`""^(227)` Ac has a half-life of 22.0 years with respect to radioactive decay. The decay follows two parallel paths, one leading to `""^(222)Th` and the other to `""^(223)Fr`. The percentage yields to these two daughter nuclides are 2.0 and 98.0 respectively. What are the decay constants `(lambda)` for each of the separate paths ?

Options

A. `k_(1) = 0.00063 y^(-1), k_(2) = 0.03087 y^(-1)`
B. `k_(1) = 0.0063 y^(-1), k_(2) = 0.003087 y^(-1)`
C. `k_(1) = 0.063 y^(-1), k_(2) = 0.04087 y^(-1)`
D. `k_(1) =0.00063 y^(-1) , k_(2) = 0.003087 y^(-1)`

Answer

A

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 reaction 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_(1) = 0.00063 y^(-1), k_(2) = 0.03087 y^(-1)`