Question

You are an early 20th-century experimental physicist and do not know the value of Planck's constant. By a suitable plot of the following data, and using Einstein's explanation of the photoelectric effect $\mathbf{(}\mathbf{KE}\mathbf{=}\mathbf{W}\mathbf{-}\mathbf{\varphi }$. where $\mathbf{h}$is not known), determine Planck's constant.

Short answer

Planck's constant$\mathrm{h}=6.665\times {10}^{-34}\text{J}\cdot \text{s}$

Step-by-step solution

Concept  and  Formula  

Energy $\mathbf{E}\mathbf{=}\mathbf{hf}\mathbf{-}\mathbf{\varphi }$

Here, $\mathbf{E}$ is the energy,$\mathbf{f}$ is the frequency and $\mathbf{\varphi }$is the work function.

$\mathbf{E}\mathbf{=}{\mathbf{eV}}_{\mathbf{0}}$. Here,$\mathbf{E}$ is the kinetic energy, is the electronic charge and is the stopping potential.

$\mathbf{c}\mathbf{=}\mathbf{f\lambda }$Here, $\mathbf{c}$is the speed of light, $\mathbf{f}$is the frequency and $\mathbf{\lambda }$is the wavelength.

Electronic charge$\mathbf{=}\mathbf{1.602}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{\text{C}}$

Speed of light $\mathbf{=}\mathbf{2.99}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathbf{\text{m}}\mathbf{\cdot }{\mathbf{\text{s}}}^{\mathbf{-}\mathbf{1}}$ .

Evaluate energy

Energy is given as,

$\begin{array}{c}\mathrm{E}=\mathrm{hf}-\mathrm{\varphi }\\ \mathrm{E}={\mathrm{eV}}_0\\ {\mathrm{eV}}_0=\mathrm{hf}-\mathrm{\varphi }\\ {\mathrm{V}}_0=\frac{\mathrm{hf}}{\mathrm{e}}-\frac{\mathrm{\varphi }}{\mathrm{e}}\end{array}$

Substituting $\mathrm{c}=\mathrm{f\lambda }$

$\begin{array}{l}{\mathrm{V}}_0=\frac{\mathrm{hc}}{\mathrm{e\lambda }}-\frac{\mathrm{\varphi }}{\mathrm{e}}\\ {\mathrm{V}}_0=\frac{\mathrm{hc}}{\mathrm{e}}\times \frac{1}{\mathrm{\lambda }}-\frac{\mathrm{\varphi }}{\mathrm{e}}\end{array}$

Plot the graph

Plotting a graph of$\frac{1}{\mathrm{\lambda }}$ versus ${\mathrm{V}}_0$ using the given data,

Calculate Planck’s constant

From the graph, gradient $=1244\text{V}\cdot \text{nm}$and intercept $=-2.2017\text{V}$

As gradient $=\frac{\mathrm{hc}}{\mathrm{e}}$

$\begin{array}{c}1244=\frac{\mathrm{hc}}{\mathrm{e}}\\ \mathrm{h}=1244\times \frac{\mathrm{e}}{\mathrm{r}}\end{array}$

Substitute values for e and $\mathrm{c}$

$\begin{array}{l}\mathrm{h}=1244\text{V}\cdot \text{nm}\times \frac{1.602\times {10}^{-19}\text{C}}{2.99\times {10}^8\text{m}\cdot {\text{s}}^{-1}}\\ \mathrm{h}=1244\times {10}^{-9}\text{V}\cdot \text{m}\times \frac{1.602\times {10}^{-19}\text{C}}{2.99\times {10}^8\text{m}\cdot {\text{s}}^{-1}}\\ \mathrm{h}=6.665\times {10}^{-34}\text{J}\cdot \text{s}\end{array}$