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Find the enrgy of the reaction N^(14)(alpha,p)O^(17) if the kinetic enrgy of the incoming alpha-pariticle is T_(alpha)=40MeV and the proton outgoing at an angle theta=60^(@) to the motion direction of the alpha-particle has a kinetic energy T_(P)= 2.09 MeV. |
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Answer» Solution :The reaction is `N^(14)(alpha,p)O^(17)` It is given that (in the Lab frmae where `N^(14)` is at rest)`T_(alpha)=4.0MeV` The momentum of incident `alpha` particle is `sqrt(2m_(alpha)T_(alpha))hat(i)=sqrt(2eta_(alpha)m_(0)T_(alpha))hat(i)` The momentum of OUTGOING proton is `sqrt(2m_(p)T_(p)) (COS theta hat(i)+ sin theta hat(j))` `=sqrt(2eta_(p)m_(0))(cos theta hat(i)+sin theta hat(j))` where `eta_(p)=(m_(p))/(m_(0)), eta_(alpha)=(m_(alpha))/(m_(0))` , and `m_(0)` is the mass of `O^(17)`. Teh momentum of `O^(17)` is `(sqrt(2 eta_(alpha)m_(0)T_(alpha))-sqrt(2eta_(p)m_(0)T_(p)cos theta))hat(i)` `-sqrt(2m_(0)eta_(p)T_(p))sin theta hat(j)`  By energy conservation( conservation of enrgy incliding rest mass energy and kinetic energy) `M_(14)c^(2)+M_(alpha)c^(2)+T_(alpha)` `=M_(p)c^(2)+T_(p)+M_(17)c^(2)` `+[(sqrt(eta_(alphaT_(alpha)))-sqrt(eta_(p)T_(p)) cos theta)^(2)+eta_(p)T_(p)sin^(2) theta+eta_(p)T_(p) sin^(2) theta]` Hence by defination of the `Q` of reaction `Q=M_(14)c^(2)+M_(alpha)c^(2)-M_(p)c^(2)-M_(17)c^(2)` `=T_(p)+eta_(alpha)T_(alpha)+eta_(p)T_(p)-2sqrt(eta_(p)eta_(a)T_(alpha)T_(p))xx cos theta-T_(alpha)` `=(1+eta_(p))T_(p)+T_(alpha)(1-eta_(alpha))` `-2sqrt(eta_(p)eta_(alpha)T_(alpha)T_(p))cos theta= -1.19MeV` |
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