Answer
(a) greater than
(b) II
Work Step by Step
(a) We know that $ m_ev_e=m_pv_p $ ...eq(1)
That is, the momentum of an electron is equal to the momentum of the proton, as there is no external force on the system.
Now $ K_e=\frac{1}{2}m_ev_e^2$
$\implies K_e=\frac{1}{2}m_e(\frac{m_pv_p}{m_e})^2$ (From eq(1) we plug in the value of $ v_e $)
$\implies K_e=\frac{1}{2}(\frac{m_p}{m_e})m_pv_p^2$
$\implies K_e=\frac{1}{2}(\frac{m_p}{m_e})K_p $
$\implies K_e=(1835)K_p $ (because $ m_p=1835m_e $)
Thus, the electron has a greater kinetic energy compared to the proton.
(b) We know that the best explanation is option (II) -- that is, the two particles experience the same force, but the light electron moves farther than the massive proton. Therefore, the work done on the electron, and hence its kinetic energy, is greater.
