Question

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.

• A. $(0,0)$
• B. $(0,1)$
• C. $(.89,.28)$
• D. $(.28,.89)$

Answer 1

The correct option is C $(.89,.28)$

Case 1 : Collision of Neutron with deuterium : $e=1$
Momentum conservation $mu+0=mv+2m{v}^{\prime }$
$⟹u=v+2v^{\prime }$ .......(1)
Newton's equation of collision $e=-1=\frac{v^{\prime }-v}{0-u}$
$⟹v^{\prime }=u+v$ .......(2)
Solving (1) and (2), we get $v=-\frac{u}{3}$
Fractional loss in kinetic energy of neutron $P_d=\frac{\frac{1}{2}m{u}^2-\frac{1}{2}m{v}^2}{\frac{1}{2}m{u}^2}$
$⟹$ $P_d=\frac{\frac{1}{2}m{u}^2-\frac{1}{2}m(\frac{-u}{3}{)}^2}{\frac{1}{2}m{u}^2}=0.89$
Case 2 : Collision of Neutron with Carbon : $e=1$
Momentum conservation $mu+0=m{v}_1+12m{v}_1^{\prime }$
$⟹u=v_1+12v_1^{\prime }$ .......(3)
Newton's equation of collision $e=-1=\frac{v_1^{\prime }-v_1}{0-u}$
$⟹v_1^{\prime }=u+v_1$ .......(4)
Solving (3) and (4), we get $v_1=-\frac{11u}{13}$
Fractional loss in kinetic energy of neutron $P_c=\frac{\frac{1}{2}m{u}^2-\frac{1}{2}m{v}_1^2}{\frac{1}{2}m{u}^2}$
$⟹$ $P_c=\frac{\frac{1}{2}m{u}^2-\frac{1}{2}m(\frac{-11u}{13}{)}^2}{\frac{1}{2}m{u}^2}=0.28$