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 `Delta E =` energy of exitation.
From conservation of linear momentum
`mv =mv_1 +mv_(2)` .......(1)
From conservation of energy
`(1)/(2) mv^(2) =(1)/(2) mv_(1)^(2)+(1)/(2) mv_(2)^(2) +Delta E` ...........(2)
from 1 and 2 As `v_(1) - v_(2)` must be real
`v^(2) -4 (Delta E)/(m) ge 0`
`rArr (1)/(2) mv^(2) = 2 Delta E = 2xx 10.2 =20.4 e V` .