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The mean lives of a radioactive substance are 1620 year and 405 year for $\alpha$ -emission and $\beta$ -emission respectively. Find the time during which three-fourth of a sample will decay if it is decaying both by $\alpha$ -emission and $\beta$ -emission simultaneously.
Option: 1
249 years

Option: 2
449 years

Option: 3
133 years

Option: 4
99 years

Answers (1)

best_answer

The decay constant $\lambda$ is the reciprocal of the mean life $\tau$ .

Thus, $\lambda_\alpha=\frac{1}{1620}$ per year

and $\lambda_\beta=\frac{1}{405}$ per year

$\therefore \text { Total decay constant, } \lambda=\lambda_\alpha+\lambda_\beta$

or $\lambda=\frac{1}{1620}+\frac{1}{405}=\frac{1}{324}$ per year

We know that $\mathrm{N}=\mathrm{N}_0 \, \mathrm{e}^{-\lambda t}$

When $\frac{3}{4} \text { th$ part of the sample has disintegrated, $N=N_0 / 4$

$\therefore \quad \frac{\mathrm{N}_0}{4}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}$

or $e^{\lambda t}=4$

Taking logarithm of both sides, we get

$\lambda t=\log _e 4$

or $\mathrm{t}=\frac{1}{\lambda} \log _{\mathrm{e}} 2^2=\frac{2}{\lambda} \log _{\mathrm{e}} 2$

$=2 \times 324 \times 0.693=449 \text { year }$

Posted by

SANGALDEEP SINGH

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