Question

`Ac^(227)` has a half`-` life of `22.0` years with respect to one leading . To `Th^(227)` and the other to `Fr^(227)`. The percentage yields of these two daughter nuclides are `2.0` and `98.0`, respectively . What are the decay constants `(lambda)` for each of the separate paths ?

Answer 1

For the radioactive decay of `A` into `B` and `C` by two parallel paths
So, `(-d[A])/(dt)=lambdaN ....(i)`
`(+d[B])/(dt)=lambda_(1)N .....(ii)`
`(+d[C])/(dt)=lambda_(2)N ....(iii)`
where `lambda,lambda_(1) , `and `lambda_(2)` are radioactive decay constants, respectively, and `N` is the number of atoms of `A` at any given time.
Thus, `(+d[A])/(dt)=(d[B])/(dt)+(d[C])/(dt)`
`:. lambdaN=lambda_(1)N+lambda_(2)N`
`:. lambda=lambda_(1)+lambda_(2)N`
From eqs. `(ii)` and `(iii)`, we get
`(d[B])/(d[C])=(lambda_(1))/(lambda_(2))`
On integration ,we get
`([B])/([C])=(lambda_(1))/(lambda_(2))`
For decay of `Ac^(227)` into `Th^(227)` and `Fr^(223)` , on the basis of given data
`(lambda_(1))/(lambda_(2))=(2.0)/(98.0) ....(iv)`
and `lambda=(0.693)/(t_(1//2))=(0.693)/(22)=0.0315year^(-1)` ltbr. So,` 0.0315=lambda_(1)+lambda_(2) ...(v)`
On solving Eqs. `(iv)` and `(v)`, we get
`lambda_(1)=6.3xx10^(-4)year^(-1)`
and `lambda_(2)=0.03087year^(-1)`