QJEEM24Phef1b: JEE Main 2024 Physics - Class 12 JEEM24Phef1b Solution: A proton and an electron have the same de Broglie wavelength.…

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:

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

where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, and $p$ is the momentum of the particle.

The momentum $p$ of a particle is given by:

$p = \sqrt{2mK}$

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:

$\lambda_{\mathrm{p}} = \lambda_{\mathrm{e}}$

This implies the momenta should be the same:

$p_{\mathrm{p}} = p_{\mathrm{e}}$

Thus, we can write:

$\sqrt{2 m_{\mathrm{p}} K_{\mathrm{p}}} = \sqrt{2 m_{\mathrm{e}} K_{\mathrm{e}}}$

Squaring both sides to eliminate the square roots:

$2 m_{\mathrm{p}} K_{\mathrm{p}} = 2 m_{\mathrm{e}} K_{\mathrm{e}}$

$m_{\mathrm{p}} K_{\mathrm{p}} = m_{\mathrm{e}} K_{\mathrm{e}}$

Rearranging to solve for $K_{\mathrm{p}}$ in terms of $K_{\mathrm{e}}$ :

$K_{\mathrm{p}} = \frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} K_{\mathrm{e}}$

Since the mass of a proton $m_{\mathrm{p}}$ is much greater than the mass of an electron $m_{\mathrm{e}}$ :

$m_{\mathrm{p}} \gg m_{\mathrm{e}}$

This means:

$\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} \ll 1$

Therefore, it implies:

K_{\mathrm{p}} < K_{\mathrm{e}}

So, the correct option is:

Option C: \mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}