Question

A proton is bombarded on a stationary lithium nucleus. As a result of the collision two $\alpha$ -particles are produced. If the direction of motion of the $\alpha$ -particles with the initial direction of motion makes an angle ${\cos }^{-1}\left(\frac{1}{4}\right)$ , then the kinetic energy of the striking proton is [Given, binding energies per nucleon of $L{i}^7=5.60MeV$ and $H{e}^4=7.60MeV$ , $m_{proton}\approx m_{neutron}$ ]

• A. $\text{17.28}MeV$
• B. $\text{17.36}MeV$
• C. $\text{17.58}MeV$
• D. $\text{17.44}MeV$

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 the conservation of linear momentum (As there is no external force) ....(ii)
$[HereP=\sqrt{2mk}]$
$\Rightarrow K_p=16K_{\alpha }{\cos }^2\theta =\left(\begin{array}{ll}16& K_{\alpha }\end{array}\right){\left(\frac{1}{4}\right)}^2($ as $m_{\alpha }=4m_p)$
∴ K_α =K_{p ....(iii)}
Solving eqs. (i) and (iii) with Q = 17.28 MeV

we get K_p = 17.28 MeV