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NEETPhysicsDual Nature of Matter

A photon of wavelength is incident on a photosensitive surface of negligible work function. If the emitted photoelectron has a de-Broglie wavelength _d , what is the ratio of the momentum of the incident photon to the momentum of the emitted photoelectron?

Options

  1. A2mc _d h
  2. Bh 2mc _d
  3. Ch mc _d
  4. Dmc _d h

Correct answer

B. h 2mc _d

Step-by-step solution

The momentum of the incident photon is p_ photon = h . The momentum of the emitted photoelectron is p_ electron = h _d . From Einstein's photoelectric equation with negligible work function, the maximum kinetic energy of the photoelectron equals the energy of the incident photon: K_ max = hc Expressing kinetic energy in terms of momentum, K_ max = p_ electron ^2 2m . Equating the two expressions for energy gives: hc = p_ electron ^2 2m Since h = p_ photon , we can write: p_ photon c = p_ electron ^2 2m Rearranging

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