JEE MainPhysicsDual Nature of Matter
Photons of two different energies illuminate a metal surface having a work function . The energy of the first incident photon is 2 and that of the second incident photon is 10 . The ratio of the de Broglie wavelength of the fastest emitted photoelectron in the first case to that in the second case is
Options
- A1 : 3
- B3 : 1
- C9 : 1
- D5 : 1
Correct answer
B. 3 : 1
Step-by-step solution
From Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by K_ max = E - For the first incident photon of energy E₁ = 2 : K₁ = 2 - = For the second incident photon of energy E₂ = 10 : K₂ = 10 - = 9 The de Broglie wavelength of a particle is related to its kinetic energy by = h 2mK Therefore, the ratio of the de Broglie wavelengths of the fastest photoelectrons is ₁ ₂ = K₂ K₁ Substituting the values of kinetic energies: ₁ ₂ = 9 = 9 = 3 The ratio is 3 : 1 . Answer: 3 :