A neutron moving with a speed `v` makes a head-on collision with a hydrogen in ground state kept at rest which inelastic collision will be take place is (assume that mass of photon is nearly equal to the mass of neutron)

Options

A. `10.2eV`
B. `20.4eV`
C. `12.1eV`
D. `16.8eV`

Answer

B

Explanation

Let V=speed of neutron before collision
`v_(1)`=speed of neutron after collision
` v_(2)`= speed of proton or hydrogen atom after
collision and `DeltaE`= energy of exitation.
From conservation of linear momentum `mv=mv_(1)+mv_(2)` ...(1)
From conservation of energy
` 1/2mv^(2)=1/2mv_(1)^(2)+1/2mv_(2)^(2)+DeltaE` ...(2)
from `1`and `2`As `v_(1)-v_(2)` must be real
`v^(2)-4(DeltaE)/mge0`
` Rightarrow1/2mv^(2)=2DeltaE=2xx10.2=20.4eV`