Consider two hydrogen atoms `H_(A)andH_(B)` in ground state. Assume that hydrogen atom `H_(A)` is at rest and hydrogen atom `H_(B)` is moving with a speed and make head-on collide on the stationary hydrogen atom `H_(A)`. After the strike, both of them move together. What is minimum value of the kinetic energy of the moving hydrogen atom `H_(B)`, such that any one of the hydrogen atoms reaches one of the excitation state.

Explanation

Collision between hydrogen `H_(A)` and hydrogen `H_(B)` atom will be inelastic if a part of kinetic energy is used to excite atom
If `u_(1)` and `u_(2)` are speed of `H_(A)` and `H_(B)` atom after collision then
mu = m`u_(1)` + m`u_(2)` ...(1)
`(1)/(2)` m`u^(2) = (1)/(2)` m`u_(1)^(2) + (1)/(2)` m`u_(2)^(2)` + Delta E ....(2)
`u^(2) = u_(1)^(2) + (u -u_(1))^(2) + (2 Delta E)/(m)`
`u_(1)^(2) - uu_(1) + (2 DeltaE)/(m) = 0`
For `u_(1)` to be real
`u^(2) - (4 Delta E)/(m) ge 0`
(m`u^(2))/(2) ge 2 xx Delta E`
`Delta E = 10.2 eV`
Thus (1)/(2) m`u^(2)_("min") = 2 xx 10.2 eV`
`((1)/(2)` m`u^(2))_("min") = 20.4 eV`
The minimum K.E of the moving hydrogen atom `H_(B) is 20.4 eV`