7. If the distance between the two directrices of an ellipse is twice the distance between the two foci, then its eccentricity $e=$ $\qquad$

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7. $\frac{\sqrt{2}}{2}$.

Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, with foci at $(-c, 0),(c, 0)$, and the equations of the directrices be $x=-\frac{a^{2}}{c}, x=\frac{a^{2}}{c}$. The distance between the two directrices is $2 \cdot \frac{a^{2}}{c}$, and the focal distance is $2 c$. According to the given condition, $2 \cdot \frac{a^{2}}{c}=4 c \Rightarrow e=\frac{c}{a}=\frac{\sqrt{2}}{2}$.