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
Answer & explanation
Correct answer: option 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 $\frac{\lambda}{2}$