Answer
$v = 0.333c$
Work Step by Step
The electron was accelerated from rest by a potential difference of 31000 V, so it has a kinetic energy of 31000 eV. Use equation 26–5b.
$$KE=31000 eV=(\gamma-1)mc^2=(\gamma-1)(511000eV)$$
$$\frac{1}{\sqrt{1-v^2/c^2}}-1=\frac{31000MeV}{511000eV}$$
$$v=c\sqrt{1-\frac{1}{(1+31000/511000)^2}}=0.333c$$
