Two $\mathrm{H}$ atoms in the ground state collide inelastically. What is the minimum and maximum amount by which their combined kinetic energy is reduced?

In an inelastic collision of elementary particles and atoms, kinetic energy is not conserved and no energy can be lost as heat because we can not consider the deformation of any lattice in the colliding body. So, the energy lost can only be absorbed by the atom involved in the collision and may get excited or ionized by this energy loss.

Let us assume both the hydrogen atoms are moving in opposite direction towards each other having kinetic energy K.

Initial kinetic energy = 2K

Final kinetic energy = 0 ( atoms will come to rest after collision as per momentum conservation)

Reduction in kinetic energy = Initial Kinetic energy – final kinetic energy = 2K

This energy loss (2K) should be utilized in exciting the electrons from its ground state in hydrogen atoms.

Thus, 2K = exciting energy

Minimum exciting energy will occur when electron in one of the atom jumps from n = 1 to n = 2

Exciting energy in this case will be the difference of total energy before excitation and after excitation

Total energy before excitation = -13.6 eV -13.6 eV = -27.2 eV

Total energy after excitation =  $-13.6-\frac{13.6}{2^2}$ = -17eV

Exciting energy = -17 - (-27.2 ) = 10.2 eV

Energy loss = exciting energy = 10.2 eV.

Now, 2K (= exciting energy) will be maximum when both the electrons from both atoms jumps from n = 1 to n = ∞.

Total energy after excitation = 0

Total ionization energy = 0 - (-27.2 eV) = 27.2 eV.

Maximum energy loss, 2K = 27.2 eV