5. In a regular tetrahedron $P-ABC$, $AB=1, AP=2$, a plane $\alpha$ passing through $AB$ bisects its volume. Then the cosine of the angle formed by edge $PC$ and plane $\alpha$ is $\qquad$.
5. $\frac{3 \sqrt{5}}{10}$.

Let the midpoints of $AB$ and $PC$ be $K$ and $M$, respectively. Then it is easy to prove that the plane $ABM$ is the plane $\alpha$.
By the median length formula, we have
$$
\begin{array}{l}
A M^{2}=\frac{1}{2}\left(A P^{2}+A C^{2}\right)-\frac{1}{4} P C^{2} \\
=\frac{1}{2}\left(2^{2}+1^{2}\right)-\frac{1}{4} \times 2^{2}=\frac{3}{2} \\
\Rightarrow K M=\sqrt{A M^{2}-A K^{2}}=\frac{\sqrt{5}}{2} .
\end{array}
$$

Since the projection of line $PC$ on plane $\alpha$ is line $MK$, and given $CM=1, KC=\frac{\sqrt{3}}{2}$, we get
$$
\cos \angle K M C=\frac{K M^{2}+M C^{2}-K C^{2}}{2 K M \cdot M C}=\frac{3 \sqrt{5}}{10} .
$$

Therefore, the cosine of the angle formed by the edge $PC$ and the plane $\alpha$ is
$$
\text { } \frac{3 \sqrt{5}}{10} \text { . }
$$