Answer
$ 2.1\times 10^{6}m/s$
Work Step by Step
Proton velocity, $v_{p}=5 \times 10^{4} m/s$
For proton:
By the conservation of energy;
Loss in energy = Gain in energy
Thus;
$q\Delta V=\frac{1}{2}m_{p}v_{p}^{2}$ ...(1)
Same potential difference applied for an electron.
Thus, Loss in energy = Gain in energy
$q\Delta V=\frac{1}{2}m_{e}v_{e}^{2}$ ...(2)
From equations (1) and (2):
$v_{e}=\sqrt {\frac{m_{p}}{m_{e}}}v_{p}$
$v_{e}=\sqrt {\frac{1.67\times 10^{-27}}{9.11 \times 10^{-31}}} \times 5
\times 10^{4}$
$v_{e}= 2.1\times 10^{6}m/s$
