Question

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.

Short answer

(a) The value of the Plank’s constant is, $4.11\times {10}^{-15}\text{eVs}$.

(b) The value of the work function is, $2.34\text{eV}$.

Step-by-step solution

Write the given data from the question.

The wavelengths

Wavelength (nm)	433.9	404.7	365	312.5	253.5
Stopping potential (V)	0.55	0.73	1.09	1.67	2.57

Determine the formulas to calculate the planks constant and work function for lithium.

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.

Step 3(a): Calculate the value of the plank’s constant

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,

Wavelength (nm)	Stopping voltage (V)	$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.$
433.9	0.55	0.002305
404.7	1.09	0.002471
365	1.09	0.002740
312.5	1.67	0.0032
253.5	2.57	0.003945

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.

Step 4(b): Calculate the value of the work function

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 $2.34\text{eV}$.