A neutron moving with a speed v strikes a hydrogen atom in ground state moving toward it with the same speed. Find the minimum speed of the neutron for which ineleastic (completely or partially) collision may take place .The mass of neutron = mass of hydrogen `= 1.67 xx 10^(-27)kg`

Explanation

Suppose the neutron and hydrogen atom move at speed `v_(1)` and `v_(2)` after the collision. The collision on will be inelastic if a part of the kinetic energy is used to excite the atom. Suppose an energy `DeltaE` is used in this way. Using conservation of linear momentum and energy.
`mv=mv_(1)+mv_(2)`.....(i)
and `(1)/(2)mv^(2)=(1)/(2)mv_(1)^(2)+(1)/(2)mv_(2)^(2)+DeltaE`......(ii)
From (i), `v^(2)=v_(1)^(2)+v_(2)^(2)+2v_(1)v_(2)`,
From (ii) `v^(2)=v_(1)^(2)+v_(2)^(2)+(2DeltaE)/(m)`
thus,`2v_(1)v_(2)=(2DeltaE)/(m)`
Here, `(v_(1)-v_(2))^(2)-4v_(1)v_(2)=v^(2)-(4DeltaE)/(m)`
As `v_(1)-v_(2)` must be real, `v^(2)-(4DeltaE)/(m) gt 0`
or `(1)/(2)mv^(2) gt 2DeltaE`
The minimum energy that can be absorbed by the hydrogen atom in ground state to go in an excited state is `10.2 eV`. Thus, the minimum kinetic energy of the nerutron needed for an inelastic collision is
`(1)/(2)mv_(min)^(2)=2xx10.2eV=20.4 eV`