Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface?

First let us recall the two quantitative relations that govern the photo-electric effect.

1. The Einstein photo-electric equation states

$$K_{\max}=h\nu-\phi,$$

where $$K_{\max}$$ is the maximum kinetic energy of the emitted electrons, $$h$$ is Planck’s constant, $$\nu$$ is the frequency of the incident light and $$\phi$$ is the work function of the metal.

2. The number of photo-electrons emitted per unit time is proportional to the number of incident photons, that is, to the intensity of the light. For a given intensity it is independent of frequency as long as $$\nu\gt\nu_{0}$$, where $$\nu_{0}=\dfrac{\phi}{h}$$ is the threshold frequency. Below this threshold no electron is emitted.

With these facts in mind we examine each proposed graph.

Graph (1): K.E. of electrons vs Frequency of Light

Starting from the photo-electric equation we can rearrange it as

$$K_{\max}=h(\nu-\nu_{0}).$$

This is the straight-line form $$y=m(x-x_{0})$$ with slope $$m=h$$, but the line intersects the frequency axis at $$\nu=\nu_{0}$$, not at the origin. Therefore a straight line that begins from the origin and rises linearly is incorrect; the correct graph must cut the frequency axis at the positive value $$\nu_{0}$$ and pass through the point $$(\nu_{0},0)$$. Hence the graph shown in option (1) does not represent the true relationship.

Graph (2): K.E. of electrons vs Energy (hν) of Light

If we choose the horizontal axis as the photon energy $$E=h\nu$$, the Einstein equation becomes

$$K_{\max}=E-\phi.$$

The slope is now $$1$$ and the line intersects the energy axis at $$E=\phi$$. A straight line of unit slope that crosses the horizontal axis at $$\phi$$ is correct, so a line of this form is acceptable.

Graph (3): Number of electrons vs Frequency of Light

For fixed intensity the photon arrival rate, and hence the electron emission rate, is constant for all $$\nu\gt\nu_{0}$$ and falls to zero for $$\nu\le\nu_{0}$$. Thus a step-like graph—zero up to $$\nu_{0}$$ and constant thereafter—is physically accurate, so the sketch provided in this option is consistent with theory.

Graph (4): K.E. of electrons vs Intensity of Light

The kinetic energy expression $$K_{\max}=h\nu-\phi$$ contains no term involving intensity. Hence $$K_{\max}$$ is independent of intensity, giving a horizontal line parallel to the intensity axis; this is exactly what option (4) shows, so it is correct.

We have found that only the first proposed graph conflicts with the photo-electric equation because it starts from the origin instead of the threshold frequency. All other graphs are compatible with the established relations.

Hence, the correct answer is Option 1.