A neutron moving with speed v makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of the neutron for which inelastic collision takes place is

Step-by-step solution

Let speed of neutron before collision = V Speed of neutron after collision = V ₁ Speed of proton or hydrogen atom after collision = V ₂ Energy of excitation = E From the law of conservation of linear momentum, mv = mv ₁+ mv ₂ And for law of conservation of energy, 1 2 mv ²= 1 2 mv ₁²+ 1 2 mv ₂² From squaring eq. (i), we get v²=v₁²+v₂²+2 v₁ v₂ From squaring eq. (ii), we get v²=v₁²+v₂²+ 2 E m From eqn (3) & (4) aligned 2 v ₁ v ₂ &= 2 E m ( v ₁- v ₂ )² &= ( v ₁+ v ₂ )²-4 v ₁ v ₂= v ²- 4 E m aligned As, v₁-v₂ must be r