Chapter 27.1, Problem 20SSC

To determine

To Calculate: The Work function (in eV) of metal.

Answer to Problem 20SSC

  $0\cdot 89\text{eV}$

Explanation of Solution

Given data:

Wavelength of incident green light, $\lambda =532\text{nm}$

Stopping Potential, $V=1.44\text{V}$

Formula used:

Energy of the incident light:

  $E=\frac{1240\text{eV}\cdot \text{nm}}{\lambda }$

Here, $\lambda$ is the wavelength of the incident light.

Work function of metal = Energy of incident light − Maximum kinetic energy of photoelectron.

  $\begin{array}{l}\varphi =E-K_{\max }\\ K_{\max }=eV\end{array}$

Here, e is the charge of electron and V is the stopping potential.

Calculation:

Energy of the incident light:

  $\begin{array}{l}E=\frac{1240\text{eV}\cdot \text{nm}}{\lambda }\\ E=\frac{1240\text{eV}\cdot \text{nm}}{532\text{nm}}\\ E=2.33\text{eV}\end{array}$

  $\begin{array}{l}{K}_{\max }=1.44\text{eV}\\ \varphi =E-K_{\max }\\ \varphi =2.33\text{eV}-1.44\text{eV}\\ \varphi \text{ = 0}\text{.89 eV}\end{array}$

Conclusion:

Thus, the work function of the metal is $0\cdot 89\text{eV}$ .