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

Vedclass · Accepted Answer

$449$

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(C) The decay constant for $\alpha$ decay is $\lambda_{\alpha} = \frac{1}{1620} 	ext{ year}^{-1}$.The decay constant for $\beta$ decay is $\lambda_{\beta} = \frac{1}{405} 	ext{ year}^{-1}$.The effective 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 activity $A$ at time $t$ is given by $A = A_0 e^{-\lambda t}$.Given $\frac{A}{A_0} = \frac{1}{4}$,we have $e^{-\lambda t} = \frac{1}{4}$,which implies $e^{\lambda t} = 4$.Taking the natural logarithm on both sides: $\lambda t = \ln(4) = 2 \ln(2)$.Substituting $\lambda = \frac{1}{324}$ and $\ln(2) \approx 0.693$:$t = 324 	imes 2 	imes 0.693 = 648 	imes 0.693 \approx 449.06 	ext{ years}$.Thus,the time is approximately $449 	ext{ years}$.

$A$ radioactive element emits $\alpha$ and $\beta$ particles. Its mean lives for $\alpha$ and $\beta$ decay are $1620$ years and $405$ years,respectively. After how many years will the activity be reduced to $1/4$ of its initial value?

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