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
$\sqrt{2m_{p}K{}_p}=2\sqrt{2m{}_{\alpha }K{}_{\alpha }}\text{cos}\theta$ ....(ii)
$[HereP=\sqrt{2mk}]$
∴ ${\left(K\right)}_p=16{\left(K\right)}_{\alpha }{\left(\text{cos}\right)}^2\theta =\left(16{\left(K\right)}_{\alpha }\right){\left(\frac{1}{4}\right)}^2\left(\text{as}{\left(m\right)}_{\alpha }=4{\left(m\right)}_p\right)$
∴ K_{α =} K_{p ....(iii)}
Solving eqs. (i) and (iii) with Q = 17.28 MeV

we get K_p = 17.28 MeV