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A rock is 1.5 xx 10^(9) years old. The r...

A rock is `1.5 xx 10^(9)` years old. The rock contains `.^(238)U` which disintegretes to form `.^(236)U`. Assume that there was no `.^(206)Pb` in the rock initially and it is the only stable product fromed by the decay. Calculate the ratio of number of nuclei of `.^(238)U` to that of `.^(206)Pb` in the rock. Half-life of `.^(238)U` is `4.5 xx 10^(9). years. `(2^(1//3)=1.259)` .

A

2.85

B

4.85

C

0.85

D

3.85

Text Solution

Verified by Experts

The correct Answer is:
D
Number of half lives, `n = (1.5 xx 10^(9))/(4.5 xx 10^(9)) = (1)/(3)`
Say initial number of active `.^(238)U` nuclei were `N_(0)`.
Number of nuclei fo `.^(238)U` left after `1.5 xx 10^(9)` years,
`N = N_(0) ((1)/(2))^(1//3)`
Number of Pb-atoms formed in `1.5 xx 10^(9)` years,
`N' = N_(0) - N = N_(0) ((1)/(2))^(1//3)`
Ratio of number of nuclei of `.^(238)U` to that of `.^(206)Pb`.
`(N)/(N') = (N_(0) ((1)/(2))^(1//3))/(N_(0) - N_(0) ((1)/(2))^(1//3)) = (1)/(2^(1//3) - 1) = (1)/(1.259 - 1) = 3.85`
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