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Uranium .(92)U^(238) decayed to .(82)Pb^...

Uranium `._(92)U^(238)` decayed to `._(82)Pb^(206)`. They decay process is `._(92)U^(238) underset((x alpha, y beta))(rarr ._(82)Pb^(206))`
`t_(1//2)` of `U^(238) = 4.5 xx 10^(9)` years
The analysis of a rock shows the relative number of `U^(238)` and `Pb^(206)` atoms `(Pb//U = 0.25)` The age of rock will be

A

`(2.303)/(0.693) (4.5 xx 10^(9)) log ((5)/(4))`

B

`(2.303)/(0.693) (4.5 xx 10^(9)) log ((1)/(4))`

C

`(2.303)/(0.693) (4.5 xx 10^(9)) log (4)`

D

`(2.303)/(0.639) (4.5 xx 10^(9)) log ((4)/(5))`

Text Solution

Verified by Experts

The correct Answer is:
a

`(Pb)/(U) = 0.25, :. 1 + (Pb)/(U) = 1.25`
`(U + Pb)/(U) = 1.25`
`(N_(0))/(N) = 1.25 = (5)/(4)`
`(0.693)/(t_(1//2)) = (2.303)/(t) log ((N_(0))/(N))`
`(0.693)/(4.5 xx 10^(9)) = (2.303)/(t_("age")) log ((3)/(4))`
`r_("age") = (2.303)/(0.693) (4.5 xx 10^(9)) log ((5)/(4))`
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