Correct Option: D\n\n**Step-by-step Solution:**\n\n1. **De Broglie Wavelength Formula:**\n The de Broglie wavelength \\(\\lambda\\) associated with a particle of momentum \\(p\\) is given by:\n \n \\(\\lambda = \\frac{h}{p}\\)\n \n where \\(h\\) is Planck's constant.\n\n2. **Momentum and Kinetic Energy Relation:**\n The kinetic energy \\(K\\) of a particle with mass \\(m\\) and momentum \\(p\\) is given by:\n \n \\(K = \\frac{p^2}{2m}\\)\n \n From this, the momentum can be expressed as:\n \n \\(p = \\sqrt{2mK}\\)\n\n3. **Kinetic Energy from Acceleration through Potential Difference:**\n When a charged particle with charge \\(q\\) is accelerated from rest through a potential difference \\(V\\), its kinetic energy \\(K\\) is given by:\n \n \\(K = qV\\)\n\n4. **De Broglie Wavelength in terms of \\(m, q, V\\):**\n Substituting \\(K = qV\\) into the momentum equation, we get:\n \n \\(p = \\sqrt{2m(qV)}\\)\n \n Now, substitute this expression for \\(p\\) into the de Broglie wavelength formula:\n \n \\(\\lambda = \\frac{h}{\\sqrt{2mqV}}\\)\n\n5. **Applying to Electron and Proton:**\n Both the electron and the proton are accelerated through the same potential difference \\(V\\) and carry the same magnitude of charge \\(e\\) (i.e., \\(q_e = q_p = e\\)).\n\n For the electron:\n \\(\\lambda_e = \\frac{h}{\\sqrt{2m_e e V}}\\)\n\n For the proton:\n \\(\\lambda_p = \\frac{h}{\\sqrt{2m_p e V}}\\)\n\n6. **Calculating the Ratio \\(\\lambda_e / \\lambda_p\\):**\n Divide the expression for \\(\\lambda_e\\) by the expression for \\(\\lambda_p\\):\n \n \\(\\frac{\\lambda_e}{\\lambda_p} = \\frac{\\frac{h}{\\sqrt{2m_e e V}}}{\\frac{h}{\\sqrt{2m_p e V}}}\\)\n \n \\(\\frac{\\lambda_e}{\\lambda_p} = \\frac{h}{\\sqrt{2m_e e V}} \\times \\frac{\\sqrt{2m_p e V}}{h}\\)\n \n Cancel out common terms \\(h\\), \\(2\\), \\(e\\), and \\(V\\):\n \n \\(\\frac{\\lambda_e}{\\lambda_p} = \\sqrt{\\frac{m_p}{m_e}}\\)\n\nThus, the ratio of the de Broglie wavelengths of the electron to the proton is \\(\\sqrt{m_p / m_e}\\).\n\nThe final answer is \\(\\boxed{\\text{D}}\\\).