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Nucleus A decays to B with decay constant \lambda_1 and \mathrm{B} decays to \mathrm{C} with decay constant \lambda_2. Initially at t=0 number of nuclei of A and B are 2 \, \, N_0 and N_0 respectively. At t=t_0, no. of nuclei of B is \frac{3 \mathrm{~N}_0}{2} and nuclei of B stop changing. Find  t_0 ?

Option: 1

\frac{1}{\lambda_1} \ln \left(\frac{2 \lambda_1}{3 \lambda_2}\right)


Option: 2

\frac{1}{\lambda_1} \ln \left(\frac{8 \lambda_1}{3 \lambda_2}\right)


Option: 3

\frac{1}{\lambda_1} \ln \left(\frac{7 \lambda_1}{3 \lambda_2}\right)


Option: 4

\frac{1}{\lambda_1} \ln \left(\frac{4 \lambda_1}{3 \lambda_2}\right)


Answers (1)

best_answer

\frac{\mathrm{dN}_{\mathrm{B}}}{\mathrm{dt}}=\lambda_1 \mathrm{~N}_{\mathrm{A}}-\lambda_2 \mathrm{~N}_{\mathrm{B}}

0=\lambda_1\left(2 \mathrm{~N}_0\right) \mathrm{e}^{-\lambda t_0}-\lambda_2\left(\frac{3 \mathrm{~N}_0}{2}\right)

\mathrm{t}_0=\frac{1}{\lambda_1} \ln \left(\frac{4 \lambda_1}{3 \lambda_2}\right)

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