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How much in per cent, does the threshold energy of gamma quantum exceed the bindign energy of a deuteron (E_(b)=2.2 MeV) in the reaction gamma+H^(2) rarrn+p? |
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Answer» <P> Solution :The formula of problem 6.271 does not APPLY here because the photon is always reletivistic. At threshold, the energy of the photon `E_(gamma)` implies a momentum `(E_(gamma))/(c )`. The velocity of center of mass with respect to the rest frame of initial `H^(2)` is`(E_(gamma))/((m_(n)+m_(p))c)` Since both `n & P` are at rest in `CM` frame at threshold, we write `E_(gamma)=(E_(gamma)^(2))/(2(m_(n)+m_(p))c^(2))+E_(B)` by conservation of energy. Since the first term is a small CORRECTION, we have `E_(gamma)~=E_(b)+(E_(b)^(2))/(2(m_(n)+m_(p))c^(2))` Thus `(deltaE)/(E_(b))=(E_(b))/(2(m_(n)+m_(p))c^(2))=(2.2)/(2xx2xx938)= 5.9xx10^(-4)` or nearly `0.06%` |
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