The electric field of a light wave at a point is given by E = 100 [ (3 10¹⁵ t) + (6 10¹⁵ t)] V/m This light falls on a metal surface, and the maximum kinetic energy of the emitted photoelectrons is observed to be 1.46 eV . Assuming h 2 = 6.6 10⁻¹⁶ eV s , the work function of the metal is:

Step-by-step solution

The given electric field is a superposition of two waves with angular frequencies ₁ = 3 10¹⁵ rad/s and ₂ = 6 10¹⁵ rad/s . The maximum kinetic energy of the photoelectrons is determined by the most energetic photons in the incident light, which correspond to the highest angular frequency, _ max = 6 10¹⁵ rad/s . The energy of the most energetic photon is: E_ ph = h = h 2 _ max E_ ph = (6.6 10⁻¹⁶ eV s ) (6 10¹⁵ s ⁻¹) = 3.96 eV According to Einstein's photoelectric equation: K_ max = E_ ph - Where is the work function