The distance between the foci of an ellipse is equal to half of its minor axis then eccentricity is

$$\begin{gathered} Let ellipse be \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ Given that SS\text{'}=\frac{1}{2}(Length of minor axis) \\ \Rightarrow 2\mathrm{ae}=\frac{1}{2}\left(2\mathrm{b}\right) \\ \Rightarrow \mathrm{ae}=\mathrm{b} \\ \Rightarrow {\mathrm{a}}^2{\mathrm{e}}^2={\mathrm{b}}^2={\mathrm{a}}^2(1-{\mathrm{e}}^2) \\ \Rightarrow 2{\mathrm{e}}^2=1 \\ \Rightarrow \mathrm{e}=\frac{1}{\sqrt{2}} \end{gathered}$$