9. A hemispherical container with a base contains three small balls that are pairwise externally tangent. If the radii of these three small balls are all 1, and each small ball is tangent to the base and the spherical surface of the hemisphere, then the radius of the hemisphere $R=$


9. $\frac{3+\sqrt{21}}{3}$.

The centers of three small spheres $O_{1}, O_{2}, O_{3}$ form an equilateral triangle with a side length of 2, then the radius of the circumcircle is $\frac{2 \sqrt{3}}{3}$.

Let the center of the hemisphere be $O$, and the small sphere $O_{1}$ touches the bottom of the hemisphere at point $A$. As shown in Figure 7, a section of the hemisphere is made through points $O, O, A$, with the semicircle $\odot O$ having radius $OC \perp OA$, and $O_{1}B \perp OC$ at point $B$. Then
$$
OA = O_{1}B = \frac{2 \sqrt{3}}{3}.
$$

In the right triangle $\triangle OAO_{1}$, by
$$
(R-1)^{2} = \left(\frac{2 \sqrt{3}}{3}\right)^{2} + 1^{2} \Rightarrow R = \frac{3+\sqrt{21}}{3}.
$$