A radio isotope `X` with a half-life `1.4 xx 10^9` years decays of `Y` which is stable. A sample of the rock from a cave was found to contain `X` and `Y` in the ratio `1 : 7`. The age of the rock is.

Options

A. `8.40 xx 10 ^(9) `years
B. `1.96 xx 10 ^(9)` years
C. `3.92 xx 10 ^(9)` years
D. `4.20 xx 10 ^(9)` years

Answer

D

Explanation

`{:(, X , to, Y),("No of nuclei" ,, N_(0), 0),("No. of nuclei at time t",, N_(0)-x, x):}`
As per question `(N _(0) -x)/(x) =1/7`
`7N_(0) - 7x =x or x = 7/8 N_(0)`
`therefore` Remaining nuclei of istope X
`= N_(0) - x = N _(0) - 7/8 N_(0) = 1/8 N_(0) = ((1)/(2))^(3) N _(0)`
So, three half lives would have been passed
`therefore t = n T _(1//2) = 3 xx 1.4 xx 10 ^(9) ` years
`= 4.2 xx 10 ^(9)` years