(11) For the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{b^{2}}=1(b>0)$, the circle with the eccentricity as the radius and the right focus as the center is tangent to the asymptotes of the hyperbola. Then the eccentricity of the hyperbola is $\qquad$
$\qquad$.
11 It is known that $a=2, c=\sqrt{4+b^{2}}, e=\frac{\sqrt{4+b^{2}}}{2}$. The equations of the asymptotes are $\frac{x}{2} \pm$ $\frac{y}{b}=0$, or $b x \pm 2 y=0$. The distance $d=b$ from the right focus $\left(\sqrt{4+b^{2}}, 0\right)$ to the asymptote, thus we get the equation $b=\frac{\sqrt{4+b^{2}}}{2}$. Solving the equation, we get $b=\frac{2}{3} \sqrt{3}, e=\frac{\sqrt{4+\frac{4}{3}}}{2}=$ $\frac{2 \sqrt{3}}{3}$. Therefore, the answer is $\frac{2 \sqrt{3}}{3}$.