If the KE of free electron doubles, its deBroglie wavelength changes by a factor

The relationship between de Broglie wavelength and kinetic energy is given by

$\lambda = \frac{h}{\sqrt{2m\left(KE\right)}}$

If kinetic energy (KE) is doubled, 

$$\begin{gathered} \lambda \text{'} = \frac{h}{\sqrt{2m\left(2KE\right)}} \\ \Rightarrow \lambda \text{'} = \frac{h}{\sqrt{2\left(2m\right(KE)}} \\ \Rightarrow \lambda \text{'} = \frac{\lambda }{\sqrt{2}} \end{gathered}$$

Its de Broglie wavelength changes by a factor of $\frac{1}{\sqrt{2}}.$