It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is `p_(d)`, while for its similar collision with carbon nucleus at rest, fractional loss of energy is `p_(c )`. The values of `p_(d)` and `p_(c )` are respectively :

Options

A. (0,0)
B. (0,1)
C. (0.89,0.28)
D. (0.28,0.89)

Answer

C

Explanation

Case - I Let ass of neutron be m and speed V

Lt `v_(1) and V_(2)` be the velocities of neutron and deutrium respectively after collision .

According to conservation of linear momentum
`mV_(1) +2mV_(2) = mV `
`rArr 2V_(2)+V_(1) = V " " ....`
From coefficient of restitution , e = 1 ( for perfectly elastic collision )
`- 1 = (V_(2)-V_(1))/(0-V) `
`V_(2) - V_(1)=V " " ...(ii)`
From equation (i) and (ii)
`V_(2) = (2V)/3 , V_(1) = - V/3 `
Negative velocity means that direction of velocity is opposite to intial velocity shown in diagram .
Fractional loss of energy of neutron is
`P_(d) = (Delta K.E)/((K.E)`
`= (1/2 mV^(2) - 1/2mV_(1)^(2))/(1/2mV^(2)) = 8/9 = 0.89`
Case - II
State of masses before collision is shown in figure .

Let `V_(1) and V_(2)` be the velocities of neutron and carbon atoms respectively after collision

According to conservation of linear momentum
`mV_(1) +12m V_(2) = mV `
`rArr " " 12V_(2)+V_(1)=V " " ....(i)`
Using definition of coefficient of restituton ( in this case e = 1 )
` -1 = (V_(2)-V_(1))/(0-V) `
`rArr V_(2) - V_(1) = V " " ...(ii)`
From above two equations
`V_(2) = (2V)/13 , V_(1)=-(11V)/13`
Fractional loss of energy of neutron is
`P_(c ) = (DeltaK.E)/(K.E) = (1/2 mV^(2)-1/2mV_(1)^(2))/(1/2mV^(2))`
`= 48/169 = 0.28`