Metals contain free electrons, yet these electrons do not spill out of the surface on their own. A minimum amount of energy is needed to pull an electron free against the attraction of the positive ions. This minimum energy is the work function $\phi_0$ of the metal, usually measured in electron-volts ($1\,\text{eV} = 1.6\times10^{-19}\,\text{J}$). Different metals have different work functions: caesium ($\approx 2.1\,\text{eV}$) lets electrons go easily, while platinum ($\approx 5.6\,\text{eV}$) holds them tightly.
When light of suitable frequency falls on a metal surface, electrons are ejected from it. This is the photoelectric effect, and the emitted electrons are called photoelectrons. The phenomenon was first noticed by Heinrich Hertz (1887), who saw that a spark jumped more easily across electrodes when ultraviolet light was shone on them. Wilhelm Hallwachs and Philipp Lenard then studied it carefully and established the experimental laws that any correct theory must obey.
In a typical experiment, light passes through a quartz window onto a photosensitive plate inside an evacuated tube. A collector plate is kept at a positive potential to attract the emitted electrons, and the resulting photoelectric current is read on a microammeter. A variable voltage and a reversible polarity let us either accelerate or retard the photoelectrons.
The experiments reveal four key laws of photoelectric emission. First, for a given metal and a fixed frequency above threshold, the photoelectric current is directly proportional to the intensity of incident light — more intense light ejects more electrons per second. Second, there exists a threshold frequency $\nu_0$ below which no emission occurs, however intense the light or however long it shines. Third, the maximum kinetic energy of the photoelectrons depends only on the frequency of the light (and the metal), not on its intensity; this maximum energy increases linearly with frequency. Fourth, the emission is instantaneous — the time lag between illumination and emission is less than $10^{-9}\,\text{s}$, even for very weak light.
These observations cannot be explained by the wave theory of light. On the wave picture, energy is spread continuously over the wavefront, so a weak beam should need a long time for an electron to accumulate enough energy — yet emission is instantaneous. The wave theory also predicts that the kinetic energy should grow with intensity and that emission should occur at any frequency given enough time. Both predictions flatly contradict the threshold frequency and the frequency-dependence of kinetic energy. A new idea was needed.
The work function of caesium is $2.14\,\text{eV}$. Find the threshold frequency $\nu_0$ for photoelectric emission. (Take $h = 6.63\times10^{-34}\,\text{J s}$.)
Solution- At threshold, the photon energy just equals the work function: $h\nu_0 = \phi_0$.
- Convert: $\phi_0 = 2.14\times 1.6\times10^{-19} = 3.42\times10^{-19}\,\text{J}$.
- $\nu_0 = \dfrac{\phi_0}{h} = \dfrac{3.42\times10^{-19}}{6.63\times10^{-34}}$.
- $\nu_0 = 5.16\times10^{14}\,\text{Hz}$.
Answer: The threshold frequency is about $5.16\times10^{14}\,\text{Hz}$.
Light of frequency $8\times10^{14}\,\text{Hz}$ falls on a metal whose threshold frequency is $5\times10^{14}\,\text{Hz}$. Will photoemission occur? Justify.
Solution- Photoemission occurs only if the incident frequency exceeds the threshold frequency: $\nu > \nu_0$.
- Here $\nu = 8\times10^{14}\,\text{Hz}$ and $\nu_0 = 5\times10^{14}\,\text{Hz}$.
- Since $8\times10^{14} > 5\times10^{14}$, the condition is satisfied.
Answer: Yes, photoemission occurs because the incident frequency is greater than the threshold frequency.
When the intensity of incident light is doubled (frequency unchanged and above threshold), how do the photoelectric current and the maximum kinetic energy of photoelectrons change?
Solution- Photoelectric current is directly proportional to intensity, so doubling intensity doubles the number of photoelectrons emitted per second.
- Hence the photoelectric current doubles.
- Maximum kinetic energy depends only on frequency (and the metal), not on intensity.
- So the maximum kinetic energy of the photoelectrons stays unchanged.
Answer: The photoelectric current doubles, but the maximum kinetic energy of the photoelectrons remains the same.
The threshold wavelength for a metal is $5000\,\text{\u00c5}$. Calculate its work function in eV. (Take $hc = 1240\,\text{eV nm}$.)
Solution- Threshold wavelength $\lambda_0 = 5000\,\text{\u00c5} = 500\,\text{nm}$.
- Work function $\phi_0 = \dfrac{hc}{\lambda_0}$.
- $\phi_0 = \dfrac{1240\,\text{eV nm}}{500\,\text{nm}}$.
- $\phi_0 = 2.48\,\text{eV}$.
Answer: The work function of the metal is about $2.48\,\text{eV}$.
Explain why the wave theory of light cannot account for the instantaneous nature of photoelectric emission.
Solution- On the wave theory, light energy is spread continuously over the entire wavefront.
- A single electron would therefore absorb energy only gradually from the small portion of the wavefront falling on it.
- For weak light this accumulation would take an appreciable time (calculations give minutes to hours) before an electron gains $\phi_0$.
- Experiment shows emission begins within $10^{-9}\,\text{s}$, contradicting the wave prediction.
Answer: The wave theory predicts a long time lag for weak light, but emission is actually instantaneous, so the wave theory fails.
Two metals A and B have work functions $2\,\text{eV}$ and $4\,\text{eV}$. Which one emits photoelectrons for visible light (energy range about $1.8$ to $3.1\,\text{eV}$)?
Solution- Photoemission requires the photon energy to be at least equal to the work function.
- For metal A, $\phi_0 = 2\,\text{eV}$; visible photons up to $3.1\,\text{eV}$ exceed this, so emission occurs.
- For metal B, $\phi_0 = 4\,\text{eV}$; even the most energetic visible photon ($3.1\,\text{eV}$) is below this, so no emission.
Answer: Only metal A emits photoelectrons with visible light; metal B needs ultraviolet light.