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(C) The decay constant for $\alpha$-emission is $\lambda_{\alpha} = \frac{1}{1620} \ year^{-1}$.The decay constant for $\beta$-emission is $\lambda_{\

Vedclass · Accepted Answer

$449$

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(C) The decay constant for $\alpha$-emission is $\lambda_{\alpha} = \frac{1}{1620} \ year^{-1}$.The decay constant for $\beta$-emission is $\lambda_{\beta} = \frac{1}{405} \ year^{-1}$.Since the emissions are simultaneous,the total decay constant is $\lambda = \lambda_{\alpha} + \lambda_{\beta} = \frac{1}{1620} + \frac{1}{405} = \frac{1 + 4}{1620} = \frac{5}{1620} = \frac{1}{324} \ year^{-1}$.The law of radioactive decay is given by $N = N_{0} e^{-\lambda t}$,where $\frac{N}{N_{0}} = \frac{1}{4}$.Taking the natural logarithm on both sides: $\ln(\frac{N_{0}}{N}) = \lambda t$.$\ln(4) = \lambda t \Rightarrow t = \frac{\ln(4)}{\lambda} = \frac{2 \ln(2)}{\lambda}$.Substituting the values: $t = 324 	imes 2 	imes 0.6931 \approx 449 \ years$.

For a substance,the average life for $\alpha$-emission is $1620 \ years$ and for $\beta$-emission is $405 \ years$. After how much time will $\frac{1}{4}$ of the material remain due to simultaneous emission?

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