A radioactive sample of U^{238} decays to Pb | vedclass

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(B) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{1.5 	imes 10^9}{4.5 	imes 10^9} = \frac{1}{3}$.The number of remaining $

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$0.26$

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(B) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{1.5 	imes 10^9}{4.5 	imes 10^9} = \frac{1}{3}$.The number of remaining $U^{238}$ nuclei $(N_U)$ is given by $N_U = N_0 \left(\frac{1}{2}ight)^n = N_0 \left(\frac{1}{2}ight)^{1/3}$.Given $2^{1/3} = 1.26$,we have $N_U = \frac{N_0}{1.26}$.The number of $Pb$ nuclei formed is $N_{Pb} = N_0 - N_U = N_0 - \frac{N_0}{1.26} = N_0 \left(1 - \frac{1}{1.26}ight) = N_0 \left(\frac{0.26}{1.26}ight)$.The ratio of the number of nuclei of $Pb$ to $U^{238}$ is $\frac{N_{Pb}}{N_U} = \frac{N_0 (0.26 / 1.26)}{N_0 / 1.26} = 0.26$.

$A$ radioactive sample of $U^{238}$ decays to $Pb$ through a process for which the half-life is $4.5 \times 10^9 \text{ years}$. What is the ratio of the number of nuclei of $Pb$ to $U^{238}$ after a time of $1.5 \times 10^9 \text{ years}$? (Given $2^{1/3} = 1.26$)

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