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(A) The decay constants for the two processes are given by $\lambda_1 = \frac{0.693}{T_1}$ and $\lambda_2 = \frac{0.693}{T_2}$.Given $T_1 = 1620 	ext

Vedclass · Accepted Answer

$1080$

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(A) The decay constants for the two processes are given by $\lambda_1 = \frac{0.693}{T_1}$ and $\lambda_2 = \frac{0.693}{T_2}$.Given $T_1 = 1620 	ext{ years}$ and $T_2 = 810 	ext{ years}$.The effective decay constant is $\lambda_{	ext{net}} = \lambda_1 + \lambda_2$.$\lambda_{	ext{net}} = 0.693 \left( \frac{1}{1620} + \frac{1}{810} ight) = 0.693 \left( \frac{1 + 2}{1620} ight) = 0.693 \left( \frac{3}{1620} ight) = \frac{0.693}{540} 	ext{ year}^{-1}$.The effective half-life is $T_{	ext{net}} = \frac{0.693}{\lambda_{	ext{net}}} = 540 	ext{ years}$.For the substance to be reduced to one-fourth of its initial amount,the time $t$ required is given by $N = N_0 (1/2)^n$,where $n$ is the number of half-lives.Since $1/4 = (1/2)^2$,we have $n = 2$.Therefore,$t = n 	imes T_{	ext{net}} = 2 	imes 540 = 1080 	ext{ years}$.

$A$ radioactive substance emits two particles with half-lives of $1620$ years and $810$ years respectively. After how much time will one-fourth of the substance remain?

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