Electrons in an experiment are accelerated by a voltage of 36 kV. If the voltage is increased to 144 kV, then the de Broglie wavelength associated with the electron would
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → decreased by 2-times
An electron accelerated through a potential difference \( V \) gains kinetic energy \( eV \), so its momentum is \( \sqrt{2meV} \) and its de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{\sqrt{2meV}} \;\Rightarrow\; \lambda \propto \frac{1}{\sqrt{V}} \)
\( \displaystyle \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{36}{144}} = \frac{1}{2} \)
The wavelength becomes half, so it decreases by 2 times.