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Protons accelerated by a potential difference of 6.8 MV bombard a stationary lithium target. The collision of a proton with a nucleus of Li^(7) isotope results in the birth of two alpha-particles which separate symmetrically with respect to the direction of the proton beam. Find the kinetic energy and the separation angle of the alpha-particles. |
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Answer» `2K_(alpha)=(M_(H)+M_(Li)-2M_(alpha))xx931.5+K_(H)=24.2MeV` SInce the kinetic energy of the alpha-particle TURNS out to be MUCH lower than its rest energy, the alpha-particles PRODUCED in the reaction are nonrelativistic. The separation angle is found from the law of conservation of momentum `p_(H)=2p_(alpha)"cos"theta/2` But for a nonrelativistic particle `p=sqrt(2mK)`, so `"cos"theta/2=1/2sqrt((m_(H)K_(H))/(m_(alphaK_(alpha))))`
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