A neutron makes a head-on elastic collision with a stationary deuteron. The fraction energy loss of the neutron in the collision is

Options

A. `16//81 `
B. `8//9 `
C. `8 //27`
D. `2//3`

Answer

B

Explanation

Let mass of neutron = m, then mass of deutron = 2m

By Momentum conservation
mu = `mv _(1) + 2 mV_(2) implies u =V _(1) + 2v_(2) …. (2) `
( Elasic collision ) `u= v_(2) - V_(1) ….. (2)`
by equation (1) & (2) `V_(1) =- ( u )/(3) `
Fractional loss of KE of neutron =` (KE _(i) -KE _(f))/( KE _(i))`
`=((1)/(2)m u ^(2) -(1)/(2)m ((u ) /(3) ) ^(2)) /( (1)/(2) m u ^(2) ) = 1-(1)/(9) =(8) /(9) `