By successive decay `""_(92)^(238)U` changed to `""_(82)Pb^(206)`, when a sample of uranium ore was analysed. It was found that it contains 1g of `U^(238)` and 0.1 of `Pb^(206`, considering that all the `Pb^(206)` had accumulatd due to the decay of `U^(238)`
Calculate the age of the ore. (Half life of `U^(238)=4.5xx10^(9)yrs)`

Options

A. `0.1155xx10^(8)` yrs
B. `7.099xx10^(8)` yrs
C. `0.154xx10^(-9)` yrs
D. `7.099xx10^(10)` yrs

Answer

B

Explanation

Considering that the whole of `Pb^(206)` comes from `U^(238)` amount of `U^(238)` decayed for =0.1 g `Pb^(206)` `=(0.1xx238)/206g U^(238)`
`=0.1155gU^(238)`
Hence initial amount of `U^(238)=(1+0.1155)g U^(238)`. Thus value of disintegration constant can be calculated as:
Now `lamda=0.693/(T_(1//2))=0.693/(4.5xx10^(9)yrs)`
`=0.154xx10^(-9)yrs^(-1)`
`t=2.303/(lamda)"log"(N_(0))/N`
`t=2.303/(0.154xx10^(-9))log(1.1135/1)`
`=7.099xx10^(8)`yrs