The mean lives of a radioactive substance for alpha and beta emissions are 1620 years and 405 years respectively. After how much time will the activity be reduced to one fourth. A. 405 years B. 1620 years C. 449 years D. None of these
Question
The mean lives of a radioactive substance for α and β emissions are 1620 years and 405 years respectively. After how much time will the activity be reduced to one fourth.
A. 405 years
B. 1620 years
C. 449 years
D. None of these
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Answer
The correct option is C: 449 years.
First, let's determine the decay constants for the $\alpha$ and $\beta$ emissions.

For $\alpha$ emission, the decay constant is given by: $$ \lambda_{\alpha} = \frac{1}{1620} \text{ per year} $$

For $\beta$ emission, the decay constant is: $$ \lambda_{\beta} = \frac{1}{405} \text{ per year} $$
The fraction of the remaining activity is given by: $$ \frac{A}{A_{0}} = \frac{1}{4} $$
The total decay constant $\lambda$ is the sum of the individual decay constants: $$ \lambda = \lambda_{\alpha} + \lambda_{\beta} = \frac{1}{1620} + \frac{1}{405} = \frac{1}{324} \text{ per year} $$
Using the formula for activity decay, we have: $$ A = A_{0} e^{\lambda t} $$
Rearranging the equation to solve for $t$, we get: $$ t = \frac{1}{\lambda} \log_{e} \frac{A_{0}}{A} $$
Given that: $$ \frac{A_{0}}{A} = 4 $$
So: $$ t = \frac{1}{\lambda} \log_{e} 4 = \frac{2}{\lambda} \log_{e} 2 $$
Substituting $\lambda = \frac{1}{324}$ and $\log_{e} 2 \approx 0.693$: $$ t = 324 \times 2 \times 0.693 \approx 449 \text{ years} $$
Thus, the activity will be reduced to onefourth after approximately 449 years.
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