Question

A proton is bombarded on a stationary lithium nucleus. As a result of the collision two particles are produced. If the direction of motion of the \[\alpha \] particle with the initial direction of motion makes an angle ${\cos ^{ - 1}}\left( {\dfrac{1}{4}} \right)$. Find the kinetic energy of the striking proton. Given binding energies per nucleon of $L{i^7}$ and $H{e^4}$ are $5.60\,MeV$ and $7.06\,MeV$ respectively. (Assume mass of proton $ \approx $ mass of neutron).

Answer 1

Hint: Firstly we need to calculate the Q value of this particular reaction. Then on comparing the two equations formed by the law of conservation of energy and the law of conservation of linear momentum, we will get a relation between the Q value and the kinetic energy of the striking proton.

Complete step by step answer:
The Q value of this particular reaction is,
$Q = 2{K_\alpha } - {K_p}$
On putting the required values in the above equation, we get,
$Q = 2 \times 4 \times 7.06 - 7 \times 5.6$
\[\Rightarrow Q = 56.48 - 39.2\]
$\Rightarrow Q = 17.28\,MeV$
On applying the law of conservation of energy on this collision, we get,
${K_p} + Q = 2{K_\alpha }.......(1)$
Where, ${K_p}$ is the kinetic energy of the given proton and ${K_\alpha }$ is the kinetic energy of the given alpha particle.

According to the law of conservation of the linear momentum,
$\sqrt {2{m_p}{K_p}} = 2\sqrt {2{m_\alpha }{K_\alpha }} \cos \theta .....(2)$
On squaring both the sides, we get,
$2{m_p}{K_p} = 4{\cos ^2}\theta (2{m_\alpha }{K_\alpha })$
In this question, ${m_\alpha } = 4{m_p}$
${m_p}{K_p} = 2{\cos ^2}\theta (8{m_p}{K_\alpha })$
On cancelling ${m_p}$ on both the sides,
${K_p} = 16{\cos ^2}\theta {K_\alpha }$

On putting the value of angle as ${\cos ^{ - 1}}\left( {\dfrac{1}{4}} \right)$, we get,
${K_p} = 16{\left( {\cos {{\cos }^{ - 1}}\dfrac{1}{4}} \right)^2}{K_\alpha }$
$\Rightarrow {K_p} = Q$
$\Rightarrow {K_p} = 16{\left( {\dfrac{1}{4}} \right)^2}{K_\alpha }$
On further solving,
${K_p} = 16\left( {\dfrac{1}{{16}}} \right){K_\alpha }$
$\Rightarrow {K_p} = {K_\alpha }.....(3)$
On putting the value of equation (3) in equation (1),
${K_p} + Q = 2{K_p}$
$\Rightarrow {K_p} = Q$
We know that Q value is $Q = 17.28\,MeV$
$\therefore {K_p} = 17.28\,MeV$

Therefore, the kinetic energy of the striking proton is ${K_p} = 17.28\,MeV$.

Note:The Q value for a particular reaction is defined as the amount of energy which is either absorbed or released during that particular nuclear reaction. This value relates to the enthalpy of a reaction or the energy released by the radioactive decay products. Q value can be found out from the masses of reactants and the masses of products.