Deriving h and W from the photoelectric graph

Problem

Photoelectric effect. The graph shows the maximum kinetic energy $K_\text{max}$ of the photoelectrons extracted from a metal as a function of the frequency $\nu$ of the incident light. From the slope find $h$, and from the intercept find the work function $W$.

Solution

Step 1 — Einstein’s line. The law $K_\text{max} = h\nu - W$ is a straight line $K_\text{max}(\nu)$: the slope is $h$, the intercept on the $\nu$ axis (where $K_\text{max}=0$) is the threshold frequency $\nu_0 = W/h$.

Step 2 — Slope $\to h$. From the graph the line passes through $(\nu_0, 0)$ with $\nu_0 = 5{,}0\cdot 10^{14}\;\text{Hz}$ and rises to $K_\text{max} \approx 3{,}3\;\text{eV}$ at $\nu = 15\cdot 10^{14}\;\text{Hz}$. The slope is

$h = \frac{\Delta K_\text{max}}{\Delta\nu} = \frac{3{,}3\;\text{eV}}{(15 - 5)\cdot 10^{14}\;\text{Hz}} = 3{,}3\cdot 10^{-15}\;\text{eV·s} = 3{,}3\cdot 10^{-15}\cdot 1{,}6\cdot 10^{-19}\;\text{J·s}$

$\ev{h \approx 6{,}6\cdot 10^{-34}\;\text{J·s}}$

in excellent agreement with the tabulated value $6{,}63\cdot 10^{-34}\;\text{J·s}$.

Step 3 — Intercept $\to W$. The work function is $W = h\nu_0$:

$\ev{W = h\nu_0 = 6{,}63\cdot 10^{-34}\cdot 5{,}0\cdot 10^{14} \approx 3{,}3\cdot 10^{-19}\;\text{J} \approx 2{,}1\;\text{eV}}$

The slope is universal (the same for every metal: it is $h$); only the intercept $\nu_0$, i.e. $W$, changes, depending on the material.

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Topics: Quantum physics Concepts: Photoelectric effect · Photon Skills: Reading graphs