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Question:

In a photoelectric experiment, a parallel beam of monochromatic light with a power of 200W is incident on a perfectly absorbing cathode of work function 6.25eV. 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%

Solution:

Power = nhv n = number of photons per second
Since KE = 0, hv = Φ
200=n[6.25 × 1.6 × 10⁻¹⁹ Joule]
n=200/(1.6 × 10⁻¹⁹ × 6.25)
As photon is just above threshold frequency, KEmax is zero and they are accelerated by potential difference of 500V.
KEf = qΔV
P²/2m = qΔV ⇒ P = √(2mqΔV)
Since efficiency is 100%