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

Answer

B

Explanation

Let `v = ` speed of neutron before collision,
`v_(1) = ` speed of neutron after collision,
`v_(2) = ` speed of photon or hydrogen atom after collision, and `Delta E=` energy of excitation
From conservartion of linear momentum,
`mv = mv_(1) + mv_(2)` (i)
From conservation of energy
`(1)/(2) mv^(3) = (1)/(2) mv_(1)^(2) + (1)/(2) mv_(2)^(2) + Delta E` (ii)
From eq. (i),
`v^(2) = v_(1)^(2) + v_(2)^(2) + 2 v_(1) v_(2)`
From eq. (ii),
`v^(2) = v_(1)^(2) + v_(2)^(2) + (2 Delta E)/(m)`
`:. 2 v_(1) v_(2) = (2 Delta E)/(m)`
`:. (v_(1) - v_(2))^(2) = (v_(1) + v_(2))^(2) - 4v_(1) v_(2)`
`implies (v_(1) - v_(2))^(2) = V^(2) - 4 (Delta E)/(m)`
As `v_(1) - v_(2)` must be real, therefore
`v^(2) - 4 (Delta E)/(m) ge 0`
or
`(1)/(2) mv^(2) ge 2 Delta E`
The minimum energy that can be obsorbed by hydrogen atom in the ground state go to into excited is `10.2 eV`. Therefore,
`(1)/(2) mv_(min)^(2) = 2 xx 10.2 eV`
`= 20.4 e V`