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

Options

A. `(2.303)/(0.693) xx 4.5 xx 10^(9) "log" 1.25`
B. `(2.303)/(0.693) xx 4.5 xx 10^(9) "log" 0.25`
C. `(2.303)/(0.693) xx 4.5 xx 10^(9) "log" 4`
D. `(2.303)/(0.693) xx 4.5 xx 10^(9) "log" 1.25`

Answer

A

Explanation

`(0.693)/(t_(1//2)(U^(238))) = (2.303)/(t_("age")) "log" ((N_(0))/(N))`
`(0.693)/(4.5 xx 10^(9)) = (2.303)/(t_("age")) "log" 1.25`
`:.t_("age") = (2.303 xx 4.5 xx 10^(9))/(0.693) "log" 1.25`