In the figure , what type of collision can be possible , if `K = 14 eV, 20.4 eV,22 eV, 24.18 eV,(elastic // inelastic // perfectly inelastic).

Explanation

Loss in `(Delta E)` during the collision will be need to excite the atom of electron from one level to another. Acceleration to Bohr 's theory, for hydrogen atom.
`Delta E = (0, 10.2 e V, 12.09 e V, …. 13.6 e V)`
Acceleration to Newtonion machanics, minimum loss `= 0` for elestic collision, For maximum loss, collision will be perfectly inelastic.

If neutron collides perfectly inelastically , then applying momentum conservation ,
`m v_(0) = 2 m v_(f) implies v_(r) = (v_(0))/(2)`
Final `KE = (1)/(2) xx 2 m xx (v_(0)^(2))/(4) = ((1)/(2) m v_(0)^(2))/(2) = (K)/(2)`
So, maximum loss = ` (K)/(2)`
According to classical mechaincs, energy loss lies in the range :
`Delta E =[0,(K)/(2)]`
a. `K = 14 e V`:According to quantum mechaincs,
`Delta E = {0, 10.2 e V, 12.09 e V}`
Acceleration to classic machanics `Delta E = [0.7 e V]`
So, possible less `=0`, hence it is an classic collision.
b. `K = 20.4 e V`:According to quantum mechaincs,
`Delta E = {0, 10.2 e V, 12.09 e V ....}`
Acceleration to classic machanics `Delta E = [10.2 e V]`
If possible less `= 0` `implies` classic collision ,
If less `= 10.2 e V` `implies` inelastic collision (not perfectly),
and If loss `= 12.09 e Vimplies` perfectly inclastic collision.