Answer
The fraction of the total energy that goes into the kinetic energy of the proton is $\frac{1}{1837}$.
Work Step by Step
Let $m_p$ be the mass of the proton, and let $m_e$ be the mass of the electron. Let $v_p$ be the velocity of the proton. We can use conservation of momentum to find the velocity of the electron $v_e$
$m_p~v_p + m_e~v_e = 0$
$m_e~v_e = -m_p~v_p$
$v_e = \frac{-m_p~v_p}{m_e}$
$v_e = \frac{-1836~m_p~v_p}{m_p}$
$v_e = -1836~v_p$
We can find the total kinetic energy.
$K = \frac{1}{2}m_p~v_p^2+\frac{1}{2}m_e~v_e^2$
$K = \frac{1}{2}m_p~v_p^2+\frac{1}{2}(\frac{m_p}{1836})~ (-1836~v_p)^2$
$K = \frac{1}{2}(1837)~m_p~v_p^2$
We can find the fraction of the total energy that goes into the kinetic energy of the proton.
$\frac{K_p}{K} = \frac{\frac{1}{2}m_p~v_p^2}{\frac{1}{2}(1837)~m_p~v_p^2}$
$\frac{K_p}{K} = \frac{1}{1837}$
The fraction of the total energy that goes into the kinetic energy of the proton is $\frac{1}{1837}$.
