Answer
The electron would attain a speed of $~~0.745~c$
Work Step by Step
We can find the kinetic energy $K$ if the electron is accelerated across a potential difference of $256~keV$:
$K = \vert q \vert~V$
$K = (1~e)~(256~keV)$
$K = 256~kV$
We can find $\gamma$:
$K = mc^2(\gamma-1)$
$\frac{K}{mc^2} = \gamma-1$
$\gamma = \frac{K}{mc^2}+1$
$\gamma = \frac{256~keV}{511~keV}+1$
$\gamma = 1.50$
We can find $\beta$:
$\gamma = \frac{1}{\sqrt{1-\beta^2}}$
$\sqrt{1-\beta^2} = \frac{1}{\gamma}$
$1-\beta^2 = \frac{1}{\gamma^2}$
$\beta^2 = 1-\frac{1}{\gamma^2}$
$\beta = \sqrt{1-\frac{1}{\gamma^2}}$
$\beta = \sqrt{1-\frac{1}{1.50^2}}$
$\beta = 0.745$
The electron would attain a speed of $~~0.745~c$
