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A thin foil of certain stable isotope is irradiated by thermal neutrons falling normally on its surface. Due to the capature of neutrons a radioonuclide with decay constant lambda appears. Find the law describing, accumulation of that radionuclide N(t) per unit area of the foil's surface. The neutron flux density is J, the number of nuclei per unit area of the foil's surface is n, and effective cross-section of formation of active nuclei is sigma. |
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Answer» SOLUTION :Rate of FORMATION of the radionuclide is `n.J.sigma` PER unit area per.sec. Rate of DECAY is `lambda N`. Thus`(dN)/(dt)=n.J.sigma-lambda N` per unit area per second Then `((dN)/(dt)+lambdaN)E^(lambda t)=n.J.sigma^(lambda t)` or `(d)/(dt)(Ne^(lambda t))=n.J.sigma.e^(lambda t)` Hence `Ne^(lambda t)= Const+(n.J.sigma)/(lambda)e^(lambda t)` The number of radionuclide at `t=0` whenthe process starts is zero. So constant`=-(n.J.sigma)/(lambda)` Then `N=(n.J.sigma)/(lambda)(1-e^(-lambda t))` |
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