Photoelectric effect: Figure 38-20 gives the stopping voltage V versus the wavelength $\mathit{\lambda }$of light for three different materials. Rank the materials according to their work function, greatest first.

The electrons are emitted from the surface by photoelectric effect within the metal when the light of high enough frequency falls on a clean metal surface. The resultant relation is given by,

$\mathit{h}\mathit{f}\mathbf{=}{\mathit{K}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{+}\mathit{\varphi }$

Here, hf is photon energy, ${\mathit{K}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$is the kinetic energy of the most energetic emitted electron, and role="math" localid="1663071925969" $\mathit{\varphi }$is the work function of the target material.

The kinetic energy is given by,

$K_{\max }=eV$

The frequency is given by,

$f=\frac{c}{\lambda }$

Substitute all the values in the equation .

$hf=K_{\max }+\Phi$

$$\begin{gathered} \frac{hc}{\lambda }=eV+\Phi \\ V=\frac{hc}{e}\left(\frac{1}{\lambda }\right)-\frac{\Phi }{e} \end{gathered}$$

A graph between $\lambda$ and V is shown in the given figure, and the stopping potential is zero at the cutoff wavelength.

Substitute $V=0$in the above equation.

$$\begin{gathered} 0=\frac{hc}{e}\left(\frac{1}{\lambda }\right)-\frac{\Phi }{e} \\ \Phi =\frac{hc}{\lambda } \end{gathered}$$

From the above equation, it can be observed that the material with smaller cutoff wavelength has the greater work function. The rank will be as follows.

$$\begin{gathered} {\lambda }_1>{\lambda }_2>{\lambda }_3 \\ \frac{1}{{\lambda }_1}<\frac{1}{{\lambda }_2}<\frac{1}{{\lambda }_3} \\ \frac{hc}{{\lambda }_1}<\frac{hc}{{\lambda }_2}<\frac{hc}{{\lambda }_3} \\ {\Phi }_1<{\Phi }_2<{\Phi }_3 \end{gathered}$$

Therefore, the rank of the material is ${\Phi }_1<{\Phi }_2<{\Phi }_3$.