Question

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.

Short answer

The rank of the material is ${\Phi }_1<{\Phi }_2<{\Phi }_3$

Step-by-step solution

Describe the Photoelectric effect

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.

Rank the materials according to their work function, greatest first

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 ${\Phi }_1<{\Phi }_2<{\Phi }_3$.