7. The equation of the hyperbola $C$ is $x^{2}-\frac{y^{2}}{3}=1$, with the left and right foci being $F_{1}$ and $F_{2}$, respectively. A line is drawn through point $F_{2}$ intersecting the right branch of the hyperbola $C$ at points $P$ and $Q$, such that $\angle F_{1} P Q=90^{\circ}$. Then the radius of the incircle of $\triangle F_{1} P Q$ is $\qquad$.
Consider the triangle $\triangle F_{1} P Q$ with its incircle touching the sides $P F_{1}, Q F_{1}, P Q$ at points $A, B, D$ respectively. Then we have:
$$
\begin{array}{l}
\left|P F_{1}\right|-\left|P F_{2}\right|=\left|Q F_{1}\right|-\left|Q F_{2}\right| \\
\Rightarrow\left|Q F_{1}\right|-\left|P F_{1}\right|=\left|Q F_{2}\right|-\left|P F_{2}\right| \\
\Rightarrow|Q B|-|P A|=\left|Q F_{2}\right|-\left|P F_{2}\right| \\
\Rightarrow|Q D|-|P D|=\left|Q F_{2}\right|-\left|P F_{2}\right| \\
\Rightarrow|Q D|-\left|Q F_{2}\right|=|P D|-\left|P F_{2}\right| \\
\Rightarrow-\left|D F_{2}\right|=\left|D F_{2}\right| \Rightarrow D \text { coincides with } F_{2}.
\end{array}
$$

Notice that point $P$ lies on the circle with diameter $F_{1} F_{2}$. Without loss of generality, assume point $P$ is in the first quadrant. Solving the system of equations:
$$
\left\{\begin{array}{l}x^{2}-\frac{y^{2}}{3}=1, \\ x^{2}+y^{2}=4\end{array} \Rightarrow\left\{\begin{array}{l}x^{2}=\frac{7}{4}, \\ y^{2}=\frac{9}{4}\end{array}\right.\right.
$$
$\Rightarrow P\left(\frac{\sqrt{7}}{2}, \frac{3}{2}\right)$. Therefore, the inradius of $\triangle F_{1} P Q$ is
$$
r=\left|P F_{2}\right|=\sqrt{\left(\frac{\sqrt{7}}{2}-2\right)^{2}+\frac{9}{4}}=\sqrt{8-2 \sqrt{7}}=\sqrt{7}-1 .
$$