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%