Question

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).

Answer 1

Loss in the energy `(DeltaE)` during the collision will be used to excite the atom from one level to another.
`DeltaE={0,10.2 eV, 12.09 eV,......13.6 eV)`
According to Newtonion mechanics
minimum loss = 0. (elastic collision)

for maximum loss collision will be perfectly inelasitc if nuutron colliedrs perfectly inelastically then, Applying momentum conservation
`mv_(0)=2mv`,
`v_(f)=(v_(0))/(2)`
final `K.E=(1)/(2)xx2mxx(v_(0)^(2))/(4)=((1)/(2)mv_(0)^(2))/(2)=(K)/(2)`
maximum loss `=(K)/(2)`
According to classical mechanics `(DeltaE)=[0,(K)/(2)]`
(a) If `K=14 eV` According to quantum meachics
`(DeltaE)={0,10.2eV,12.09eV}`
According to classical mechanics
`Delta E=[0.7eV]`
loss `= 0`, hence it is elastic collision speed of particle changes.
(b) If `K=20.4 eV`
According to classical mechanics
loss `=[0,10.2 eV]`
According to quantum mechanics
loss `={0,10.2eV,12.09 eV,.....}`
loss `= 0` elastic collision.
loss `=10.2 eV` perfectly inelastic collision
(b) If `K=22 eV`
Classical mechanics `DeltaE=[0,11]`
Quantum mechanics `DeltaE={0,10.2eV,12.09eV,....}`
loss `= 0` elastic collision
loss `=10.2 eV` elastic collision
(d) If `K=24.18eV`
According to classical mechanics `DeltaE=[0,12.09eV]`
According to quantum mechanics `DeltaE={0,10.2eV,12.09eV,....13.6eV}`
loss `=0` elastic collision
loss `=10.2 eV` inelastic collision
loss `=12.09eV` perfectly in elastic collision