A right pyramid has regular hexagon $ABCDEF$ with side length $1$ as its base and apex $V$. Segments $\overline{AV}$ and $\overline{CV}$ are perpendicular. What is the volume of the pyramid?

1. Let $O$ be the center of the base and $h=VO$. The regular hexagon of side $1$ has circumradius $OA=1$, and $\angle AOC=120^\circ$, so $AC^2=1^2+1^2-2\cos120^\circ=3$.
2. By symmetry $VA=VC=\sqrt{1+h^2}$, and $\overline{AV}\perp\overline{CV}$ gives $AC^2=VA^2+VC^2=2(1+h^2)$.
3. So $3=2+2h^2$, $h^2=\frac12$, $h=\frac{\sqrt2}2$.
4. The base has area $6\cdot\frac{\sqrt3}4=\frac{3\sqrt3}2$, so the volume is $\frac13\cdot\frac{3\sqrt3}2\cdot\frac{\sqrt2}2=\boxed{\dfrac{\sqrt6}4}$.