A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:

(Assume h = 6.63 $$\times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg}$$ and $\mathrm{m}_{\mathrm{p}}=1836$ times $\mathrm{m}_{\mathrm{e}}$ )

<p>To solve for the ratio of the kinetic energies of a proton and an electron with the same de-Broglie wavelength, let us first recall the relationship between kinetic energy, momentum, and the de-Broglie wavelength.</p> <p>The de-Broglie wavelength $\lambda$ is given by:</p> <p>$\lambda = \frac{h}{p}$</p> <p>where $h$ is Planck’s constant and $p$ is the momentum.</p> <p>Since the proton and the electron have the same de-Broglie wavelength, their momenta must be equal:</p> <p>$$\lambda_{\text{electron}} = \lambda_{\text{proton}} \implies \frac{h}{p_{\text{e}}} = \frac{h}{p_{\text{p}}} \implies p_{\text{e}} = p_{\text{p}}$$</p> <p>Next, the kinetic energy $K$ of a particle is related to its momentum $p$ and mass $m$ by the following formula:</p> <p>$K = \frac{p^2}{2m}$</p> <p>Given that the momentum $p$ is the same for both the proton and the electron, we can write the kinetic energies as:</p> <p>$K_{\text{e}} = \frac{p^2}{2m_{\text{e}}}$</p> <p>$K_{\text{p}} = \frac{p^2}{2m_{\text{p}}}$</p> <p>The ratio of the kinetic energies is therefore:</p> <p>$$\frac{K_{\text{e}}}{K_{\text{p}}} = \frac{\frac{p^2}{2m_{\text{e}}}}{\frac{p^2}{2m_{\text{p}}}} = \frac{m_{\text{p}}}{m_{\text{e}}}$$</p> <p>Given that the mass of the proton $m_{\text{p}}$ is 1836 times the mass of the electron $m_{\text{e}}$, we have:</p> <p>$\frac{m_{\text{p}}}{m_{\text{e}}} = 1836$</p> <p>Thus, the ratio of the kinetic energies is:</p> <p>$\frac{K_{\text{e}}}{K_{\text{p}}} = 1836$</p> <p>Therefore, the correct answer is:</p> <p>Option C</p> <p>$1: 1836$</p>