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. `1.96xx10^(9)` years
B. `3.92xx10^(9)` years
C. `4.20xx10^(9)` years
D. `8.40xx10^(9)` years

Explanation

`{:(,X,rarr,Y),("Number of nuclei at = 0",N_(0),,0),("Number 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 isotope X"`
`=N_(0)-x=N_(0)-(7)/(8)N_(0)=(1)/(8)N_(0)=((1)/(2))^(3)N_(0)`
`=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=nT_(1/2)=3xx1.4xx10^(9)" years"=4.2xx10^(9)" years"`
Hence, the age of hte rock is `4.2xx10^(9)" years"`