If the kinetic energy of an electron is increased four times, the wavelength of the de-Broglie wave associated with it would become

For any particle of mass $$m$$ moving with momentum $$p$$, the de-Broglie wavelength is

$$\lambda = \frac{h}{p}$$

For an electron whose speed is much less than the speed of light, the classical relation between kinetic energy $$K$$ and momentum $$p$$ is

$$K = \frac{p^{2}}{2m} \; \; \Longrightarrow \; \; p = \sqrt{2mK}$$

Substituting this $$p$$ in the de-Broglie formula,

$$\lambda = \frac{h}{\sqrt{2mK}}$$

Thus, the wavelength is inversely proportional to the square root of the kinetic energy:

$$\lambda \propto \frac{1}{\sqrt{K}}$$

If the kinetic energy is increased four times, i.e. $$K' = 4K$$, then

$$\lambda' = \frac{\lambda}{\sqrt{4}} = \frac{\lambda}{2}$$

Therefore, the de-Broglie wavelength becomes one-half of its original value.

Option B which is: half