1. Tardigrade
  2. Question
  3. Physics
  4. 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 Li7=5.60 MeV and He4=7.60 MeV , mproton≈ mneutron ]

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Q. 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 $Li^{7}=5.60 \, MeV$ and $He^{4}=7.60 \, MeV$ , $m_{proton}\approx m_{neutron}$ ]

NTA AbhyasNTA Abhyas 2020Nuclei

Solution:

Q value of the reaction is,
Q = (2 × 4 × 7.06 - 7 × 5.6) MeV = 17.28 MeV
Solution
Applying conservation of energy for collision,
Kp + Q = 2 Kα ....(i)
(Here, Kp and Kα are the kinetic energies of proton and α - particle respectively)
From the conservation of linear momentum (As there is no external force) ....(ii)
$\left[\right.Here P=\sqrt{2 mk}\left]\right.$
$\Rightarrow K_p=16 K_\alpha \cos ^2 \theta=\left(\begin{array}{ll}16 & K_\alpha\end{array}\right)\left(\frac{1}{4}\right)^2 \quad\left(\right.$ as $\left.m_\alpha=4 m_p\right)$
∴ Kα =Kp ....(iii)
Solving eqs. (i) and (iii) with Q = 17.28 MeV

we get Kp = 17.28 MeV