14. Two spheres $O_{1}$ and $O_{2}$, both with a radius of 1, are tangent to each other and are also tangent to the two half-planes of the right dihedral angle $\alpha-l-\beta$. There is another smaller sphere $O$ with radius $r (r<1)$ that is also tangent to the two half-planes of the dihedral angle, and is externally tangent to both spheres $O_{1}$ and $O_{2}$. Then the value of $r$ is $\qquad$
$14 \cdot 3-\sqrt{7}$. 
It is known that the centers of the three spheres $O, O_{1}, O_{2}$ are all on the bisecting plane $\gamma$ of the right dihedral angle $\alpha-l-\beta$, as shown in Figure 8.

In the plane $\gamma$, draw $O_{1} A \perp l, O_{2} B \perp l, O C \perp l$, with the feet of the perpendiculars being $A, B, C$ respectively. Let the spheres $O_{1}$ and $O_{2}$ be tangent at point $T$. Then
$$
\begin{array}{l}
O O_{1}=O O_{2}=r+1, O_{1} A=O_{2} B=\sqrt{2}, \\
O C=\sqrt{2} r, O_{1} T=O_{2} T=1 .
\end{array}
$$

Thus, $O T=\sqrt{2}-\sqrt{2} r$.
In the right triangle $\triangle O T O_{1}$, by $O O_{1}^{2}=O T^{2}+O_{1} T^{2}$, we get $(r+1)^{2}=(\sqrt{2}-\sqrt{2} r)^{2}+1$.
Noting that $r<1$, solving yields $r=3-\sqrt{7}$.