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Thermal neutrons fall normally on the surface of a thin gold foil consisting of stable Au^(197) nuclide. The neutron flux density is J=1.0.10^(10) "part"//(1s,cm^(2)). The mass of the foil is m= 10 mg. Then neuteron capture produces beta-active Au^(198) nuclei with half-life T= 2.7 days. The effective capture cross-section is sigma= 98 b. Find : (a) the irradiation time after which the number of Au^(197) nuclei decreases by eta= 1.0%, (b) the maximum number of Au^(198) nuclei that can be formed during protracted irradiation. |
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Answer» Solution :(a) Suppose `N_(0)=` no. of `Au^(197)` nuclei in the foil. Then the number of `Au^(197)` nuclei transformed in time `t` is `N_(0).J.sigma.t` For this to equal `etaN_(0)`, we must have `t= eta//(J. sigma)= 323 YEARS` (b) Rate of formation of the `Au^(198)` nuclei is `N_(0).J.sigma` per sec and rate of decay is `lambda n,` where `n` is the number of `Au^(198)` at any instant. Thus `(dn)/(dt)=n_(0).J.sigma-lambda n` The maximum number of `Au^(198)` is clearly `n_(max)=(N_(0).J.sigma)/(lambda)=(N_(0).J.sigma.T)/(In 2)` beacuse if `n` is smaller, `(dn)/(dt) gt 0` and `n` will increase further and if `n` is larger `(dn)/(dt) lt 0` and `n` decrease. (ACTUALLY `n_(max)` is approached steadily as `trarr oo`) Substitution gives using `N_(0)= 3.05xx10^(19), n_(max)= 1.01xx10^(13)` |
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