The analysis of a rock shows that relative no. of `""^(206)Pb` and `""^(238)U` atoms is Pb/U =0.25. If `t_(0.5) ""^(238)U rarr ""^(206)Pb` is `4 xx 10^(9)` years, the age of the rock is

Options

A. `(2.303)/(0.693)xx(4xx10^(9)) "log"1.25`
B. `(2.303)/(0.693)xx(4xx10^(9))"log"0.25`
C. `(2.303)/(0.693)xx(4xx10^(9))"log" 4`
D. `(2.303)/(4xx10^(9))xx0.693 xx "log" 4`

Answer

A

Explanation

`("Number of atoms Pb")/("Number of atoms of U")=0.25`
`=(25)/(100)=(1)/(4)`
Therefore, the rock contains one atom of Pb for every 4 atom U.
`implies N_(0)1+4=5 " and " N=4`
`t_(0.5)=4xx10^(9)` years
`lambda =(0.693)/(t_(0.5))=(0.693)/(4xx10^(9)) Yr^(-1)`
`t=(2.303)/(lambda) log (N_(0))/(N)`
`=(2.303xx4xx10^(9))/(0.693) log (5)/(4)`
`t=(2.033xx4xx10^(9))/(0.693)=log 1.25`