A proton and an electron have the same de Broglie wavelength. If $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{e}}$ be the kinetic energies of proton and electron respectively, then choose the correct relation :

<p>To determine the correct relation between the kinetic energies of a proton ($\mathrm{K}_{\mathrm{p}}$) and an electron ($\mathrm{K}_{\mathrm{e}}$) when they have the same de Broglie wavelength, we need to use the de Broglie wavelength formula:</p> <p> <p>$\lambda = \frac{h}{p}$</p> </p> <p>where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, and $p$ is the momentum of the particle.</p> <p>The momentum $p$ of a particle is given by:</p> <p> <p>$p = \sqrt{2mK}$</p> </p> <p>where $m$ is the mass of the particle and $K$ is its kinetic energy. For a proton and an electron with the same de Broglie wavelength:</p> <p> <p>$\lambda_{\mathrm{p}} = \lambda_{\mathrm{e}}$</p> </p> <p>This implies the momenta should be the same:</p> <p> <p>$p_{\mathrm{p}} = p_{\mathrm{e}}$</p> </p> <p>Thus, we can write:</p> <p> <p>$\sqrt{2 m_{\mathrm{p}} K_{\mathrm{p}}} = \sqrt{2 m_{\mathrm{e}} K_{\mathrm{e}}}$</p> </p> <p>Squaring both sides to eliminate the square roots:</p> <p> <p>$2 m_{\mathrm{p}} K_{\mathrm{p}} = 2 m_{\mathrm{e}} K_{\mathrm{e}}$</p> </p> <p>$m_{\mathrm{p}} K_{\mathrm{p}} = m_{\mathrm{e}} K_{\mathrm{e}}$ </p> <p>Rearranging to solve for $K_{\mathrm{p}}$ in terms of $K_{\mathrm{e}}$:</p> <p> <p>$K_{\mathrm{p}} = \frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} K_{\mathrm{e}}$</p> </p> <p>Since the mass of a proton $m_{\mathrm{p}}$ is much greater than the mass of an electron $m_{\mathrm{e}}$:</p> <p> <p>$m_{\mathrm{p}} \gg m_{\mathrm{e}}$</p> </p> <p>This means:</p> <p> <p>$\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} \ll 1$</p> </p> <p>Therefore, it implies:</p> <p> <p>$K_{\mathrm{p}} < K_{\mathrm{e}}$</p> </p> <p>So, the correct option is:</p> <p>Option C: $\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}$</p>