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

After First half hours,
`N=N_(0)(1)/(2)`
For`t=(1)/(2)h" to "t=1(1)/(2),1h` = four half-lives
Hence, `N=(N_(0)(1)/(2))[(1)/(2)]^(2)=N_(0)((1)/(2))^(5)`
For `t=(1)/(2)" to "t=2h`
`[" for both" `A` and `B`, `(1)/(t_(1//2))=(1)/(t_(1//2)) + (1)/(t_(1)//4)=2 +4 =6 rArr t_(1//2) =(1)/(6)]`
`(1)/(2) h=` half -lives
`:. N=[(N_(0)(1)/(2))^(5)](1)/(2^(3))=N_(0)(1)/(2^(8))`.