7. Given the circle $\Gamma:(x-2)^{2}+y^{2}=3$, let the line $l_{1}: x-m y-1=0$ intersect the circle $E$ at points $A, C$, and the line $l_{2}: m x+y-m=0$ intersect the circle $E$ at points $B, D$. Segments $A B$ and $C D$ are located above and below the $x$-axis, respectively. When the slope of $C D$ is -1, the length of segment $A B$ is $\qquad$ .
7. $2 \sqrt{2}$.

Take the midpoint $M$ of $C D$, and connect $F M, E M, E C$. Since the slope of $C D$ is -1, we have $\angle F E M=45^{\circ}$. Applying the cosine rule in $\triangle F E M$, we get
$$
F M^{2}=F E^{2}+E M^{2}-2 F E \cdot E M \cos 45^{\circ},
$$

Also, $F M^{2}=E C^{2}-E M^{2}$. Substituting the data, we get $E M=\sqrt{2}, F M=1$. Thus, $F M \perp F E$, so $\angle F M D=45^{\circ}$, and hence $\angle F C D=22.5^{\circ}$. Therefore, $F C=2 \cos 22.5^{\circ}, F D=2 \sin 22.5^{\circ}$. According to the power of a point theorem, we have
$$
D F \cdot F B=C F \cdot F A=E C^{2}-E F^{2}=2,
$$

Thus,
$$
A B^{2}=A F^{2}+B F^{2}=\frac{1}{\sin ^{2} 22.5^{\circ}}+\frac{1}{\cos ^{2} 22.5^{\circ}}=\frac{4}{\sin ^{2} 45^{\circ}}=8 .
$$

Therefore, the length of segment $A B$ is $2 \sqrt{2}$.