Answer
$8.35 \times 10^{4}$ volt
Work Step by Step
Initial speed of an $He^{+}$ ion = 0
Thus;
Loss in energy = Gain in energy
$e\Delta V=KE_{f}-KE_{i}$
$e\Delta |V|=\frac{1}{2}m_{He^{+}}v_{He^{+}}^{2}-0$
$\Delta |V|=\frac{\frac{1}{2}m_{He^{+}}v_{He^{+}}^{2}}{e}$
$=\frac{{\frac{1}{2}4m_{p}v_{He^{+}}^{2}}}{1.6 \times 10^{-19}}$
$=\frac{1}{2}\times \frac{{ 4\times 1.67 \times 10^{-27} \times(2\times 10^{6})^{2}}}{1.6 \times 10^{-19}}$
$=8.35 \times 10^{4} volt$
