5. The necessary and sufficient condition for line segments of lengths $1, a, a, a, a, a$ to be the 6 edges of the same tetrahedron is ( ).
(A) $0\frac{\sqrt{3}}{3}$
(D) $\frac{\sqrt{3}}{3}<a<\sqrt{3}$
5.C.

In tetrahedron $ABCD$, $\triangle ABC$ and $\triangle BCD$ are both equilateral triangles with side length $a$, and $AD=1$. Taking the midpoint $E$ of $BC$, we have
$$
AE=DE=\frac{\sqrt{3}}{2} a.
$$

Clearly, the necessary and sufficient condition for the lengths $1, a, a, a, a, a$ to be the 6 edges of tetrahedron $ABCD$ is $AE + DE > AD$, i.e., $\sqrt{3} a > 1$, which gives $a > \frac{\sqrt{3}}{3}$.