If the minimum value of $f\left(x\right)=\frac{5x^2}{2}+\frac{\alpha }{x^5}, x>0,$is 14, then the value of $\alpha$ is equa to :

$$\begin{gathered} \frac{\frac{x^2}{2}+\frac{x^2}{2}+\frac{x^2}{2}+\frac{x^2}{2}+\frac{x^2}{2}+\frac{\alpha }{2x^5}+\frac{\alpha }{2x^5}}{7}\ge {\left({\left(\frac{x^2}{2}\right)}^5{\left(\frac{\alpha }{2x^5}\right)}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$7$}\right.} \\ \Rightarrow \frac{x^2}{2}+\frac{x^2}{2}+\frac{x^2}{2}+\frac{x^2}{2}+\frac{x^2}{2}+\frac{\alpha }{2x^5}+\frac{\alpha }{2x^5}\ge 7{\left(\frac{{\alpha }^2}{2^7}\right)}^{\frac{1}{7}} \\ \frac{7·{\left(\alpha \right)}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}}}{2}=14 \\ \Rightarrow {\left({\alpha }^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}=2^2 \\ \Rightarrow \alpha ={\left(2^2\right)}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=2^7 \\ \Rightarrow \alpha =128 \end{gathered}$$