Correct Option: D **Step 1: Recall the formula for de Broglie wavelength for a charged particle.** The de Broglie wavelength ($\lambda$) of a particle is given by: $$ \lambda = \frac{h}{p} $$ where $h$ is Planck's constant and $p$ is the momentum of the particle. **Step 2: Relate momentum to kinetic energy.** The kinetic energy ($K$) of a particle is given by $K = \frac{p^2}{2m}$, where $m$ is the mass of the particle. From this, the momentum can be expressed as: $$ p = \sqrt{2mK} $$ Substituting this into the de Broglie wavelength formula: $$ \lambda = \frac{h}{\sqrt{2mK}} $$ **Step 3: Relate kinetic energy to potential difference.** When a charged particle with charge $q$ is accelerated from rest through a potential difference $V$, its kinetic energy gained is $K = qV$. Substituting this into the de Broglie wavelength formula: $$ \lambda = \frac{h}{\sqrt{2mqV}} $$ **Step 4: Apply the formula to the electron.** For an electron, the charge is $e$ and its mass is $m_e$. When accelerated through potential difference $V$, its de Broglie wavelength is $\lambda_e$: $$ \lambda_e = \frac{h}{\sqrt{2m_e e V}} \quad (1) $$ **Step 5: Apply the formula to the proton.** For a proton, the charge is also $e$ (magnitude) and its mass is $m_p$. When accelerated through the same potential difference $V$, its de Broglie wavelength is $\lambda_p$: $$ \lambda_p = \frac{h}{\sqrt{2m_p e V}} \quad (2) $$ **Step 6: Determine the relationship between $\lambda_e$ and $\lambda_p$.** Divide equation (1) by equation (2): $$ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_e e V}}}{\frac{h}{\sqrt{2m_p e V}}} $$ $$ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{2m_p e V}{2m_e e V}} $$ $$ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} $$ **Step 7: Substitute the given mass relationship.** We are given that $m_p \approx 1836 m_e$. $$ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{1836 m_e}{m_e}} = \sqrt{1836} $$ **Step 8: Solve for $\lambda_p$.** $$ \lambda_p = \frac{\lambda_e}{\sqrt{1836}} $$ Thus, the de Broglie wavelength of the proton is $\frac{1}{\sqrt{1836}}$ times the de Broglie wavelength of the electron. The final answer is $\boxed{\text{D}}$