1. Given ellipse

   $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b$

   Its eccentricity is

   $e=\frac{1}{\sqrt{2}}$

   For an ellipse,

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

   So,

   $\frac{1}{2}=1-\frac{b^2}{a^2}$

   which gives

   $\frac{b^2}{a^2}=\frac{1}{2}$

   Hence,

   $b^2=\frac{a^2}{2}$

2. Use latus rectum length

   For the ellipse, length of latus rectum is

   $\frac{2b^2}{a}$

   Given,

   $\frac{2b^2}{a}=\sqrt{14}$

   Substitute $b^2=\frac{a^2}{2}$:

   $\frac{2\cdot \frac{a^2}{2}}{a}=\sqrt{14}$

   $a=\sqrt{14}$

   Therefore,

   $a^2=14, \qquad b^2=\frac{14}{2}=7$

3. Now consider the hyperbola

   $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

   For this hyperbola, eccentricity $E$ satisfies

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

   Using $a^2=14$ and $b^2=7$:

   $E^2=1+\frac{7}{14}=1+\frac{1}{2}=\frac{3}{2}$

4. Check options

   $\frac{3}{2}$

   corresponds to Option C.

Final Answer: $\boxed{\frac{3}{2}}$