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A radio isotope X with a half life 1.4xx...

A radio isotope `X` with a half life `1.4xx10^(9)` yr 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

Text Solution

Verified by Experts

The correct Answer is:
`3`
Use the law of radioactivity
`N_(A)=N_(A)^(0)e^((-lambda_(A)t))`
`N_(B)=N_(B)^(0)e^((-lambda_(e )t))`
Divide `(2)` by
`(N_(B))/(N_(A))=e^((lambda_(A)-lambda_(B))t` (Because `N_(A)^(0)=N_(B)^(0))`
`t=(1)/((lambda_(A)-lambda_(B))ln((N_(B))/(N_(A))))`
Given `(N_(B))/(N_(A))=(2.2)/(0.53)=4.15`
`lambda_(A)=(0.693)/(T_(1//2))(A)=(0.693)/(12)=0.05775 "years"^(-1)`
`lambda_(B)=(0.693)/(T_(1//2))(B)=(0.693)/(18)=0.0985"years"^(-1)`
we find age of alloy `t=74 "years"`
`7xx10+4`
`M=7`
`n=4`
`M-n=3`
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