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Deuteron is a bound state of a neutron and a proton with a binding energy `B=2.2MeV. A gamma`-ray of energy E is aimed at a deuteron nucleus to try to breack it into a (neutron +proton) such that the n and p move in the direction of the incident `gamma`-rays. If E=B, show that this can not happen. Hence, calculate how much bigger than B must E be for such a process to happen. |
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Answer» Applying the principal of conservation of energy in the given situation, `E-B=K_(n)+K_(p)=(p_(n)^(2))/(2m)+(p_(p)^(2))/(2m).........(i)` for the principal of conservation of linear momentum, `p_(n)+p_(p)=E//c.......(ii)` If `E=B`, eqn. (i) would give `p_(n)=p_(p)=0`. Hence, the process cannot take place. for the process to take place, let `E=B+lambda`, where `lambda lt lt B`. form (i), `lambda=(p_(n)^(2))/(2m)+(p_(p)^(2))/(2m)` using (ii), `lambda=1/(2m) [p_(p)^(2)+(p_(p)-E//c)^(2)]` `:. 2p_(p)^(2)-(2E)/c p_(p)+((E^(2))/(c^(2))-2mlambda)=0 or p_(p)=(2E//c+-sqrt(4E^2//c^2-8((E^(2))/(c^(2))-2mlambda)))/4` for `p_(p)` to be real, the determinant on RHS must be positive. `:. (4E^2)/c^2-8((E^(2))/(c^(2))-2mlambda)ge0 or 16mlambda=(4E^2)/(c^2)` `lambda=(E^(2))/(4mc^(2))~=(B^(2))/(4mc^(2))` Hence, E must be bigger than B by `lambda=B^(2)//4mc^(2)`, for the given process to happen. |
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