Answer
$0.77c$
Work Step by Step
We can find the required speed of the electron as follows.
First, we find the kinetic energy of the proton:
$K.E_p=m_pc^2(\frac{1}{\sqrt{1-v^2/c^2}}-1)$
$K.E_p=1836m_e c^2(\frac{1}{\sqrt{1-(0.0250)^2}-1})=0.574m_e c^2$
The kinetic energy of the electron is equal to the proton:
$\implies K.E_p=K.E_e=m_e c^2(\frac{1}{\sqrt{1-v^2/c^2}}-1)$
This simplifies to:
$v=c\sqrt{1-(K.E_p/m_e c^2+1)^{-2}}$
We plug in the known values to obtain:
$v=c\sqrt{1-(0.574+1)^{-2}}=0.77c$
