5. In a regular tetrahedron $A B C D$ with edge length 1, $M, N$ are the midpoints of $A D, B C$ respectively. Then the minimum area of the section cut by the plane through $M N$ in this tetrahedron is $\qquad$ .
5. $\frac{1}{4}$.

First, the plane passing through $M N$ must intersect a pair of edges at one point each, let's assume it intersects $A B$ at $E$ and $C D$ at $F$. Note that $S_{\triangle E M N}=\frac{1}{2} M N \cdot h_{E-M N}$, which means $h_{E-M N} \geqslant h_{A B-M N}$.

Let $E_{1}$ be the midpoint of $A B$, and $O$ be the midpoint of $M N$. It is easy to see that $O E_{1}$ is the common perpendicular line between $A B$ and $M N$. Thus, $h_{A B-M N}=O E_{1}=\frac{\sqrt{2}}{4}$. Therefore, $S_{\triangle E M N} \geqslant \frac{1}{2} M N \cdot O E_{1}=\frac{1}{8}$. Similarly, $S_{\triangle F M N} \geqslant \frac{1}{8}$. Hence, $S_{\text {quadrilateral } E M F N} \geqslant \frac{1}{4}$. When $E, F$ are the midpoints of $A B, C D$ respectively, the equality holds. At this time, $E, M, F, N$ are coplanar, which meets the condition.