Question

A neutron with velocity $v$ strikes a stationary deuterium atom, its KE changes by a factor of:

• A. $15/16$
• B. $1/2$
• C. $2/1$
• D. None of these

Answer 1

Answer: (D)

None of these

Neutron velocity = v, mass = m
Deuteron contains 1 neutron and 1 proton = 2m
In elastic collision, both momentum and KE are conserved
$Pi=Pf$
$mv={mv}_1+2{mv}_2$ ....... (1)
By kinetic energy
$\frac{1}{2}{mv}^2=\frac{1}{2}{mv}_1^2+\frac{1}{2}\left(2m\right)v_2^2$ ........ (2)

(Considering first mass as $m_1$ and other mass is $m_2$)
By solving equation (1) and (2)
$v_1=\frac{m_1-m_2}{m_1+m_2}v+\frac{2m_2}{(m_1+m_2)}v$
$v_1=\frac{m_1-2m}{3m}$
$v_1=-\frac{v}{3}$
$K_i=\frac{1}{2}{mv}^2$
$K_f=\frac{1}{2}{mv}_1^2$
$\frac{K_i-K_f}{K_i}=1-\frac{v_1^2}{v^2}$
$=1-\frac{1}{9}$
$=\frac{8}{9}$ fractional change in KE