An electron accelerated with potential V has a de-Broglie wavelength λ associated with it. If the potential is changed to 4 V, the wavelength becomes

$λ/2$

The correct answer is Option (2) → $λ/2$

The de-Broglie wavelength of an electron is

$\lambda = \frac{h}{\sqrt{2 m e V}}$

If potential is changed from $V$ to $4V$:

$\lambda' = \frac{h}{\sqrt{2 m e (4V)}} = \frac{h}{2\sqrt{2 m e V}}$

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

Answer: The wavelength becomes $\frac{\lambda}{2}$