An electron accelerated with potential V has a de-Broglie wavelength \( \lambda \) 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) → \( \lambda/2 \)
An electron accelerated through a potential \( V \) gains kinetic energy \( eV \), so its momentum is \( p = \sqrt{2meV} \) and its de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}} \;\Rightarrow\; \lambda \propto \frac{1}{\sqrt{V}} \)
Changing \( V \) to \( 4V \):
\( \displaystyle \frac{\lambda'}{\lambda} = \sqrt{\frac{V}{4V}} = \frac{1}{2} \;\Rightarrow\; \lambda' = \frac{\lambda}{2} \)
So four times the potential halves the wavelength.