已知幂函数$y=f\left( x \right)$的图象过点$\left( 2,\sqrt{2} \right)$，则$f\left( 16 \right)=$                 ．

由题意令$y=f(x)={{x}^{a}}$，由于图象过点$(2,\sqrt{2})$，

得$\sqrt{2}={{2}^{a}}$，$a=\dfrac{1}{2}$

$\therefore y=f(x)={{x}^{\tfrac{1}{2}}}$

$\therefore f(16)={16}^{\tfrac{1}{2}}=4$．

故答案为：$4$