Question

If a photon and an electron have same de-Broglie wavelength, then

• A. both have same kinetic energy
• B. proton has more K.E. than electron
• C. electron has more K.E. than proton
• D. both have same velocity

Answer 1

Answer: (C)

$\lambda =\frac{h}{m_p{v}_p}=\frac{h}{m_e{v}_e}$; then $m_p{v}_p=m_e{v}_e$ or $\frac{v_p}{v_e}=\frac{m_e}{m_p}$
$\frac{E_p}{E_e}=\frac{\frac{1}{2}{m}_p{v}_p^2}{\frac{1}{2}{m}_e{v}_e^2}=\frac{m_p}{m_e}\times {\left(\frac{m_e}{m_p}\right)}^2=\frac{m_e}{m_p}<1$
$\therefore E_p<E_e$