A neutron collides, head-on with a deuterium at rest. What fraction of the neutron's energy would be transferred to the deuterium?

Options

A. `89%`
B. `11%`
C. `79%`
D. `21%`

Answer

A

Explanation

According to law of conservation of linear momentum , we get
`m_n v_(ni)+m_d xx 0 =m_n v_(nf)+m_d v_(df)`
where `v_(ni)` is the initial velocity of the neutron before collision and `v_(nf)` and `v_(df)` are the velocities of the neutron and deuterium after collision
`m_n v_(ni)=m_(n) v_(nf)+m_d v_(df)`....(i)
According to conservation of kinetic energy , we get
`1/2 m_n v_(ni)^2 =1/2m_n v_(nf)^2 + 1/2 m_d v_(df)^2`
`m_n v_(ni)^2 =m_n v_(nf)^2 + m_d v_(df)^2`....(ii)
Both Eqs. (i) and (ii), it follows that
`m_n v_(ni)(v_(df)-v_(ni))=m_n v_(nf)(v_(df)-v_(nf))`
`v_(df)(v_(ni)-v_(nf))=v_(ni)^2 - v_(nf)^2 , v_(df)=v_(ni)+v_(nf)`
Substituting this in Eq.(i) , we get
`v_(nf)=((m_n - m_d))/(m_n+m_d) v_(ni)`...(iii)
and `v_(df)=(2m_n v_(ni))/(m_n+m_d)`...(iv)
The initial kinetic energy of the neutron is `K_(ni) = 1/2 m_n v_(ni)^2`
Final kinetic energy of the deuterium is
`K_(df)=1/2m_d v_(df)^2 =1/2 m_d ((2m_n v_(ni))/(m_n + m_d))`...(iv)
The initial kinetic energy of the neutron is `K_(ni)=1/2 m_n v_(ni)^2`
Final kinetic energy of the deuterium is `K_(df)=1/2 m_d v_(df)^2 =1/2m_d ((2m_n v_(ni))/(m_n + m_d))^2` (Using (iv))
Fraction of neutron's energy transferred to deuterium is `f=K_(df)/K_(ni)=(4m_n m_d)/((m_n + m_d)^2)`
For deuterium , `m_d=2m_n`
`therefore f=(4(m_n)(2m_n))/((m_n+2m_n)^2)=8/9=89%`