A hydrogen atom having kinetic energy `E` collides with a stationary hydrogen atom. Assume all motions are taking place along the line of motion of the moving hydrogen atom. For this situation, mark out the correct statement (s).

Options

A. For `E ge 20.4 eV` only, collision would be elastic.
B. For `E ge 20.4 eV` only, collision would be inelastic.
C. For `E = 2.4 eV` only, collision would be perfectly inelastic.
D. For `E = 18 eV` the `KE` of initially moving hydrogen atom after collisition is zero.

Answer

B, C, D

Explanation

Let collisition between two atoms be an elastic one . From momentum conservation, `mv_(0) = mv_(1) + mv_(2)`
from energy conservation,
`(mv_(1)^(2))/(2) + (mv_(2)^(2))/(2) - (mv_(0)^(2))/(2) = - Delta E`
where `Delta E` is the energy absorbed by the initally stationary atom to changes its state.
Solving above equation, we get
`(v_(1) - v_(2))^(2) = v_(0)^(2) = - (4 Delta E)/(m)`
For collision to be inelastic collisition, `(v_(1) = v_(2))^(2)` ge 0: a real quantity [equal to sign for perfect inelastic collision.]
The minimum value of `Delta E is 10.2 eV`, so for collision inelastic `E ge 20.2 eV`
For perfectly inelastic collision `v_(1) = v_(2)` and hence `E = 20.4 eV` for `E = 18 eV`, the collisition is elastic one and as masses are the same , velocity would be interchanged during collision.