2. Given a unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the midpoints of edges $A B$, $A_{1} D_{1}$, $A_{1} B_{1}$, and $B C$ are $L$, $M$, $N$, and $K$ respectively. Then the radius of the inscribed sphere of the tetrahedron $L M N K$ is $\qquad$
2. $\frac{\sqrt{3}-\sqrt{2}}{2}$.

Let the radius of the inscribed sphere of tetrahedron $L M N K$ be $r$. Since $\angle N L K=\angle L N M=90^{\circ}$, we have $S_{\triangle L M N}=S_{\triangle L N K}=\frac{1}{2} \times 1 \times \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{4}$.
Also, $\angle M L K=\angle M N K=90^{\circ}$, so $S_{\triangle C M K}=S_{\triangle M N K}=\frac{1}{2} \times \frac{\sqrt{6}}{2} \times \frac{\sqrt{2}}{2}=\frac{\sqrt{3}}{4}$. Therefore, the surface area of tetrahedron $L M N K$ is $S=2\left(\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{4}\right)=\frac{\sqrt{2}+\sqrt{3}}{2}$.
Since $K L \perp$ the base $L M N$, we have
$$
\begin{array}{l}
V_{\text {tetrahedron } L M N K}=\frac{1}{3} K L \cdot S_{\triangle L M N} \\
=\frac{1}{3} \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{4}=\frac{1}{12} .
\end{array}
$$

From $V_{\text {tetrahedron } L M N K}=\frac{1}{3} r S \Rightarrow r=\frac{\sqrt{3}-\sqrt{2}}{2}$.