Physics · Matter waves: wave nature of particle, de-Broglie relation
If wavelength of photon and electron is same then ratio of total energy of elect
If wavelength of photon and electron is same then ratio of total energy of electron to total energy of photon would be:
- A. \( \frac{\text { Velocity of electron }}{\text { Light's speed }} \)
- B. \( \frac{\text { Light's speed }}{\text { Electon's speed }} \)
- C. \( \frac{\text { Light's speed }}{\text { Velocity of electron }} \)
- D. \( \frac{\text { Electron's speed }}{\text { Light's speed }} \)
Step-by-step solution
For a photon, energy E_ph = hc/λ. For an electron with de Broglie wavelength λ, momentum p = h/λ. Relativistically, total energy E_e = √(p²c² + m₀²c⁴) = γ m₀ c². Using p = γ m₀ v, we get E_ph = hc/λ = γ m₀ v c. Therefore, E_e/E_ph = (γ m₀ c²)/(γ m₀ v c) = c/v, i.e., light speed divided by electron speed.
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