QJEEM24Ph857b: JEE Main 2024 Physics - Class 12 JEEM24Ph857b Solution: A proton and an electron are associated with same de-Broglie…

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.

The de-Broglie wavelength $\lambda$ is given by:

$\lambda = \frac{h}{p}$

where $h$ is Planck’s constant and $p$ is the momentum.

Since the proton and the electron have the same de-Broglie wavelength, their momenta must be equal:

$\lambda_{\text{electron}} = \lambda_{\text{proton}} \implies \frac{h}{p_{\text{e}}} = \frac{h}{p_{\text{p}}} \implies p_{\text{e}} = p_{\text{p}}$

Next, the kinetic energy $K$ of a particle is related to its momentum $p$ and mass $m$ by the following formula:

$K = \frac{p^2}{2m}$

Given that the momentum $p$ is the same for both the proton and the electron, we can write the kinetic energies as:

$K_{\text{e}} = \frac{p^2}{2m_{\text{e}}}$

$K_{\text{p}} = \frac{p^2}{2m_{\text{p}}}$

The ratio of the kinetic energies is therefore:

$\frac{K_{\text{e}}}{K_{\text{p}}} = \frac{\frac{p^2}{2m_{\text{e}}}}{\frac{p^2}{2m_{\text{p}}}} = \frac{m_{\text{p}}}{m_{\text{e}}}$

Given that the mass of the proton $m_{\text{p}}$ is 1836 times the mass of the electron $m_{\text{e}}$ , we have:

$\frac{m_{\text{p}}}{m_{\text{e}}} = 1836$

Thus, the ratio of the kinetic energies is:

$\frac{K_{\text{e}}}{K_{\text{p}}} = 1836$

Therefore, the correct answer is:

Option C

$1: 1836$