If , and are the kinetic energies of an electron, an -particle and a proton, with same de-Broglie wa — Atomic Structure Chemistry Question

💡 Solution & Explanation

Kinetic energy $E = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$. For constant $\lambda$, $E \propto \frac{1}{m}$. Since $m_{electron} < m_{proton} < m_{\alpha}$, it follows that $E_{electron} > E_{proton} > E_{\alpha}$. Thus $E_1 > E_3 > E_2$. Therefore, correct answer is A.