A neutron of energy 4.6 MeV collides successively with stationary protons. In each collision the neutron is deflected through 45°. Find the number of collisions required to reduce its energy to 0.23 eV.
Physics · Nuclear physics
JEE Mains Advanced

Question

A neutron of energy 4.6 MeV collides successively with stationary protons. In each collision the neutron is deflected through 45°. Find the number of collisions required to reduce its energy to 0.23 eV.

Physics · Nuclear physics

G1 Physics · Accepted Answer

Step 1: For elastic scattering of equal masses with the projectile deflected by 45°, the neutron retains half its kinetic energy in each collision. Step 2: Thus Eₙ=4.6 MeV×(1/2)^N. Setting this to 0.23 eV gives (1/2)^N=5×10⁻⁸, yielding N≈24.