In about 1916, R. A. Millikan found the following stopping potential data for lithium in his photoelectric experiments:

Wavelength (nm)	433.9	404.7	365.0	312.5	253.5
Stopping potential (V)	0.55	0.73	1.09	1.67	2.57

Use these data to make a plot like Fig. 38-2 (which is for sodium) and then use the plot to find (a) the Planck constant and (b) the work function for lithium.

The wavelengths

The expression for the photoelectric effect is given follows.

$hf=\Phi +K_{\max }$ …(i)

Here, f is the incident frequency, $\Phi$is the work function and $K_{\max }$is the maximum kinetic energy.

The express to calculate the incident frequency is given as follows.

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

Here, $\lambda$ is the wavelength and c is the speed of the light.

The expression to calculate the maximum kinetic energy of the electron is given as follows,

$K_{\max }=eV$

Here, e is the elementary charge and V is the stopping potential.

Derive the expression for the $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.$.

Substitute eVs for $K_{max}$, and $\raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.$ forfinto equation (i).

localid="1663060320209" $$\begin{gathered} \frac{hc}{\lambda }=\varphi +eV \\ \frac{1}{\lambda }=\frac{\varphi }{hc}+\frac{1}{hc}\mathrm{eV} \end{gathered}$$ …(ii)

Calculate the value of $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.$to draw the graph between $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.$ and V.

Calculate the value for 433.9 nm

$\begin{array}{rcl}\frac{1}{\lambda }& =& \frac{1}{433.9\mathrm{nm}}\\ & =& 0.002305{\mathrm{nm}}^{-1}\end{array}$

Calculate the value for 404.7 nm

$\begin{array}{rcl}\frac{1}{\lambda }& =& \frac{1}{404.7\mathrm{nm}}\\ & =& 0.002471{\mathrm{nm}}^{-1}\end{array}$

Calculate the value for 365 nm

$\begin{array}{rcl}\frac{1}{\lambda }& =& \frac{1}{365\mathrm{nm}}\\ & =& 0.002740{\mathrm{nm}}^{-1}\end{array}$

Calculate the value for 312.5 nm

$\begin{array}{rcl}\frac{1}{\lambda }& =& \frac{1}{312.5\mathrm{nm}}\\ & =& 0.0032{\mathrm{nm}}^{-1}\end{array}$

Calculate the value for 253.5 nm

$\begin{array}{rcl}\frac{1}{\lambda }& =& \frac{1}{253.5\mathrm{nm}}\\ & =& 0.003945{\mathrm{nm}}^{-1}\end{array}$

Determine the table as,

Determine the equation of the straight line,

$\begin{array}{rcl}\frac{1}{\lambda }-0.002305& =& \frac{0.003945-0.002305}{2.57-0.55}\left(V-0.55\right)\\ \frac{1}{\lambda }-0.002305& =& \frac{0.00164}{2.02}\left(V-0.55\right)\\ \frac{1}{\lambda }& =& 0.0081V-0.00044+0.002305\\ \frac{1}{\lambda }& =& \left(0.0081V+0.0019\right){\text{nm}}^{-1}\end{array}$ …(iii)

Compare the equation (ii) and (iii)

$\begin{array}{rcl}0.0081\times {10}^9{\text{m}}^{-1}& =& \frac{eV}{hc}\\ h& =& \frac{1}{0.0081\times {10}^9{\text{m}}^{-1}\times c}eV\end{array}$

Substitute $3\times {10}^8\mathrm{m}/\mathrm{s}$ for c into above equation.

$\begin{array}{rcl}h& =& \frac{1}{0.0081\times {10}^9{\mathrm{m}}^{-1}\times 3\times {10}^8\mathrm{m}/\mathrm{s}}\mathrm{eV}\\ & =& \frac{1}{0.0243\times {10}^{17}}\mathrm{eVs}\\ & =& 411.5\times {10}^{-17}\\ & =& \times {10}^{-15}\mathrm{eVs}\end{array}$

Hence the value of the Plank’s constant is $4.11\times {10}^{-15}\text{eVs}$.

Draw the graph between the andV.

Again, compare the equation (ii) and (iii),

$\begin{array}{rcl}\frac{\Phi }{hc}& =& 0.0019\times {10}^9\\ \Phi & =& 0.0019\times {10}^{9}hc\end{array}$

Substitute for hand $3\times {10}^8\mathrm{m}/\mathrm{s}$ for c into above equation.

$\begin{array}{rcl}\varphi & =& 0.0019\times {10}^9{\mathrm{m}}^{-1}\times 4.11\times {10}^{-15}\mathrm{eV}\times 3\times {10}^8\mathrm{m}/\mathrm{s}\\ & =& 0.0234\times {10}^2\mathrm{eV}\\ & =& 2.34\mathrm{eV}\end{array}$

Hence the value of the work function is $2.34\text{eV}$.