If $$OB$$ is the semi-minor axis of an ellipse, $$F_1$$ and $$F_2$$ are its foci and the angle between $$F_1B$$ and $$F_2B$$ is a right angle, then the square of the eccentricity of the ellipse is:

Consider the standard ellipse centered at the origin with the equation $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, where $$a$$ is the semi-major axis and $$b$$ is the semi-minor axis. Therefore, $$OB = b$$. The foci are at $$F_1(-c, 0)$$ and $$F_2(c, 0)$$, where $$c = \sqrt{a^2 - b^2}$$. The eccentricity $$e$$ is given by $$e = \frac{c}{a}$$, so $$c = e a$$.

Point $$B$$ is one end of the minor axis, so its coordinates are $$(0, b)$$. The vector from $$F_1$$ to $$B$$ is $$\overrightarrow{F_1B} = (0 - (-c), b - 0) = (c, b)$$. The vector from $$F_2$$ to $$B$$ is $$\overrightarrow{F_2B} = (0 - c, b - 0) = (-c, b)$$.

The angle between $$\overrightarrow{F_1B}$$ and $$\overrightarrow{F_2B}$$ is a right angle, so their dot product is zero:

$$\overrightarrow{F_1B} \cdot \overrightarrow{F_2B} = (c) \cdot (-c) + (b) \cdot (b) = -c^2 + b^2 = 0$$

This gives:

$$b^2 - c^2 = 0 \quad \Rightarrow \quad b^2 = c^2$$

For an ellipse, $$c^2 = a^2 - b^2$$. Substituting $$b^2 = c^2$$:

$$b^2 = a^2 - b^2$$

Adding $$b^2$$ to both sides:

$$b^2 + b^2 = a^2 \quad \Rightarrow \quad 2b^2 = a^2 \quad \Rightarrow \quad \frac{b^2}{a^2} = \frac{1}{2}$$

The square of the eccentricity is $$e^2 = 1 - \frac{b^2}{a^2}$$. Substituting $$\frac{b^2}{a^2} = \frac{1}{2}$$:

$$e^2 = 1 - \frac{1}{2} = \frac{1}{2}$$

Hence, the square of the eccentricity is $$\frac{1}{2}$$. Comparing with the options, Option C is $$\frac{1}{2}$$. Therefore, the correct answer is Option C.