Question
A proton is bombarded on a stationary Lithium nucleus. As a result of collision two $$\alpha $$-particles are produced. The direction of motion of the $$\alpha $$-particles with the initial direction of motion makes an angle $${\cos ^{ - 1}}\frac{1}{4}.$$ If B.E/Nucleon for $$L{i^7}$$ and $$H{e^4}$$ are $$5.60\,MeV$$ and $$7.06\,MeV$$ respectively, then :
A.
Kinetic energy of striking proton is $$17.28\,MeV$$
B.
Kinetic energy of striking proton is $$8.64\,MeV$$
C.
Kinetic energy of striking proton is $$4.32\,MeV$$
D.
Kinetic energy of striking proton is $$2.16\,MeV$$
Answer :
Kinetic energy of striking proton is $$17.28\,MeV$$
Solution :
$$Q$$ value of the reaction
$$\eqalign{
& Q = \left( {2 \times 4 \times 7.06 - 7 \times 5.6} \right) = 17.28\,MeV\,.......\left( {\text{i}} \right) \cr
& {K_P} + Q = 2{K_\alpha }\,.......\left( {{\text{ii}}} \right) \cr
& \sqrt {2{m_P}{K_P}} = 2\sqrt {2{m_\alpha }{K_\alpha }} \cos \alpha \cr
& {K_P} = {K_\alpha }\,.......\left( {{\text{iii}}} \right) \cr
& {\text{So}}\,\,{K_P} = 17.28\,MeV. \cr} $$