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Find the minimum kinetic energy of an alpha - particle to cause the reaction ""^(14)N(alpha.p)""^(17)O. The masses of ""^(14)N,""^(4)He,""^(1)H and ""^(17)O are respectively 14.000307u, 4.00262u,1.00783u, and 16.11913u. |
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Answer» Solution :Since, the masses are given in ATOMIC mass UNIT , it is easiest to proceed by finding the mass DIFFERENCE between REACTANTS and products in the same unit and then MULTIPLYING by 931. 5 Me V/u Thus , we have `Q=(14.00307u + 4.00260u - 1.00783 u - 16.99913u)` `(931.5 (MeV)/u) =-1.20 MeV` Q value is negative ,it means reaction is endothermic So , the minimum kinetic energy of `alpha` - particle to initiate this reaction would be `K_(min)=|Q|((m_(alpha))/(m_N)+1)=(1.20)((4.00260)/(14.00307)+1)=1.54 MeV` |
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