Let v = speed of neutron before collision v1=speed of neutron after collision v2= speed of hydrogen atom after collision and ΔE= energy of the excitation From the conservation of linear momentum, mv=mv1+mv2 ...(i) From the conservation of energy,
1
2
mv2=
1
2
mv12+
1
2
mv22+ΔE ...(ii) from Eq. (i), v2=v12+v22+2v1v2 From Eq. (ii), v2=v12+v22+
2ΔE
m
∴2v1v2=
2ΔE
m
∴(v1−v2)2=(v1+v2)2−4v1v2 =v2−
4ΔE
m
As, v1−v2 must be real ⇒v2−
4ΔE
m
≥0 or
1
2
mv2≥2ΔE The maximum energy that can be absorbed by hydrogen atom in ground state to go into excited state is 10.2 eV. Therefore, the minimum kinetic energy required is