A) 550 years
B) 300 years
C) 615 years
D) 655 years
Correct Answer: B
Solution :
\[N={{N}_{0}}{{e}^{-\lambda t}}\] \[\frac{{{N}_{0}}}{4}\,={{N}_{0}}{{e}^{-\lambda t}}\] \[\lambda t=\,\ln 4\] \[t=\frac{1}{\lambda }\,\ln \,4=\frac{1}{\lambda }\ln \,{{2}^{2}}=\frac{2}{\lambda }In\,2=312\] \[\lambda =\frac{2\ln 2}{312}\,\times 4.443\,\times {{10}^{-3}}\] \[=\frac{1}{225}\] \[\lambda ={{\lambda }_{\alpha }}\,+{{\lambda }_{B}}\] \[\frac{1}{225}\,=\frac{1}{900}\,+{{\lambda }_{\beta }}\] \[{{\lambda }_{\beta }}\,=\frac{1}{225}\,-\frac{1}{900}=\frac{1}{300}\] Decay constant is reciprocal of mean life. So, mean life for \[\beta \] - emission = 300 years.You need to login to perform this action.
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