NEETPhysicsDual Nature of Matter
A photon and an electron have the exact same de Broglie wavelength. If p_p and p_e are the magnitudes of the momenta of the photon and the electron respectively, and E_p and E_e are their respective energies, which of the following relationships is correct?
Options
- Ap_p = p_e and E_p = E_e
- Bp_p = p_e and E_p < E_e
- Cp_p = p_e and E_p > E_e
- Dp_p > p_e and E_p > E_e
Correct answer
C. p_p = p_e and E_p > E_e
Step-by-step solution
The de Broglie wavelength for any particle or photon is related to its momentum by = h p . Since both the photon and the electron have the same wavelength , their momenta must be equal. Therefore, p_p = p_e = p . The energy of the photon is given by E_p = pc . The kinetic energy of the electron is given by E_e = p^2 2m = p ( v 2 ) , where v is the velocity of the electron. Since the electron is a massive particle, its velocity v must be strictly less than the speed of light c ( v Comparing the two energies: E_p = p