A sample of radioactive material decays simultaneously by two processes A and B with half-lives `1/2 and 1/4` hours respectively . For first half-hour, it decays with process A, next one hour with process B and for a further half and hour with both A and B. If originally there were `N_0` nuclei, find the number of nuclei after 2 hours of such decay-

Options

A. `(N_(0))/((2)^(8))`
B. `(N_(0))/((2)^(4))`
C. `(N_(0))/((2)^(6))`
D. `(N_(0))/((2)^(5))`

Answer

A

Explanation

After first half hrs `N = N_(0) (1)/(2)`
for `t = (1)/(2)` to `t = 1 (1)/(2)`
`N = (N_(0) (1)/(2)) [(1)/(2)]^(4) = N_(0) ((1)/(2))^(5)`
For `t = 1 (1)/(2)` to `t = 2 hrs`
[for both A and B `(1)/(t_(1//2)) = (1)/(1//2)+(1)/(1//4)`
`= 2+4 = 6`
`t_(1//2) = 1//6` hrs]
`N'=[N_(0)((1)/(2))^(5)]((1)/(2))^(3)=N_(0)((1)/(2))^(2)`