1. We are given $f(x)=\frac{5x^2}{2}+\frac{\alpha}{x^5},\qquad x>0$ and its minimum value is $14$.

We need to find $\alpha$.

2. For minimum, differentiate: $f'(x)=5x-5\alpha x^{-6}=5x-\frac{5\alpha}{x^6}$

Set $f'(x)=0$: $5x-\frac{5\alpha}{x^6}=0$ $x=\frac{\alpha}{x^6}$ $x^7=\alpha$

So at the critical point, $\alpha=x^7$

3. Substitute this into $f(x)$: $f(x)=\frac{5x^2}{2}+\frac{x^7}{x^5}$ $f(x)=\frac{5x^2}{2}+x^2$ $f(x)=\frac{7x^2}{2}$

Since this minimum is given as $14$, $\frac{7x^2}{2}=14$ $7x^2=28$ $x^2=4$

Since $x>0$, we get $x=2$

4. Now compute $\alpha$: $\alpha=x^7=2^7=128$

5. Check that this is indeed a minimum: $f''(x)=5+30\alpha x^{-7}$ At $x^7=\alpha$, $f''(x)=5+30=35>0$ So the critical point gives a minimum.

Hence, $\boxed{\alpha=128}$

Option $\boxed{\text{C}}$ is correct.