Example 14 (2003 National High School Competition Question) A tetrahedron has all edges of length $\sqrt{2}$, and its four vertices lie on the same sphere. The surface area of this sphere is ( ).
A. $3 \pi$
B. $4 \pi$
C. $3 \sqrt{3} \pi$
D. $6 \pi$
Solution: Choose A. Reason: Construct a cube with an edge length of 1, as shown in Figure 7-13. Connect $A B_{1}$, $A D_{1}$, $A C$, $C D_{1}$, $C B_{1}$, and $B_{1} D_{1}$. Then the tetrahedron $A C B_{1} D_{1}$ is the tetrahedron that meets the problem's requirements, and its circumscribed sphere's diameter is the length of the cube's diagonal.

Let the radius of this circumscribed sphere be $R$, then $2 R=A C_{1}=\sqrt{3}$, so the surface area $S$ of the circumscribed sphere of this regular tetrahedron is $4 \pi R^{2}=3 \pi$