Chapter 27.1, Problem 12PP

To determine

To Find: Threshold frequency (in Hz) and the work function of zinc (in eV).

Answer to Problem 12PP

  $\text{9}\cdot \text{66}\times {\text{10}}^{14}\text{Hz}$ ; $4.0\text{eV}$

Explanation of Solution

Given data:

Threshold wavelength, ${\lambda }_o=310\text{ nm}$

Formula used:

Work function can be obtained by:

  $\varphi =\frac{hc}{{\lambda }_o}=h{v}_o$

Here, h is the Planck’s constant, c is the speed of light, ${\lambda }_o$ is the threshold wavelength and $v_o$ is the threshold frequency.

Calculation:

Threshold frequency, $v_o=\frac{c}{{\lambda }_o}$

  $\begin{array}{l}{v}_o=\frac{3\times {10}^8\text{m/s}}{310\times {10}^{-9}\text{m}}\left(\because {\text{1nm = 10}}^9\text{m}\right)\\ v_o=\text{9}\cdot \text{66}\times {\text{10}}^{14}\text{Hz }\end{array}$

  $\varphi =\frac{hc}{{\lambda }_o}=h{v}_o$

  $\begin{array}{l}\begin{array}{l}\varphi =6.63\times {10}^{-34}\text{J}\cdot \text{s}\times 9.66\times {10}^{14}\text{Hz}\hfill \\ \varphi =\text{ 64}\text{.04}\times {\text{10}}^{-20}\text{J}\hfill \end{array}\\ \varphi =4.0\text{eV}\end{array}$

Conclusion:

Thus, threshold frequency is $\text{9}\cdot \text{66}\times {\text{10}}^{14}\text{Hz }$ and work function is equal to $4.0\text{eV}$ .