A radioactive substance `"A"` having `N_(0)` active nuclei at `t=0`, decays to another radioactive substance `"B"` with decay constant `lambda_(1)`. `B` further decays to a stable substance `"C"` with decay constant `lambda_(2)`. (a) Find the number of nuclei of `A, B` and `C` time `t`. (b) What should be the answer of part (a) if `lambda_(1) gt gt lambda_(2)` and `lambda_(1) lt lt lambda_(2)`

Explanation

The decay scheme is as shown
`A rarr_(lambda_1)(rarr)Brarr_(lambda_2)C` (stable)
`{:(t = 0, N_0,0),(t,N_1,N_2,N_3):}`
Here `N_1, N_2 and N_3` represent the nuclei of A, B and C at any time t,
For A, we can write
`N_(1)= N_(0)e^(-lambda_1t)" "(1)`
For B, we can write
`(dN_2)/(dt) = lambda_1N_1 = lambda_1 - lambda_2N_2" "(2)`
or, `(dN_2)/(dt) + lambda_2 N_2 - lambda_1N_1`
This is lenear differential equation with integrating factor
`I.F. = e^(lambda)2^(t)`
`e^(lambda_2 t)(dN_2)/(dt) +e^(lambda_2 t)lambda_2N_2 = lambda_1 N_1 e^(lambda_2 t)`
`int d(N_2e^(lambda_2 t)) = int lambda_1N_1e^(lambda_2 t)dt`
`N_2 e^(lambda_2 t) = lambda_1N_0int e^(-lambda_1 t)e^(lambda_2 t)dt" "`.......using (1)
`N_2 e^(lambda_2 t) = lambda_1N_0(e^((lambda_2-lambda_1) t))/(lambda_2-lambda_1) + C" "......(3)`
At `t = 0, N_2 = 0 " "0 = (lambda_1N_0)/(lambda_2 - lambda_1)+C`
Hence `C = (lambda_1N_0)/(lambda_1 - lambda_2)`
Using C in eqn. (3), we get
`N_2 = (lambda_1N_0)/(lambda_2-lambda_1)(e^(-lambda_- t) - e^(-lambda_2 t))`
and `N_1 + N_2 + N_3 = N_0`
`:. N_3 = N_0 - (N_1 + N_2)`