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:

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