A radioactive nucleus decays by two different processes. The half-life for the 1st process is 10s and that for the 2nd process 100s. The effective half-life of the nucleus is close to:
Option: 1 55s
Option: 2 9s
Option: 3 6s
Option: 4 12s

Answers (1)

Correct Answer: 2) 9 sec

\\ \text{Radioactive decay rate is directly proportional to number of nucleus present at any time}\\ \frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=-\lambda \mathrm{N}_{-}(\mathrm{i})$\\ where $\lambda$ is a constant related to half life as $\lambda=\frac{\ln 2}{\mathrm{T}_{\frac{1}{2}}}$ where $\mathrm{T}_{\frac{1}{2}}$ is half life time of radioactive decay \\ The total rate of decay is $\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=-\left(\lambda_{1}+\lambda_{2}\right) \mathrm{N}=\lambda_{\text {effective }} \mathrm{N}_{\ldots}($ ii $)$

\begin{aligned} &\lambda_{1}=\frac{\ln 2}{\mathrm{T}_{1}}\\ &\lambda_{2}=\frac{\ln 2}{\mathrm{T}_{2}}\\ &\lambda_{\text {effective }}=\frac{\ln 2}{\mathrm{T}_{\text {effective }}}\\ &\text { substituting in equation (ii) }\\ &\left(\frac{\ln 2}{\mathrm{T}_{1}}+\frac{\ln 2}{\mathrm{T}_{2}}\right) \mathrm{N}=\frac{\ln 2}{\mathrm{T}_{\text {effective }}} \mathrm{N}\\ &\frac{1}{\mathrm{T}}=\frac{1}{\mathrm{T}_{1}}+\frac{1}{\mathrm{T}_{2}} \end{aligned}

\frac{1}{T}=\frac{1}{10}+\frac{1}{100}=\frac{11}{100}\\ \Rightarrow T=\frac{100}{11}=9 sec

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