6. A line with slope $k$ is drawn through one of the foci of the ellipse $x^{2}+2 y^{2}=3$, intersecting the ellipse at points $A, B$. If $|A B|=2$, then $|k|=$ $\qquad$ .
Given the ellipse $\frac{x^{2}}{3}+\frac{y^{2}}{\frac{3}{2}}=1 \Rightarrow c=\frac{\sqrt{6}}{2}$, the line $A B: y=k\left(x-\frac{\sqrt{6}}{2}\right)$, substituting into the ellipse equation yields $x^{2}+2 k^{2}\left(x-\frac{\sqrt{6}}{2}\right)^{2}-3=0 \Rightarrow\left(2 k^{2}+1\right) x^{2}-2 \sqrt{6} k^{2} x+3 k^{2}-3=0$.
Thus, $|A B|=\sqrt{k^{2}+1} \cdot \frac{\sqrt{24 k^{4}-12\left(2 k^{2}+1\right)\left(k^{2}-1\right)}}{2 k^{2}+1}=\frac{2 \sqrt{3}\left(k^{2}+1\right)}{2 k^{2}+1}=2$ $\Rightarrow k^{2}=\frac{\sqrt{3}-1}{2-\sqrt{3}}=1+\sqrt{3} \Rightarrow|k|=\sqrt{1+\sqrt{3}}$