Q.An electron and a proton have exactly the same kinetic energy. Show that the de Broglie wavelength of the electron is about $42.8$ times that of the proton. (Take $m_p/m_e \approx 1836$.)

Setting up. For a particle of mass $m$ and kinetic energy $E$, momentum is $p=\sqrt{2mE}$ (from $E=p^2/2m$), so the de Broglie wavelength is $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

Taking the ratio at equal $E$. Since $h$ and $E$ are the SAME for both particles, $\frac{\lambda_e}{\lambda_p}=\frac{\sqrt{2m_pE}}{\sqrt{2m_eE}}=\sqrt{\frac{m_p}{m_e}}=\sqrt{1836}\approx42.8$ …

Express $\lambda$ in terms of mass and kinetic energy ($\lambda=h/\sqrt{2mE}$), then take the ratio for the two part …

• Using $\lambda=h/mv$ directly with an assumed common $v$ -- the two particles do NOT have the same speed here, only the same kinetic energy, so the mass-and-energy form $\lambda=h/\sqrt{2mE}$ must be used instead. …