A sample of radioactive material decays simultaneouly by two processes A and B with half-lives `(1)/(2)` and `(1)/(4)h`, respectively. For the first half hour it decays with the process A, next one hour with the proecess B, and for further half an hour with both A and B. If, origianlly, there were `N_0` nuceli, find the number of nuclei after 2 h 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

`T_(A) = 1/2hr, T_(B) = 1/4 hr, T_(A + B) = ((1)/(2) xx (1)/(4))/((1)/(2) + (1)/(4)) = 1/6 hr`
so, first `1/2 hr = 1` half live (by A)
next `1` hr = `4` half lives (by B)
thus `N = (N_(0))/(2^(R))` (`:.` Total light eight half lives)