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(A) Let the initial amount of $U-238$ be $N_0$ atoms. At time $t$,let $N_t$ be the number of $U-238$ atoms remaining and $N_{Pb}$ be the number of $Pb

Vedclass · Accepted Answer

$143$

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(A) Let the initial amount of $U-238$ be $N_0$ atoms. At time $t$,let $N_t$ be the number of $U-238$ atoms remaining and $N_{Pb}$ be the number of $Pb-206$ atoms formed.Since $1$ atom of $U-238$ produces $1$ atom of $Pb-206$,$N_0 = N_t + N_{Pb}$.The mass ratio is given as $\frac{m_{Pb}}{m_U} = 7$. Since $m = \frac{N 	imes M}{N_A}$,we have $\frac{N_{Pb} 	imes 206}{N_t 	imes 238} = 7$.Thus,$N_{Pb} = 7 	imes N_t 	imes \frac{238}{206} = N_t 	imes \frac{1666}{206} \approx 8.087 N_t$.Then $N_0 = N_t + 8.087 N_t = 9.087 N_t$.Using the radioactive decay law: $N_t = N_0 e^{-\lambda t}$,so $\frac{N_0}{N_t} = e^{\lambda t}$.$\lambda t = \ln(9.087) \approx 2.2068$.Given $T_{1/2} = 4.5 	imes 10^9$ years,$\lambda = \frac{\ln 2}{4.5 	imes 10^9} \approx \frac{0.693}{4.5 	imes 10^9}$.$t = \frac{2.2068 	imes 4.5 	imes 10^9}{0.693} \approx 14.33 	imes 10^9 = 143.3 	imes 10^8$ years.Thus,$P \approx 143$.

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