Question

A neutron with a kinetic energy of 0.6 MeV directly collides with a stationary carbon nucleus (mass number is 12). Find the kinetic energy of the carbon nucleus after the collision.
A) 1.7 MeV
B) 0.17 MeV
C) 17 MeV
D) zero

Answer 1

Hint: The momentum must be conserved before the collision and after the collision. The collision is said to be direct and so it is elastic. This implies then that the velocity of approach will be equal to the velocity of separation.

Formula Used:
1) The momentum of a system of two is given by, $p = {m_1}{v_1} + {m_2}{v_2}$ where ${m_1}$, ${m_2}$ are the masses of the two particles and ${v_1}$, ${v_2}$ are their velocities.
2) The kinetic energy of a particle is given by, $E = \dfrac{1}{2}m{v^2}$ where $m$ is the mass of the particle and $v$ is its velocity.

Complete step by step answer:
Step 1: List the known parameters in the question.
The kinetic energy of the approaching neutron is 0.6 MeV.
The mass number of the carbon nucleus is 12. This implies that the carbon nucleus has six protons and six neutrons. The masses of protons and neutrons are almost similar. So we assume that the mass of the carbon nucleus is 12 times the mass of the approaching neutron.
So, if the mass of the neutron is $m$ then mass of the carbon nucleus will be $12m$ .

Step 2: Apply the conservation of momentum theory.
The system comprises the approaching neutron and the carbon nucleus.
Let’s consider the system before and after the collision.
Before collision:
Let $v$ be the velocity of the approaching neutron before the collision. Since the carbon nucleus is stationary as the neutron approaches, its initial velocity is zero.
Then the momentum of the system will be ${p_b} = mv$
After collision:
Let ${v_1}$ be the velocity of the neutron and ${v_2}$ be the velocity of the carbon nucleus after the collision.
Then the momentum of the system will be ${p_a} = m{v_1} + 12m{v_2}$
Momentum must be conserved before and after the collision.
$\therefore {p_b} = {p_a}$
i.e., $mv = m{v_1} + 12m{v_2}$
Cancel out $m$ on both sides of the above equation to get $v = {v_1} + 12{v_2}$ ------- (1)
Step 3: Obtain an expression for the velocity of the carbon nucleus using the assumption that the collision is elastic.
For an elastic collision, the ratio of the velocity of separation to the velocity of approach is one.
Here, the velocity of approach is $v$ and the velocity of separation is ${v_2} - {v_1}$ .
Then we have, $\dfrac{{{v_2} - {v_1}}}{v} = 1$
i.e., $v = {v_2} - {v_1}$ -------- (2)
Adding equations (1) and (2) we get, $2v = 13{v_2}$ or ${v_2} = \dfrac{{2v}}{{13}}$
Thus the velocity of the carbon nucleus after the collision is ${v_2} = \dfrac{{2v}}{{13}}$ .
Step 4: Find the kinetic energy of the carbon nucleus.
The kinetic energy of a particle is given by, $E = \dfrac{1}{2}m{v^2}$ -------- (3) where $m$ is the mass of the particle and $v$ is its velocity.
Substituting the values for the velocity ${v_2} = \dfrac{{2v}}{{13}}$ and mass $12m$ of the carbon nucleus in equation (3) we get the kinetic energy of the carbon nucleus as $E = \dfrac{1}{2} \times 12m \times {\left( {\dfrac{{2v}}{{13}}} \right)^2}$
Simplifying we get, $E = \dfrac{1}{2}m{v^2} \times \dfrac{{48}}{{169}}$
Since ${E_1} = \dfrac{1}{2}m{v^2} = 0.6{\text{MeV}}$ is the kinetic energy of the approaching neutron, we get $E = 0.6 \times \dfrac{{48}}{{169}} = 0.17{\text{MeV}}$
$\therefore $ the kinetic energy of the carbon nucleus after the collision is 0.17 MeV.
Hence the correct option is B.

Note: The velocity of approach refers to how fast an object approaches another object. The velocity of approach will be the sum of the velocities of the two objects. Here, the carbon nucleus is at rest so the velocity of approach will be equal to the velocity of the neutron $v$. The velocity of separation will be the sum of the velocities of the two objects if they move in opposite directions and will be the difference of their velocities if they move in the same direction. Here, the neutron and the carbon nucleus move in the same direction and so it is ${v_2} - {v_1}$ .