The side length of the square $A B C D$ and the square $A B E F$ is 1, and the planes on which these squares lie form a $120^{\circ}$ angle. Points $M, N$ are on the diagonals $A C, B F$ respectively, and $A M=F N$. Then the range of values for $M N$ is
(A) $\left(\frac{\sqrt{2}}{2}, 1\right]$
(B) $\left(\frac{\sqrt{3}}{2}, 1\right]$
(C) $\left[\frac{\sqrt{2}}{2}, 1\right]$
(D) $\left[\frac{\sqrt{3}}{2}, 1\right]$
5. (D).

Passing through $N$ in the plane $A B E F$ draw $N G \perp A B$, with $G$ as the foot of the perpendicular, and connect $M G$. It is easy to prove from the conditions that $M G \perp A B$, so $\angle N G M=120^{\circ}$. Let $A M=x$, then $G M=\frac{\sqrt{2}}{2} x, N G=1-\frac{\sqrt{2}}{2} x$. From this, $M N^{2}=\frac{1}{2}-x^{2}+\left(1-\frac{\sqrt{2}}{2} x\right)^{2}$ $\rightarrow \frac{\sqrt{2}}{2} x\left(1-\frac{\sqrt{2}}{2} x\right)=\frac{1}{2} x^{2}-\frac{\sqrt{2}}{2} x+1=\frac{1}{2}\left(x-\frac{\sqrt{2}}{2}\right)^{\prime}+\frac{3}{4}$. Therefore, $\frac{\sqrt{3}}{2} \leqslant M N \leqslant 1$ (equality can be achieved in both cases).