11. Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ lie on the ellipse $C$. If $x_{1}+x_{2}=\frac{1}{2}$, and $\overrightarrow{A F_{2}}=\lambda \overrightarrow{F_{2} B}$, find the value of $\lambda$.
11. From $\overrightarrow{F_{2} A}=\lambda \overrightarrow{F_{2} B}$, we know that points $A$, $B$, and $F_{2}$ are collinear.
If line $A B \perp x$-axis, then $x_{1}=x_{2}=1$, which does not meet the requirement.
If line $A B$ has a slope (let it be $k$), then the line $l_{A B}: y=k(x-1)$.
From $\left\{\begin{array}{l}\frac{x^{2}}{4}+\frac{y^{2}}{3}=1, \\ y=k(x-1)\end{array}\right.$ we get
$$
\left(4 k^{2}+3\right) x^{2}-8 k^{2} x+4 k^{2}-12=0 \text {. }
$$

From $x_{1}+x_{2}=\frac{8 k^{2}}{4 k^{2}+3}=\frac{1}{2}$, we get $k^{2}=\frac{1}{4}$.
Substituting $k^{2}=\frac{1}{4}$ into equation (1) yields
$4 x^{2}-2 x-11=0$.
Solving for $x$ gives $x=\frac{1 \pm 3 \sqrt{5}}{4}$.
$$
\begin{array}{l} 
\text { Also, } \overrightarrow{A F_{2}}=\left(1-x_{1},-y_{1}\right), \\
\overrightarrow{F_{2} B}=\left(x_{2}-1 ; y_{2}\right), \\
\overrightarrow{A F_{2}}=\lambda \overrightarrow{F_{2} B}, \lambda=\frac{1-x_{1}}{x_{2}-1}, \\
\text { Therefore, } \lambda=\frac{3 \sqrt{5}}{2} \text { or } \frac{3+\sqrt{5}}{2} .
\end{array}
$$