Correct Answer - A::B::D
At time t, let say there are N atoms of `^7Be` (radioactive). Then, net rate of formation of `^7Be` nuclei at this instant is
`(dN)/(dt)=(10^-4)/(1.6xx10^-19xx1000)-lambdaN`
or `(dN)/(dt)=6.25xx10^11-lambdaN`
or `int_0^(N_0)(dN)/(6.25xx10^11-lambdaN)=int_0^3600dt`
where, `N_0` are the number of nuclei at `t=1h` or `3600 s`.
`:.` `-1/lambda1n((6.25xx10^11-lambdaN_0)/(6.25xx10^11))=3600`
`lambdaN_0`=activity of `^7Be` at `t=1h=1.8xx10^8` disintegrations/s
`:.` -1/lambda1n((6.25xx10^11-1.8xx10^8`)/(6.25xx10^11))=3600`
`:.` `lambda=8.0xx10^-8sec^-1`
Therefore, half-life `t_(1//2)=(0.693)/(8.0xx10^-8)=8.66xx10^6s`
`=100.26` days