JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Modern Physics

The de Broglie wavelength is $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$, so for particles of equal kinetic energy $K$, $\lambda \propto \frac{1}{\sqrt{m}}$: the lighter the particle, the longer its wavelength. Since the electron is far lighter than the proton, which in turn is about a quarter the mass of the doubly ionized helium ion ($m_{{\mathrm{He}}^{++}}\approx 4m_p$), the masses order as $m_e<m_p<m_{{\mathrm{He}}^{++}}$. The wavelengths therefore order the opposite way, ${\lambda }_e>{\lambda }_p>{\lambda }_{{\mathrm{He}}^{++}}$.