13. In tetrahedron $ABCD$, $AB=2$, $AC=3$, $AD=4$, $BC=\sqrt{13}$, $BD=2\sqrt{5}$, $CD=5$. Then the volume of tetrahedron $ABCD$ is ( ).
(A) 3
(B) $2 \sqrt{3}$
(C) 4
(D) $3 \sqrt{3}$
(E) 6
13. C.

As shown in Figure 4, from the problem statement we know
$$
\begin{array}{l}
A B^{2}+A C^{2}=B C^{2} \\
\Rightarrow A B \perp A C .
\end{array}
$$

Similarly, $A B \perp A D$, $A C \perp A D$.
$$
\begin{array}{l}
\text { Therefore, } V_{\text {tetrahedron } B-A C D}=\frac{1}{3} S_{\triangle A C D} A B \\
=\frac{1}{6} A B \cdot A C \cdot A D=\frac{1}{6} \times 2 \times 3 \times 4 \\
=4 .
\end{array}
$$