Let $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a > b$$ be an ellipse, whose eccentricity is $$\frac{1}{\sqrt{2}}$$ and the length of the latus rectum is $$\sqrt{14}$$. Then the square of the eccentricity of $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ is:

Consider an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a > b$$, eccentricity $$e_1 = \frac{1}{\sqrt{2}}$$, and latus rectum length $$= \sqrt{14}$$. Since $$e_1^2 = 1 - \frac{b^2}{a^2}$$, we have $$\frac{1}{2} = 1 - \frac{b^2}{a^2}$$, which gives $$\frac{b^2}{a^2} = \frac{1}{2}$$ and hence $$b^2 = \frac{a^2}{2}$$. The length of the latus rectum is $$\frac{2b^2}{a} = \sqrt{14}$$; substituting $$b^2 = \frac{a^2}{2}$$ yields $$\frac{2\cdot (a^2/2)}{a} = a = \sqrt{14}$$. Therefore, $$a = \sqrt{14}$$, $$a^2 = 14$$, and $$b^2 = 7$$.

Now consider the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, which becomes $$\frac{x^2}{14} - \frac{y^2}{7} = 1$$ when we use the above values. Its eccentricity satisfies $$e_2^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{7}{14} = 1 + \frac{1}{2} = \frac{3}{2}$$. The answer is Option C: $$\frac{3}{2}$$.