A neutron with an energy of `4.6 MeV` collides elastically with proton and is retarded. Assuming that upon each collision the neutron is deflected by `45^(@)`, find the number of collisions which will reduce its energy to `0.23 eV`.

Explanation

To solve the problem step by step, we will analyze the elastic collision between the neutron and the proton, and how the energy of the neutron is affected by each collision.

### Step 1: Understand the Initial Conditions
The initial energy of the neutron is given as \( K_0 = 4.6 \, \text{MeV} \). We need to find the number of collisions required to reduce its energy to \( K_f = 0.23 \, \text{eV} \).

### Step 2: Energy Loss per Collision
From the video transcript, it is mentioned that after each collision, the energy of the neutron is halved. Therefore, after each collision, the energy can be expressed as:
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