The percentage of {}^{235}U presently on Eart | vedclass

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(D) Let $N_1$ and $N_2$ be the present amounts of ${}^{235}U$ and ${}^{238}U$,and $N_{01}$ and $N_{02}$ be their initial amounts.Given: $\frac{N_1}{N_

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$7 	imes 10^9$ years

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(D) Let $N_1$ and $N_2$ be the present amounts of ${}^{235}U$ and ${}^{238}U$,and $N_{01}$ and $N_{02}$ be their initial amounts.Given: $\frac{N_1}{N_2} = \frac{0.72}{99.28}$ and $\frac{N_{01}}{N_{02}} = 2.0$.The decay law is $N = N_0 e^{-\lambda t}$,where $\lambda = \frac{\ln 2}{T_{1/2}}$.Thus,$\frac{N_1}{N_2} = \frac{N_{01} e^{-\lambda_1 t}}{N_{02} e^{-\lambda_2 t}} = \frac{N_{01}}{N_{02}} e^{-(\lambda_1 - \lambda_2)t}$.Substituting the values: $\frac{0.72}{99.28} = 2.0 	imes e^{-(\lambda_1 - \lambda_2)t}$.$e^{(\lambda_1 - \lambda_2)t} = 2.0 	imes \frac{99.28}{0.72} \approx 275.78$.Taking natural log on both sides: $(\lambda_1 - \lambda_2)t = \ln(275.78) \approx 5.62$.Using $\lambda = \frac{\ln 2}{T_{1/2}}$: $(\frac{\ln 2}{7.1 	imes 10^8} - \frac{\ln 2}{4.5 	imes 10^9})t = 5.62$.$(\ln 2) 	imes (1.408 	imes 10^{-9} - 0.222 	imes 10^{-9})t = 5.62$.$0.693 	imes (1.186 	imes 10^{-9})t = 5.62$.$t = \frac{5.62}{8.219 	imes 10^{-10}} \approx 6.84 	imes 10^9 \approx 7 	imes 10^9$ years.

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