Question

If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is :

• A. Zero
• B. Less than that of a proton
• C. More than that of a proton
• D. Equal to that of a proton

Answer 1

Answer: (C)

More than that of a proton

de Broglie wavelength $\lambda =\frac{h}{\sqrt{2mK}}$

We get $K=\frac{h^2}{2m{\lambda }^2}$

$⟹$ $K\propto \frac{1}{m}$

We know that $m_p>m_e$

$\therefore$ $\frac{K_e}{K_p}=\frac{m_p}{m_e}>1$

$⟹$ $K_e>K_p$