Correct option : (2) 1.2×1015 Hz
Explanation :
Energy of a photon, of the incident radiation,
Let W be the work function of the given photosensitive surface. We then have W = 2.49 eV – 1.99 eV = 0.5 eV
∴ If v0 is the threshold frequency, we have hv0 = W = 0.5 ×1.6×10–19 J