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: Recognise the straight-line law.
Einstein's relation \(eV = h\nu - \phi\) can be written as \(V = (h/e)\nu - (\phi/e)\). Comparing with \(y = mx + c\), the stopping-potential graph against frequency is a straight line whose gradient is \(h/e\) and whose x-intercept is the threshold frequency \(\nu_0\).

Step 2: Use the intercept for the work function first.
The graph meets the frequency axis at \(\nu_0 = 1\times10^{15}\ \text{Hz}\). At this point the electron just escapes with zero kinetic energy, so all the photon energy is spent on the work function: \(\phi = h\nu_0\).

Step 3: Find the gradient from the marked point.
The line rises from \((\,1\times10^{15},\,0\,)\) to \((\,5\times10^{15},\,1.656\,)\). Its gradient is
\(\dfrac{h}{e} = \dfrac{1.656}{(5-1)\times10^{15}} = \dfrac{1.656}{4\times10^{15}} = 4.14\times10^{-16}\ \text{V s}\).

Step 4: Plug the gradient into the work function.
Since \(\phi/e = (h/e)\nu_0\), the work function in electron-volts is \((4.14\times10^{-16})(1\times10^{15}) = 0.414\ \text{eV}\), i.e. \(0.414\times1.6\times10^{-19} = 6.62\times10^{-20}\ \text{J}\).
\[\boxed{\dfrac{h}{e} = 4.14\times10^{-16}\ \text{V s},\ \phi = 0.414\ \text{eV} = 6.62\times10^{-20}\ \text{J}}\]