \quad \lambda=\ln (2)-\frac{1}{2} \ln (3) \\ & \end{aligned}}

Substituting the given values,

\mathrm{\begin{aligned} \lambda & =0.6931-\frac{1}{2} \times(1.0986) \\ & =0.14395 \mathrm{~s}^{-1} \end{aligned}}

\mathrm{\therefore \text { Mean-life } \mathrm{t}_{\text {means }} \quad=\frac{1}{\lambda}=6.95 \mathrm{sec}}

Posted by

HARSH KANKARIA

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