Photoelectric effect experiments are performed using three different metal plates p, q, and r having work functions φp = 2.0 eV, φq = 2.5 eV, and φr = 3.0 eV, respectively. A light beam containing wavelengths of 550 nm, 450 nm, and 350 nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is: (Take hc = 1240 eV nm)

The photoelectric effect can be explained by the following relation:
eVB = [(hc/λ) - φ]
VB = (1/e)[(hc/λ) - φ]
For plate p (φp = 2.0 eV):
If λ = 550 nm, then VB = (1/e)[(1240/550) - 2.0] = (1/e)[2.25 - 2.0] = 0.25/e > 0
If λ = 450 nm, then VB = (1/e)[(1240/450) - 2.0] = (1/e)[2.76 - 2.0] = 0.76/e > 0
If λ = 350 nm, then VB = (1/e)[(1240/350) - 2.0] = (1/e)[3.54 - 2.0] = 1.54/e > 0
For plate q (φq = 2.5 eV):
If λ = 550 nm, then VB = (1/e)[(1240/550) - 2.5] = (1/e)[2.25 - 2.5] = -0.25/e < 0
If λ = 450 nm, then VB = (1/e)[(1240/450) - 2.5] = (1/e)[2.76 - 2.5] = 0.26/e > 0
If λ = 350 nm, then VB = (1/e)[(1240/350) - 2.5] = (1/e)[3.54 - 2.5] = 1.04/e > 0
For plate r (φr = 3.0 eV):
If λ = 550 nm, then VB = (1/e)[(1240/550) - 3.0] = (1/e)[2.25 - 3.0] = -0.75/e < 0
If λ = 450 nm, then VB = (1/e)[(1240/450) - 3.0] = (1/e)[2.76 - 3.0] = -0.24/e < 0
If λ = 350 nm, then VB = (1/e)[(1240/350) - 3.0] = (1/e)[3.54 - 3.0] = 0.54/e > 0
Therefore, for plate p, current will flow for all three wavelengths. For plate q, current will flow for 450 nm and 350 nm. For plate r, current will flow only for 350 nm. The correct I-V graph will show the saturation current for p being highest, followed by q, and then r. The stopping potential will be different for each plate and wavelength combination.