A neutron travelling with a velocity `v` and kinetic energy `E` collides perfectly elastically head on with the nucleus of an atom of mass number `A` at rest. The fraction of the total kinetic energy retained by the neutron is

Options

A. `((A-1)/(A+1))^(2)`
B. `((A+1)/(A-1))^(2)`
C. `((A-1)/(A))^(2)`
D. `((A+1)/(A))^(2)`

Answer

To solve the problem, we need to find the fraction of the total kinetic energy retained by the neutron after it collides elastically with a nucleus at rest. Here’s a step-by-step solution:

### Step 1: Understand the initial conditions
- Let the mass of the neutron be \( m_1 \) and its initial velocity be \( u_1 = v \).
- The nucleus of the atom has a mass \( m_2 \) which is related to the mass number \( A \) (since mass number is a measure of the number of nucleons, we can approximate the mass of the nucleus as \( m_2 \approx A \cdot m_u \), where \( m_u \) is the atomic mass unit).
- The initial velocity of the nucleus \( u_2 = 0 \) (it is at rest).

### Step 2: Apply the conservation of momentum
In an elastic collision, both momentum and kinetic energy are conserved. The conservation of momentum gives us:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting \( u_2 = 0 \):
\[ m_1 v = m_1 v_1 + m_2 v_2 \]

### Step 3: Apply the conservation of kinetic energy
The conservation of kinetic energy gives us:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Again substituting \( u_2 = 0 \):
\[ \frac{1}{2} m_1 v^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]

### Step 4: Solve for final velocities
From the equations of elastic collisions, the final velocities can be expressed as:
\[ v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2m_2}{m_1 + m_2} u_2 \]
Since \( u_2 = 0 \):
\[ v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 \]
Substituting \( u_1 = v \):
\[ v_1 = \frac{m_1 - m_2}{m_1 + m_2} v \]

### Step 5: Calculate the fraction of kinetic energy retained
The initial kinetic energy of the neutron is:
\[ E = \frac{1}{2} m_1 v^2 \]
The final kinetic energy of the neutron after the collision is:
\[ E_f = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 \left(\frac{m_1 - m_2}{m_1 + m_2} v\right)^2 \]
Thus, the fraction of kinetic energy retained by the neutron is:
\[ \text{Fraction} = \frac{E_f}{E} = \frac{\frac{1}{2} m_1 \left(\frac{m_1 - m_2}{m_1 + m_2} v\right)^2}{\frac{1}{2} m_1 v^2} \]
This simplifies to:
\[ \text{Fraction} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 \]

### Step 6: Substitute \( m_2 \) in terms of \( A \)
Since \( m_2 \) can be approximated as \( A \) (the mass number):
\[ \text{Fraction} = \left(\frac{m_1 - A}{m_1 + A}\right)^2 \]

### Final Answer
Thus, the fraction of the total kinetic energy retained by the neutron after the collision is:
\[ \text{Fraction} = \left(\frac{1 - A}{1 + A}\right)^2 \]