1. For a free electron, the de Broglie wavelength is $\lambda = \frac{h}{p}$ where $h$ is Planck’s constant and $p$ is momentum.

2. For a non-relativistic electron, kinetic energy is $K = \frac{p^2}{2m}$ so $p = \sqrt{2mK}$

3. Substitute into the de Broglie relation: $\lambda = \frac{h}{\sqrt{2mK}}$

4. Thus, $\lambda \propto \frac{1}{\sqrt{K}}$

5. If the kinetic energy doubles, $K' = 2K$. Then the new wavelength is $\lambda' = \frac{h}{\sqrt{2m(2K)}} = \frac{h}{\sqrt{2}\,\sqrt{2mK}} = \frac{\lambda}{\sqrt{2}}$

6. Therefore, the wavelength changes by the factor $\frac{\lambda'}{\lambda} = \frac{1}{\sqrt{2}}$

7. Checking options:

   • A: $2$ ❌
   • B: $\frac{1}{2}$ ❌
   • C: $\sqrt{2}$ ❌
   • D: $\frac{1}{\sqrt{2}}$ ✅

Hence, the correct answer is D.