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A gold foil of mass m= 0.20 g was irradiated during t=6.0 hours by a thermal neutron flux falling normally on its suface. Following tau=12 hours after the completion of irradiation the activity of the foil became equal to A= 1.9.10^(7)dis//s. Find the neutron flux density if the effective cross-section of formation of a radioactive nucleus is sigma= 96b, and the half-life is equal to T=2.7 days |
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Answer» Solution :We apply the formula of the PREVIOUS problem except that have `N=no`. Of radio nuclide and no. of host nuclei originally. Here `n=(0.2)/(197)xx6.023xx10^(23)=6.115xx10^(20)` Then after time `t N=(n.J.sigma.T)/(In 2)(1-e^(-(tIn 2)/(T)))` T= half life of the radionuclide After the SOURCE of neutrons is cut off the ACITIVITY after time `T` will be `A=(n.J.sigma.T)/(In 2)(1-e^(-tIn2//T))e^(tauIn2//Txx(In 2)/(T))=n.J.sigma(1-e^(-tIn2//T))e^(-tauIn2//T)` Thus `J=Ae^(TAU In 2//T)//n sigma(1-e^(-tIn 2//T))= 5.92xx10^(9)part//cm^(2).s` |
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