The following graph represents the variation of photocurrent with anode potential for a metal surface. Here I₁, I₂ and I₃ represents intensities and ₁, ₂, ₃ represent frequency for curves 1,2 and 3 respectively, then

Step-by-step solution

The given figure can be shown as From figure, we can see that the stopping potential (V₀ ) is same for curve 1 and 2. But for curve 3 , it is greater than that for land 2 . As, we know that, e V₀=E_ =h - ₀ where, ₀ is the work function. Since, V₀ is same for 1 and 2, so from above equation, we have r₁=r₂ Thus, for 1,2 and 3 ₁= ₂ ₃ Also, the saturation current of curve 2 is greater than of curve 1 . But for curve 2 and 3 it is equal, so, I₁ I₂=I₃