The de Broglie wavelength of an electron having kinetic energy $$E$$ is $$\lambda$$. If the kinetic energy of electron becomes $$\frac{E}{4}$$, then its de-Broglie wavelength will be:

The de Broglie wavelength: $$\lambda = \frac{h}{\sqrt{2mE}}$$

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

If KE becomes $$E/4$$:

$$\lambda' = \frac{h}{\sqrt{2m(E/4)}} = \frac{h}{\frac{1}{2}\sqrt{2mE}} = 2 \times \frac{h}{\sqrt{2mE}} = 2\lambda$$

The new wavelength is $$2\lambda$$.

This matches option 4.

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