`""^(227)Ac` has a half-life of 22 years with respect to radioactive decay. The decay follows two prallel paths : `""^(227)Ac to ""^(227)Th " and "^(227)Ac to ""^(223)Fr.` If the percentage of the two daughter nuclides are 2.0 and 98.0, respectively, the decay constant (in `"year"^(-1)`) for `""^(227)Ac to ""^(227)Th` path is closest to

Explanation

The rate constant of the decay is
`k=(0.693)/t_(1//2)=(0.693)/22`
If `k_(1) and k_(2)` are the rate constant of the reactions leading to `""^(222)Th and ""^(223)Fr` , respectively we have
`k_(1)+k_(2)=(0.693)/22,k_(1)/k_(2)=2/98`
On solving for `k_(1) and k_(2)`, we get
`k_(2) = 0.03087 y^(–1)," " k_(1) = 0.00063 y^(–1)`