If the kinetic energy of a free electron doubles. Its de Broglie wavelength changes by the factor:

The de Broglie wavelength is given by $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$, where $K$ is the kinetic energy.Thus, $\lambda \propto \frac{1}{\sqrt{K}}$.If kinetic energy doubles ($K^{\prime }=2K$):
$${\lambda }^{\prime }=\frac{h}{\sqrt{2m(2K)}}=\frac{1}{\sqrt{2}}\left(\frac{h}{\sqrt{2mK}}\right)=\frac{\lambda }{\sqrt{2}}$$
The factor is $1/\sqrt{2}$.