An `H` atom in ground state is moving with initial kinetic energy `K` .It collides head on with `He ^(+)` ion in ground state kept at rest but free to move. Find minimum value of `K` so that both the particles can excite to their first excited state.

Explanation

Here, energy loss during the collision is used to excite the atom or ion.
Now according to quntum mechaincs, loss in energy `(Delta E)` for `h` atom
`{0,10.2 e V, 12.09 e v, …. 13.06 e V}`
For `He ^(+)` ion in the first excited state, minimum energy `= 40.8 + 10.2 = 51 e v`
Now according to Newtonion , minimum loss is `0` (for elestic collion). Maximum loss will be when there is perfectly inelastic collision.
Now, let mass of `H` atom be `m`, then mass of `He^(+) ion = 4 m` let `v_(0)` be the intial speed of `H` atom and `v_(f)` the final common speed. According to momentum conservation:
`m v_(0) = 4 m v_(f) + m v_(f) implies (v_(0))/(5)`
Kinetic energy `= (1)/(2) (5 m) = (v_(0)^(2))/(25) = (1)/(5) ((1)/(2) m v_(0)^(2)) = (K)/(5)`

Now for minimum value of `K` so that the electron excite to the first excited state of `H` atom and `He^(+)` ion.
`(4 K)/(5) = 51 e V` or ` K = (51 xx 5)/(4) e V` or ` K = 63.75 e V`