In the uranium ore, the ratio of `U^(238)` nuclei to `Pb^(206)` nuclei is `2.8`. If it is assumed that all the lead `Pb^(206)` to be a final decay product of the uranuium series, the age of the ore is `[T_(1//2)` for `U^(238)` is `4.5xx10^(9)` years]

Options

A. `4.5xx10^(9)` years
B. `2.0xx10^(9)` years
C. `3.2xx10^(9)` years
D. `6.4xx10^(9)` years

Answer

B

Explanation

`U^(238)rarrPb^(206)`
(radio nuclide) (stable nuclide)
number of `Pb^(206)` nuclei=number of decayed uranium nuclei=`N_(0)(1-e^(-lambda t))` In present sample
`(N(U^(238)))/(N(Pb^(206)))=2.8 rArr(N_(0)e^(-lambda t))/(N_(0)(1-e^(-lambda t)))=(28)/(10)`
solve for time, t.