7. Given the hyperbola $C: x^{2}-\frac{y^{2}}{3}=1$, with left and right foci $F_{1}$ 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 inradius of $\triangle F_{1} P Q$ is
7. $\sqrt{7}-1$.

As shown in Figure 4.
By the properties of the hyperbola, we have
$$
\begin{array}{l}
F_{1} F_{2}=2 \sqrt{1+3}=4, \\
P F_{1}-P F_{2}=Q F_{1}-Q F_{2}=2 . \\
\text { Given } \angle F_{1} P Q=90^{\circ} \\
\Rightarrow P F_{1}^{2}+P F_{2}^{2}=F_{1} F_{2}^{2} \\
\Rightarrow P F_{1}+P F_{2} \\
\quad=\sqrt{2\left(P F_{1}^{2}+P F_{2}^{2}\right)-\left(P F_{1}-P F_{2}\right)^{2}} \\
\quad=\sqrt{2 \times 4^{2}-2^{2}}=2 \sqrt{7} .
\end{array}
$$

Thus, the inradius of the right triangle $F_{1} P Q$ is
$$
\begin{aligned}
r & =\frac{1}{2}\left(F_{1} P+P Q-F_{1} Q\right) \\
& =\frac{1}{2}\left(P F_{1}+P F_{2}\right)-\frac{1}{2}\left(Q F_{1}-Q F_{2}\right) \\
& =\sqrt{7}-1 .
\end{aligned}
$$