Question

In the figure, the plot of stopping potential versus frequency of light used in an experiment of the photoelectric effect is shown. Find the ratio \( h/e \) and the work function.
(From the graph: the line meets the frequency axis at \( \nu_0 = 1\times10^{15}\ \text{Hz} \), and the stopping potential is \( V = 1.656\ \text{V} \) at \( \nu = 5\times10^{15}\ \text{Hz} \).)

Hint:

On a stopping-potential vs frequency graph the slope is \( h/e \) and the frequency intercept is the threshold \( \nu_0 \); the work function is \( \phi = h\nu_0 \).

Answer 1

Step 1: Einstein's photoelectric equation in terms of stopping potential.
The maximum kinetic energy of the emitted electrons equals \( eV \), where \( V \) is the stopping potential:
\[ eV = h\nu - \phi \]
Dividing by \( e \):
\[ V = \frac{h}{e}\,\nu - \frac{\phi}{e} \]
This is a straight line of the form \( V = m\nu + c \). So the slope of the \( V \) versus \( \nu \) graph is \( h/e \), and the graph cuts the frequency axis (\( V = 0 \)) at the threshold frequency \( \nu_0 \).

Step 2: Read two points from the graph.
The line crosses the \( \nu \)-axis at \( \nu_0 = 1\times10^{15}\ \text{Hz} \) (where \( V = 0 \)), and passes through \( V = 1.656\ \text{V} \) at \( \nu = 5\times10^{15}\ \text{Hz} \).

Step 3: Slope gives \( h/e \).
\[ \frac{h}{e} = \text{slope} = \frac{\Delta V}{\Delta \nu} = \frac{1.656 - 0}{(5 - 1)\times10^{15}} = \frac{1.656}{4\times10^{15}} \]
\[ \frac{h}{e} = 4.14\times10^{-16}\ \text{V s} \]

Step 4: Work function from the threshold frequency.
At the threshold frequency the stopping potential is zero, so \( \phi = h\nu_0 \). In electron-volt units,
\[ \phi\ (\text{in eV}) = \frac{h}{e}\,\nu_0 = (4.14\times10^{-16})\times(1\times10^{15}) = 0.414\ \text{eV} \]
Converting to joules,
\[ \phi = 0.414\times1.6\times10^{-19} = 6.62\times10^{-20}\ \text{J} \]

\[\boxed{\dfrac{h}{e} = 4.14\times10^{-16}\ \text{V s},\qquad \phi = 0.414\ \text{eV} = 6.62\times10^{-20}\ \text{J}}\]