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 2.20Kg of B. The age of the alloy is M x 10 + n then find M-n.

Answer 1

let $N_0$ be the intial amount of each metal and $Y$ be its age.

the number of half lives is $\frac{Y}{12}$and$\frac{Y}{18}$ for metals A and B

also $N=\frac{N_0}{2^n}$

where n = number of half lives & N= present amount

$0.53=\frac{N_0}{2^{\frac{Y}{12}}}$ $2.20=\frac{N_0}{2^{\frac{Y}{18}}}$

$⟹2.20=0.53\frac{2^{\frac{Y}{12}}}{2^{\frac{Y}{18}}}$

$4=2^{Y[\frac{1}{12}-\frac{1}{18}]}$ = $2^{\frac{Y}{36}}$

$Y$=72years

$72=7\times 10+2⟹m\times 10+n⟹m=7,n=2$

$\therefore m-n=5$