In a radioactive sample,{ }_{19}^{40} K nucle | vedclass

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Question

(A) The total decay constant $\lambda$ is the sum of the individual decay constants for the two branches:$\lambda = \lambda_1 + \lambda_2 = 4.5 	imes

Vedclass · Accepted Answer

$9.2$

Vedclass · Answer

(A) The total decay constant $\lambda$ is the sum of the individual decay constants for the two branches:$\lambda = \lambda_1 + \lambda_2 = 4.5 	imes 10^{-10} + 0.5 	imes 10^{-10} = 5.0 	imes 10^{-10} 	ext{ per year}$.Let $N_0$ be the initial number of ${ }_{19}^{40} K$ nuclei and $N$ be the number of radioactive nuclei at time $t$.The number of stable nuclei produced is $N_s = N_0 - N$.According to the problem,the ratio of stable nuclei to radioactive nuclei is $99$:$\frac{N_0 - N}{N} = 99 \Rightarrow \frac{N_0}{N} - 1 = 99 \Rightarrow \frac{N_0}{N} = 100$.Using the radioactive decay law $N = N_0 e^{-\lambda t}$,we have $\frac{N}{N_0} = e^{-\lambda t} = \frac{1}{100} = 10^{-2}$.Taking the natural logarithm on both sides:$-\lambda t = \ln(10^{-2}) = -2 \ln 10$.Given $\ln 10 = 2.3$,we get $\lambda t = 2 	imes 2.3 = 4.6$.Substituting $\lambda = 5 	imes 10^{-10} 	ext{ per year}$:$(5 	imes 10^{-10}) 	imes t = 4.6 \Rightarrow t = \frac{4.6}{5} 	imes 10^{10} = 0.92 	imes 10^{10} = 9.2 	imes 10^9 	ext{ years}$.Thus,the value of $t$ is $9.2$.

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