7. In the Cartesian coordinate system $x O y$, $F$ is the focus of the parabola $\Gamma: y^{2}=2 p x(p>0)$. Point $B$ is on the $x$-axis and to the right of point $F$. Point $A$ is on $\Gamma$, and $|A F|=$ $|B F|$. The second intersection points of lines $A F$ and $A B$ with $\Gamma$ are $M$ and $N$, respectively. If $\angle A M N=90^{\circ}$, then the slope of line $A F$ is $\qquad$
7. $\sqrt{3}$.

Assume point $A$ is in the first quadrant.
It is easy to know that $F\left(\frac{p}{2}, 0\right)$.
Then set $l_{A F}: x=m y+\frac{p}{2}$.
Let $A\left(x_{1}, y_{1}\right), M\left(x_{2}, y_{2}\right), N\left(x_{3}, y_{3}\right)$.
From $\left\{\begin{array}{l}x=m y+\frac{p}{2}, \\ y^{2}=2 p x,\end{array}\right.$ eliminating $x$ we get
$y^{2}-2 p m y-p^{2}=0$.
Thus, $y_{1} y_{2}=-p^{2}$.
By the second definition of a parabola, we know
$|A F|=x_{1}+\frac{p}{2}$.
Combining $|A F|=|B F|$, we know $B\left(x_{1}+p, 0\right)$.
Hence $k_{A B}=-\frac{y_{1}}{p}$
$\Rightarrow l_{A D}: y-y_{1}=-\frac{y_{1}}{p}\left(x-x_{1}\right)$
$\Rightarrow x=-\frac{p}{y_{1}} y+x_{1}+p$.
From $\left\{\begin{array}{l}x=-\frac{p}{y_{1}} y+x_{1}+p, \\ y^{2}=2 p x,\end{array}\right.$ eliminating $x$ we get
$y^{2}+\frac{2 p^{2}}{y_{1}} y-2 p\left(x_{1}+p\right)=0$.
Thus, $y_{1}+y_{3}=-\frac{2 p^{2}}{y_{1}}=2 y_{2}$.
Let the midpoint of segment $A N$ be $D$. Then the y-coordinate of point $D$ is the same as that of $M$.
Hence $M D / / x$ axis
$$
\begin{array}{l}
\Rightarrow \triangle A M D \backsim \triangle A F B \\
\Rightarrow|A M|=|M D| .
\end{array}
$$

In the right triangle $\triangle A M N$, $|M D|=|A D|$, so $\triangle A M D$ is a right triangle.

Therefore, $\angle A F B=\angle A M D=60^{\circ}$, which means the slope of line $A F$ is $\sqrt{3}$.