If the two particles of the same mass and same charge are accelerated with a potential difference of 120 V and 480 V respectively, the ratio of de Broglie's wavelengths associated with the two particles respectively is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2:1 \)
A particle of mass \( m \) and charge \( q \) accelerated through a potential difference \( V \) gains kinetic energy \( qV \), so its momentum is \( p = \sqrt{2mqV} \). The de Broglie wavelength is \( \lambda = h/p \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2mqV}} \;\Rightarrow\; \lambda \propto \frac{1}{\sqrt{V}} \)
The mass and charge are the same, so only the voltage changes:
\( \displaystyle \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}} = \sqrt{\frac{480}{120}} = \sqrt{4} = 2 \)
So \( \lambda_1 : \lambda_2 = 2:1 \).