A neutron moving with a speed v makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy neutron for which inelastic collision will take place is (assume that mass of proton is nearly equal to the mass of neutron)

Step-by-step solution

Let v = speed of neutron before collision v 1 = speed of neutron after collision v 2 = speed of proton or hydrogen atom after collision and ∆ E = energy of excitation. From conservation of linear momentum m v = m v 1 + m v 2 ...(i) From conservation of energy 1 2 m v 2 = 1 2 m v 1 2 + 1 2 m v 2 2 + ∆ E ...(ii) From i and ii As v 1 - v 2 must be real v 2 - 4 ∆ E m ≥ 0 ⇒ 1 2 m v 2 = 2 ∆ E = 2 × 10.2 = 20.4 e V