Question

A sealed box was found which stated to have contained alloy composed of equal parts by weight of two metals `A` and `B`. These metals are radioactive, with half lives of `12` years and `18` years. Respectively and when the container was opened it was found to contain `0.53 kg` of `A` and `2.20 kg` of `B`. The age of the alloy is `Mxx10+n` then find `M-n`

Answer 1

Correct Answer - `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`