Ratio of De Broglie Wavelengths of Electron and Proton

To find the ratio of the de Broglie wavelengths of an electron and a proton accelerated through the same potential difference $V$, we use the fundamental principles of quantum mechanics and classical kinetic energy.

1. Relation between Kinetic Energy and Potential Difference: When a charged particle carrying a charge $q$ is accelerated through a potential difference $V$, the kinetic energy $K$ gained by the particle is given by: $K = qV$

2. De Broglie Wavelength Equation: The de Broglie wavelength $\lambda$ of a particle with momentum $p$ is defined as: $\lambda = \frac{h}{p}$

Since momentum $p$ is related to kinetic energy $K$ by $p = \sqrt{2mK}$, where $m$ is the mass of the particle, we can express $\lambda$ as: $\lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$

1. Application to Electron and Proton: Both the electron and the proton carry a charge of equal magnitude, $q_e = q_p = e$.

• For the electron (mass $m_e$): $\lambda_e = \frac{h}{\sqrt{2 m_e e V}}$

• For the proton (mass $m_p$): $\lambda_p = \frac{h}{\sqrt{2 m_p e V}}$

1. Calculating the Ratio $\frac{\lambda_e}{\lambda_p}$: Taking the ratio of the two wavelengths: $\frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2 m_e e V}}}{\frac{h}{\sqrt{2 m_p e V}}} = \sqrt{\frac{m_p}{m_e}}$

Thus, the correct option is A, which corresponds to $\sqrt{\frac{m_p}{m_e}}$.