Q 12-11-040JEE MainJEE Main 2026 (4 Apr, Shift 2)Easy
The de Broglie wavelength associated with an electron accelerated through a potential difference $V$ is $\lambda_e$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_p$. If their corresponding masses are $m_e$ and $m_p$, respectively, then the ratio of their de Broglie wavelengths $\left(\dfrac{\lambda_e}{\lambda_p}\right)$ is ______.
Answer: (A) $\sqrt{\dfrac{m_p}{m_e}}$
Concept. Same accelerating voltage and same charge size: $\lambda=\dfrac{h}{\sqrt{2meV}}\propto\dfrac{1}{\sqrt m}$.
Answer. $\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{m_p}{m_e}}$ (Option A).
Solution by Sreeraj P, M.Sc Physics
