QJEEM24Ph6afc: JEE Main 2024 Physics - Class 12 JEEM24Ph6afc Solution: A proton, an electron and an alpha particle have the same…

To determine the relationship between the de-Broglie wavelengths of a proton, an electron, and an alpha particle with the same energies, we need to use the de-Broglie wavelength formula:

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

where $h$ is Planck's constant and $p$ is the momentum of the particle.

For particles with the same kinetic energy $E$ , we have:

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

Solving for $p$ :

$p = \sqrt{2mE}$

Substituting this into the de-Broglie equation, we get:

$\lambda = \frac{h}{\sqrt{2mE}}$

Since all three particles have the same energy $E$ , the de-Broglie wavelength is inversely proportional to the square root of the mass $m$ :

$\lambda \propto \frac{1}{\sqrt{m}}$

The masses of the particles are as follows:

• Mass of proton $m_p$ is approximately $1.67 \times 10^{-27}$ kg
• Mass of electron $m_e$ is approximately $9.11 \times 10^{-31}$ kg
• Mass of alpha particle $m_{\alpha}$ is approximately $4 \times 1.67 \times 10^{-27}$ kg (since it has 2 protons and 2 neutrons)

Comparatively:

• Mass of alpha particle $m_{\alpha}$ is the largest.
• Mass of proton $m_p$ is intermediate.
• Mass of electron $m_e$ is the smallest.

Therefore, the de-Broglie wavelength will be:

• Largest for the electron ( $\lambda_{\mathrm{e}}$ )
• Intermediate for the proton ( $\lambda_{\mathrm{p}}$ )
• Smallest for the alpha particle ( $\lambda_{\alpha}$ )

Hence, the correct order is:

Option B:

\lambda_\alpha<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}