A neutron moving with a speed v makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of neutron for which inelastic collision will take place is (asssume that mass of proton 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`