= Solution
An <Electron> has <electric charge> $-e$, so its gain of <kinetic energy> through an accelerating <electric potential difference> of magnitude $V$ is $eV$. Starting from rest, <conservation of energy> gives
$$
\frac12m_ev^2=eV,\qquad
v=\sqrt{\frac{2eV}{m_e}}.
$$
With $e=1.60218\times10^{-19}\ \mathrm C$, <electron mass> $m_e=9.10938\times10^{-31}\ \mathrm{kg}$, and $V=150\ \mathrm V$,
$$
\boxed{v=7.26\times10^6\ \mathrm{m\,s^{-1}}.}
$$
This is a speed; the <velocity> points along the accelerating path, opposite the <electric field>. Here $v/c\simeq0.0242$, where $c$ is the <speed of light>, so the classical <kinetic energy> approximation is accurate. As a check, <special relativity> gives $\gamma=1+eV/(m_ec^2)$ and $v=c\sqrt{1-\gamma^{-2}}\simeq7.262\times10^6\ \mathrm{m\,s^{-1}}$, agreeing to the quoted precision.
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