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Physics
If a photon and an electron have same de-Broglie wavelength, then
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Q. If a photon and an electron have same de-Broglie wavelength, then
Dual Nature of Radiation and Matter
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
Solution:
$\lambda=\frac{h}{m_p v_p}=\frac{h}{m_e v_e}$; then $m_p v_p=m_e v_e$ or $\frac{v_p}{v_e}=\frac{m_e}{m_p}$
$\frac{E_p}{E_e}=\frac{\frac{1}{2} m_p v_p^2}{\frac{1}{2} m_e v_e^2}=\frac{m_p}{m_e} \times\left(\frac{m_e}{m_p}\right)^2=\frac{m_e}{m_p}<1$
$\therefore E_p < E_e$