Correct Option: B To determine the ratio of de Broglie wavelengths for an electron and a proton accelerated through the same potential difference, we utilize the relationship between kinetic energy, momentum, and de Broglie wavelength. **Step 1: Relate kinetic energy to potential difference.** When a charged particle of charge $q$ is accelerated through a potential difference $V$, its kinetic energy $K$ is given by: $K = qV$ For an electron, the charge is $e$, so its kinetic energy $K_e$ is: $K_e = eV$ For a proton, the charge is also $e$, so its kinetic energy $K_p$ is: $K_p = eV$ Since both are accelerated through the same potential difference $V$ and have the same magnitude of charge $e$, their kinetic energies will be equal: $K_e = K_p = eV$ **Step 2: Relate kinetic energy to momentum.** The kinetic energy $K$ of a particle with mass $m$ and momentum $p$ is given by: $K = \frac{p^2}{2m}$ This implies that the momentum $p$ can be expressed as: $p = \sqrt{2mK}$ For the electron, its momentum $p_e$ is: $p_e = \sqrt{2m_e K_e} = \sqrt{2m_e (eV)}$ For the proton, its momentum $p_p$ is: $p_p = \sqrt{2m_p K_p} = \sqrt{2m_p (eV)}$ **Step 3: Apply the de Broglie wavelength formula.** The de Broglie wavelength $\lambda$ of a particle with momentum $p$ is given by: $\lambda = \frac{h}{p}$ where $h$ is Planck's constant. For the electron, its de Broglie wavelength $\lambda_e$ is: $\lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2m_e eV}}$ For the proton, its de Broglie wavelength $\lambda_p$ is: $\lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2m_p eV}}$ **Step 4: Calculate the ratio $\lambda_e / \lambda_p$.** Now, we can find the ratio of their de Broglie wavelengths: $\frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_e eV}}}{\frac{h}{\sqrt{2m_p eV}}}$ $\frac{\lambda_e}{\lambda_p} = \frac{h}{\sqrt{2m_e eV}} \times \frac{\sqrt{2m_p eV}}{h}$ $\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{2m_p eV}{2m_e eV}}$ $\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}$ Thus, the ratio of the de Broglie wavelength of the electron to that of the proton is $\sqrt{\frac{m_p}{m_e}}$. The final answer is $\boxed{\text{B}}$. **Analysis of Options:** A) $\sqrt{m_e / m_p}$: This would be the ratio if the electron had greater mass or smaller momentum, which is incorrect. B) $\sqrt{m_p / m_e}$: This is the correct ratio as derived. C) $m_e / m_p$: This ratio would be obtained if the de Broglie wavelength was inversely proportional to mass, which is incorrect as it's inversely proportional to the square root of mass (for same kinetic energy). D) $m_p / m_e$: This ratio would be obtained if the de Broglie wavelength was directly proportional to mass, which is incorrect.