In a smaple of rock, the ration `.^(206)Pb` to `.^(238)U` nulei is found to be `0.5.` The age of the rock is (given half-life of `U^(238)` is `4.5 xx 10^(9)` years).A. `2.25 xx 10^(9)year`B. `4.5 x 10^(9)` ln 3 yearC. `4.5 x 10^(9) (ln(3)/(2))/(ln2) year`D. `2.25 xx 10^(9) ln((3)/(2))` year

In a smaple of rock, the ration `.^(206)Pb` to `.^(238)U` nulei is fou

1.	In a smaple of rock, the ration `.^(206)Pb` to `.^(238)U` nulei is found to be `0.5.` The age of the rock is (given half-life of `U^(238)` is `4.5 xx 10^(9)` years).A. `2.25 xx 10^(9)year`B. `4.5 x 10^(9)` ln 3 yearC. `4.5 x 10^(9) (ln(3)/(2))/(ln2) year`D. `2.25 xx 10^(9) ln((3)/(2))` year
Answer» Correct Answer - c Suppose an initial radionuclide I decays to a final product F with a half-life `T_(1//2)` At any time, `N_(1) =N_(0) e^(- lambda `t`)` Number of product nuclei `=N_(F) =N_(0) -N_1` `(N_(F))/(N_(I))=(N_(0)- N_(I))/(N_I)=((N_(0))/(N_(I) -1))` `(N_(0))/(N_(I))=(1 +(N_(F))/(N_(I)))=1 +0.5=1.5` `e^(- lambda t)=1.5 rArr lambda t=1n 1.5` `:. (T_(1//2) In(1.5))/(1n 2) =4.5 xx 10^(9) (In ((3)/(2)))/(In2) year`.

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In a smaple of rock, the ration `.^(206)Pb` to `.^(238)U` nulei is found to be `0.5.` The age of the rock is (given half-life of `U^(238)` is `4.5 xx 10^(9)` years).A. `2.25 xx 10^(9)year`B. `4.5 x 10^(9)` ln 3 yearC. `4.5 x 10^(9) (ln(3)/(2))/(ln2) year`D. `2.25 xx 10^(9) ln((3)/(2))` year

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