Question

Two radioactive sources X and Y have half-lives of 3.0 hours and 2.0 hours, respectively. The product of the decay is a stable isotope of the element Z. Six hours ago a mixture contained the same number of atoms of both X and Y, and no other atoms.
What fraction of the mixture is now made up of atoms of Z?
(the answer is 12/16)

Answer 1

So the issue is having X,Y equal to make Z. The amount of X after t hours is
A is original amount) X(t)=Ae^-kt/3
and amount of Y after t hours is
Y(t)=Ae^-kt/2

The slower of the two is X changing, so the number of Z= X(t)
Fraction unchanged = Z/(total excess Y+unchanged atoms)
total excess Y= Y(t)-X(t)
unchanged X= X0-X(t) = A-X(t)
unchanged y=A-y(t)

fraction mixture unchanged= Z/(total excess Y+unchanged atoms)
= X(t)/ (Y(t)-X(t)+A-X(t)+A-Y(t) )
= X(t)/ ( -2X(t)+2A )=1/2 (1/(1-e-kt/3)
= 1/2 (1-1/2)=1/4
fraction changed (ie, now Z)= 1-fraction unchanged= 3/4 or 12/16 if you prefer that.