5. In a regular tetrahedron $P-ABC$, $AB=1, AP=2$. A plane $\alpha$ passing through $AB$ divides its volume equally. Then the cosine of the angle formed by edge $PC$ and plane $\alpha$ is $\qquad$
Let the midpoint of $PC$ be $D$, then the plane $\alpha$ is the plane $ABD$. Let $E$ be the midpoint of $AB$, and $O$ be the center of the base $ABC$,
$$
\begin{array}{l}
CE=\frac{\sqrt{3}}{2} \Rightarrow CO=\frac{\sqrt{3}}{3} \Rightarrow PO=\sqrt{4-\frac{1}{3}}=\frac{\sqrt{33}}{3}. \\
\cos \angle PCB=\frac{1}{4} \Rightarrow BD^{2}=1+1-2 \times \frac{1}{4}=\frac{3}{2} \\
\Rightarrow BD=\frac{\sqrt{6}}{2} \Rightarrow DE=\frac{\sqrt{5}}{2}.
\end{array}
$$

In $\triangle CDE$, draw $CF \perp DE, PO \perp$ plane $ABC$
$\Rightarrow AB \perp$ plane $CDE \Rightarrow CF \perp$ plane $ABD$.
$$
\begin{array}{l}
\text{By } S_{\triangle CDE}=\frac{1}{2} CE \cdot \frac{1}{2} PO=\frac{1}{2} DE \cdot CF \Rightarrow CF=\frac{CE \cdot \frac{1}{2} PO}{DE}=\frac{\sqrt{55}}{10} \\
\Rightarrow \sin \angle CDF=\frac{\sqrt{55}}{10} \Rightarrow \cos \angle CDF=\frac{3 \sqrt{5}}{10}.
\end{array}
$$

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