In a photoelectric experiment, a parallel beam of monochromatic light with the power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV . The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode

Step-by-step solution

Power = N h ν N = number of photons per second Since K E = 0 , h ν = ϕ 200 = N 6.25 × 1.6 × 10 - 19 J o u l e N = 200 6.25 × 1.6 × 10 - 19 As photon is just above threshold frequency K E m a x is zero and they are accelerated by potential difference of 500 V K E f = q ∆ V P 2 2 m = q ∆ V ⇒ P = 2 m q ∆ V Since efficiency is 100%, number of electrons emitted is equal to number of photons falling per second as electrons are completely absorbed, force exerted n m v = 200 6.25 × 1.6 × 10 - 19 × 2 9 × 10 - 31 × 1.6 × 10