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

Vedclass · Accepted Answer

$449$

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(C) The decay constant for $\alpha$-emission is $\lambda_{\alpha} = \frac{1}{1620} 	ext{ year}^{-1}$.The decay constant for $\beta$-emission is $\lambda_{\beta} = \frac{1}{405} 	ext{ year}^{-1}$.The total decay constant is $\lambda = \lambda_{\alpha} + \lambda_{\beta} = \frac{1}{1620} + \frac{1}{405} = \frac{1+4}{1620} = \frac{5}{1620} = \frac{1}{324} 	ext{ year}^{-1}$.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: $\ln\left(\frac{N_0}{N}ight) = \lambda t$.Substituting the values: $\ln(4) = \left(\frac{1}{324}ight) t$.$t = 324 	imes \ln(4) = 324 	imes 2 	imes \ln(2) \approx 648 	imes 0.6931 = 449.13 	ext{ years}$.Thus,the time required is approximately $449 	ext{ years}$.

For a substance,the average life for $\alpha$-emission is $1620$ years and for $\beta$-emission is $405$ years. After how much time will $1/4$ of the material remain after both $\alpha$ and $\beta$ emissions?

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