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Protons accelerated by a potential diffe...

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.

Text Solution

Verified by Experts

The correct Answer is:
`12.1MeV;158.4^(@)`.
The nuclear reaction takes the form `""_(1)H^(1)+""_(3)Li^(7)=2""_(2)He^(4)`. The kinetic energy of the alpha-particle is found from the law of conservation of energy which for this reaction is of the form
`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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