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A `^7Li` target is bombarded with a proton beam current of `10^-4` A for 1 hour to produce `^Be` of activity `1.8xx10^8` disintegrations per second. Assuming that `^7Be` radioactive nucleus is produced by bombarding 1000 protons, determine its half-life. |
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Answer» 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 |
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