The electric field at a point associated with a light wave is E = (100 V/m) sin [(3.0 × 10^15 s^-1)t] sin [(6.0 × 10^15 s^-1)t]. If this light falls on a metal surface having work function of 2.0 eV, what will be the maximum kinetic energy of the photoelectrons?

Sol. E = 100 sin[(3 * 10^15 s^–1)t] sin [6 * 10^15 s^–1)t] = 100 ½ [cos[(9 * 10^15 s^–1)t] – cos [3 * 10^15 s^–1)t] The w are 9 * 10^15 and 3 * 10^15 for largest K.E. f base max = w base max/2π = 9 * 10^15/2π E - ∅ base 0 = K.E. ⇒ hf = ∅ base 0 = K.E. ⇒ 6.63 * 10^-34 *9 * 10^15/2π * 1.6 * 10^19 – 2 KE ⇒ KE = 3.938 ev = 3.93 ev.