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

`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)