• # question_answer 27) If electron, proton and He have same energy, then their de Broglie wavelength decreases in order: A) ${{\lambda }_{e}}>{{\lambda }_{p}}>{{\lambda }_{He}}$    B) ${{\lambda }_{He}}>{{\lambda }_{p}}>{{\lambda }_{e}}$C) ${{\lambda }_{He}}>{{\lambda }_{e}}>{{\lambda }_{p}}$    D) ${{\lambda }_{p}}>{{\lambda }_{e}}>{{\lambda }_{He}}$

de-Broglie wavelength, $\lambda =\frac{h}{\sqrt{2mE}}$ Since, E is the same, Hence  $\lambda \propto \frac{h}{\sqrt{m}}$ Since, ${{m}_{He}}>{{m}_{p}}>{{m}_{e}}$ $\therefore \,\,\,{{\lambda }_{e}}>{{\lambda }_{p}}>{{\lambda }_{He}}$