Question

Find the kinetic energy of the striking proton if a proton is bombarded on a stationary lithium nucleus and upon collision two $\alpha $ - particles are produced. Given the direction of motion of the α - particles with the initial direction of motion makes an angle cos^–1 (1/4) and binding energies per nucleon of Li⁷ and He⁴ are 5.60 and 7.06 MeV respectively.
(Assume mass of proton ≈ mass of neutron).

• A. 17.28 MeV
• B. 17.38 MeV
• C. 17.29 MeV
• D. – 17.40 MeV

Answer 1

Answer: (A)

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 conservation of linear momentum
$\sqrt{2 m _{ p ⁡} K ⁡_{ p ⁡}} = 2 \sqrt{2 m ⁡_{\alpha } K ⁡_{\alpha }} \text{cos} \theta $ ....(ii)
$\left[\right.Here P=\sqrt{2 mk}\left]\right.$
∴ $\left( K \right)_{ p ⁡} = 1 6 \left( K ⁡\right)_{\alpha } \left(\text{cos}\right)^{2} \theta = \left(1 6 \left( K ⁡\right)_{\alpha }\right) \left(\frac{1}{4}\right)^{2} \left(\text{as} \left(m ⁡\right)_{\alpha } = 4 \left(m ⁡\right)_{ p ⁡}\right)$
∴ K_{α =} K_{p ....(iii)}
Solving eqs. (i) and (iii) with Q = 17.28 MeV

we get K_p = 17.28 MeV