Question

An electron and proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is

• A. Zero
• B. Infinity
• C. Equal to the kinetic energy of the proton
• D. Greater than the kinetic energy of the proton

Answer 1

The correct option is D

Greater than the kinetic energy of the proton

$\lambda =\frac{h}{\sqrt{2mE}}\Rightarrow E\propto \frac{1}{\sqrt{m}}$ $(\lambda =constant)$
$\because m_e<m_p$ so $E_e>E_p$