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(B) When a radioactive material decays through two different processes simultaneously,the effective decay constant $\lambda_{eff}$ is the sum of the i

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$24$

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(B) When a radioactive material decays through two different processes simultaneously,the effective decay constant $\lambda_{eff}$ is the sum of the individual decay constants: $\lambda_{eff} = \lambda_{\alpha} + \lambda_{\beta}$.Given $T_{\alpha} = 16$ years and $T_{\beta} = 48$ years.The decay constants are $\lambda_{\alpha} = \frac{\ln 2}{16}$ and $\lambda_{\beta} = \frac{\ln 2}{48}$.Thus,$\lambda_{eff} = \frac{\ln 2}{16} + \frac{\ln 2}{48} = \ln 2 \left( \frac{3+1}{48} \right) = \frac{\ln 2}{12}$.The effective half-life $T_{eff} = \frac{\ln 2}{\lambda_{eff}} = 12$ years.We want to find the time $t$ when $3/4$th of the material has decayed,which means $1/4$th of the material remains.Using the radioactive decay law: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T_{eff}}$.$\frac{1}{4} = \left( \frac{1}{2} \right)^{t/12} \implies \left( \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^{t/12}$.Equating the exponents: $2 = \frac{t}{12} \implies t = 24$ years.

Half-lives for $\alpha$ and $\beta$ emission of a radioactive material are $16$ years and $48$ years respectively. When the material decays by giving $\alpha$ and $\beta$ emission simultaneously,the time in which $3/4$th of the material decays is ....... years.

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