A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :

<p>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:</p> <p> <p>$\lambda = \frac{h}{p}$</p> </p> <p> <p>where $ h $ is Planck's constant and $ p $ is the momentum of the particle.</p> </p> <p>For particles with the same kinetic energy $ E $, we have:</p> <p> <p>$E = \frac{p^2}{2m}$</p> </p> <p> <p>Solving for $ p $:</p> </p> <p> <p>$p = \sqrt{2mE}$</p> </p> <p> <p>Substituting this into the de-Broglie equation, we get:</p> </p> <p> <p>$\lambda = \frac{h}{\sqrt{2mE}}$</p> </p> <p> <p>Since all three particles have the same energy $ E $, the de-Broglie wavelength is inversely proportional to the square root of the mass $ m $:</p> </p> <p> <p>$\lambda \propto \frac{1}{\sqrt{m}}$</p> </p> <p> <p>The masses of the particles are as follows:</p> </p> <ul> <li>Mass of proton $ m_p $ is approximately $ 1.67 \times 10^{-27} $ kg</li> <li>Mass of electron $ m_e $ is approximately $ 9.11 \times 10^{-31} $ kg</li> <li>Mass of alpha particle $ m_{\alpha} $ is approximately $ 4 \times 1.67 \times 10^{-27} $ kg (since it has 2 protons and 2 neutrons)</li> </ul> <p>Comparatively:</p> <ul> <li>Mass of alpha particle $ m_{\alpha} $ is the largest.</li> <li>Mass of proton $ m_p $ is intermediate.</li> <li>Mass of electron $ m_e $ is the smallest.</li> </ul> <p>Therefore, the de-Broglie wavelength will be:</p> <ul> <li>Largest for the electron ($ \lambda_{\mathrm{e}} $)</li> <li>Intermediate for the proton ($ \lambda_{\mathrm{p}} $)</li> <li>Smallest for the alpha particle ($ \lambda_{\alpha} $)</li> </ul> <p>Hence, the correct order is:</p> <p>Option B:</p> <p> <p>$\lambda_\alpha<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}$</p> </p>