A proton is bombarded on a stationary lithium nucleus. As a result of the collision two α -particles are produced. If the direction of motion of the α -particles with the initial direction of motion makes an angle cos - 1 1 4 , then the kinetic energy of the striking proton is [Given, binding energies per nucleon of Li 7 = 5 . 60 MeV and He 4 = 7 . 60 MeV , m proton ≈ m neutron ]

Step-by-step solution

Q value of the reaction is, Q = (2 × 4 × 7.06 – 7 × 5.6) MeV = 17.28 MeV Applying conservation of energy for collision, K p + Q = 2 K α ....(i) (Here, K p and K α are the kinetic energies of proton and α - particle respectively) From the conservation of linear momentum (As there is no external force) ....(ii) [ Here P = 2 mk ] ⇒ K ⁡ p ⁡ = 1 6 K ⁡ α cos 2 θ = 1 6 K ⁡ α 1 4 2 as m ⁡ α = 4 m ⁡ p ⁡ ∴ K α =K p ....(iii) Solving eqs. (i) and (iii) with Q = 17.28 MeV we get K p = 17.28 MeV