5. As shown in Figure 1, in the right triangle $\triangle ABC$, the two legs are $AC=2, BC=3$, and $P$ is a point on the hypotenuse $AB$. The triangle is folded along $CP$ to form a right dihedral angle $A-CP-B$. When $AB=\sqrt{7}$, the value of the dihedral angle $P-AC-B$ is $\qquad$.
5. $\arctan \sqrt{2}$.

As shown in Figure 2, draw $P D \perp A C$ at point $D$, draw $P E \perp C P$ intersecting $B C$ at point $E$, and connect $D E$.

Since $A-C P-B$ is a right dihedral angle, i.e., plane $A C P \perp$ plane $C P B$, therefore,
$P E \perp$ plane $A C P$.
Also, $P D \perp A C$, so by the theorem of three perpendiculars, we know
$D E \perp A C$.
Thus, $\angle P D E$ is the plane angle of the dihedral angle $P-A C-B$.
Let $\angle B C P=\theta$. Then
$$
\begin{array}{l}
\cos \angle E C D=\cos \theta \cdot \cos \left(\frac{\pi}{2}-\theta\right) \\
=\sin \theta \cdot \cos \theta .
\end{array}
$$

By the cosine rule,
$\cos \angle A C B=\frac{2^{2}+3^{2}-(\sqrt{7})^{2}}{2 \times 2 \times 3}=\frac{1}{2}$.
Thus, $\sin \theta \cdot \cos \theta=\frac{1}{2} \Rightarrow \sin 2 \theta=1$.
Since $0<2 \theta<\pi$, then $\theta=\frac{\pi}{4}$.
Let $C P=a$. Then $P D=\frac{\sqrt{2}}{2} a, P E=a$.
Therefore, $\tan \angle P D E=\frac{P E}{P D}=\sqrt{2}$
$\Rightarrow \angle P D E=\arctan \sqrt{2}$.