NEETPhysicsDual Nature of Matter
Light of wavelength is incident on a photosensitive surface with a work function W > 0 . The fastest emitted photoelectron has a de-Broglie wavelength ₁ . When the incident wavelength is changed to 2 , the fastest photoelectron has a de-Broglie wavelength ₂ . Which of the following inequalities correctly relates ₁ and ₂ ?
Options
- A₂ < ₁ 2
- B₂ = ₁ 2
- C₂ > ₁ 2
- D₂ < ₁ 2
Correct answer
A. ₂ < ₁ 2
Step-by-step solution
From Einstein's photoelectric equation, the maximum kinetic energy for the initial wavelength is: K₁ = hc - W When the incident wavelength is halved to 2 , the new maximum kinetic energy is: K₂ = hc /2 - W = 2hc - W We can express K₂ in terms of K₁ : K₂ = 2 (K₁ + W ) - W = 2K₁ + W Since the work function W > 0 , it follows that K₂ > 2K₁ . The de-Broglie wavelength is inversely proportional to the square root of the kinetic energy ( _d = h 2mK ). Therefore, for the new wavelength: ₂ = h 2mK₂ Since K₂ > 2K₁ , we have