Photoelectric effect

Three observations classical waves could not explain. Heinrich Hertz first noticed (1887) that a spark gap sparked more easily under ultraviolet light; Lenard later showed electrons were being ejected from the metal surface. Careful experiments then established:

• Below a threshold frequency $\nu_0$, characteristic of the metal, no electrons are ever emitted, however intense the light.
• Above $\nu_0$, the maximum kinetic energy of the emitted electrons depends only on the frequency of the light, not on its intensity — intensity changes only how many electrons come out per second.
• Emission is, for all practical purposes, instantaneous, with no measurable time lag even for very weak light.

A classical wave picture predicts none of this: it expects any frequency to work eventually (energy just piles up until an electron can escape) and expects a measurable delay at low intensity. Both predictions fail in the laboratory.

Einstein's photon picture. Einstein (1905) proposed that light of frequency $\nu$ is quantised into photons, each carrying a fixed energy

where: $h = 6.63\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ is Planck's constant. A photon is absorbed by a single electron in one all-or-nothing event — there is no gradual accumulation, which is why emission is instantaneous. Part of the photon's energy, the metal's work function $W_0 = h\nu_0$, is spent freeing the electron from the surface; whatever energy remains becomes the electron's kinetic energy. This gives Einstein's photoelectric equation:

If $h\nu < W_0$ (that is, $\nu < \nu_0$), no term is left over for kinetic energy and no electron escapes, regardless of how many photons (how intense the light) arrive per second — this single equation explains all three puzzling observations at once.

Stopping potential. Emitted electrons can be turned back by a retarding potential $V_s$ applied between the plate and a collector. The stopping potential is the minimum retarding voltage at which even the fastest electrons are just stopped, so $eV_s = E_{max}$:

where: $e = 1.6\times10^{-19}\ \text{C}$ is the electron charge. A graph of $V_s$ against $\nu$ is a straight line whose slope is $h/e$ (giving an experimental route to Planck's constant) and whose $x$-intercept is the threshold frequency $\nu_0$. This experiment, performed with great care by Millikan, confirmed Einstein's equation exactly and earned both scientists Nobel Prizes.

Photon momentum. Light also carries momentum, since $E = pc$ for a massless quantum and $E = h\nu = hc/\lambda$:

This momentum is what radiation-pressure experiments measure to demonstrate that light behaves as a particle stream, not just a wave.

The work function of sodium is $2.3\ \text{eV}$. Light of wavelength $400\ \text{nm}$ falls on the sodium surface. Find (a) the threshold wavelength, (b) the maximum kinetic energy of the emitted photoelectrons in eV, and (c) the stopping potential.

Step 1. Find the threshold frequency and wavelength from $W_0 = h\nu_0$. Convert $W_0$ to joules: $W_0 = 2.3\times1.6\times10^{-19} = 3.68\times10^{-19}\ \text{J}$, so $\nu_0 = W_0/h = 3.68\times10^{-19}/6.63\times10^{-34} = 5.55\times10^{14}\ \text{Hz}$, giving $\lambda_0 = c/\nu_0 = 3\times10^8/5.55\times10^{14} = 5.41\times10^{-7}\ \text{m} = 541\ \text{nm}$.

Step 2. Find the energy of the incident photon at $400\ \text{nm}$: $E = hc/\lambda = (6.63\times10^{-34})(3\times10^8)/(400\times10^{-9}) = 4.97\times10^{-19}\ \text{J} = 3.11\ \text{eV}$.

Step 3. Subtract the work function to get the maximum kinetic energy: $E_{max} = E - W_0 = 3.11 - 2.3 = 0.81\ \text{eV}$.

Step 4. Since $eV_s = E_{max}$ and the energy is already in electron-volts, the stopping potential is numerically equal: threshold wavelength $541\ \text{nm}$, maximum kinetic energy $0.81\ \text{eV}$, stopping potential $V_s = 0.81\ \text{V}$.

Competitive stretch

Using stopping potentials of $1.0\ \text{V}$ at $\nu_1 = 6\times10^{14}\ \text{Hz}$ and $2.5\ \text{V}$ at $\nu_2 = 9\times10^{14}\ \text{Hz}$, apply the derivation above: $h = \dfrac{(1.6\times10^{-19})(2.5-1.0)}{9\times10^{14}-6\times10^{14}} = \dfrac{2.4\times10^{-19}}{3\times10^{14}} = 8.0\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ (close to the accepted $6.63\times10^{-34}$; the deviation here traces entirely to the rounded input data, not the method). Then $W_0 = h\nu_1 - eV_{s1} = (8.0\times10^{-34})(6\times10^{14}) - (1.6\times10^{-19})(1.0) = 4.8\times10^{-19} - 1.6\times10^{-19} = 3.2\times10^{-19}\ \text{J} = 2.0\ \text{eV}$. Result: $h \approx 8.0\times10^{-34}\ \text{J}\,\cdot\,\text{s}$, $W_0 = 2.0\ \text{eV}$.

• Thinking that raising the intensity of light below the threshold frequency will eventually cause emission — it will not; intensity only changes the number of photoelectrons emitted per second, never their individual energy.
• Confusing the threshold frequency $\nu_0$ (a fixed property of the metal) with the frequency of the incident light $\nu$ (which you can vary) — Einstein's equation only applies once $\nu \ge \nu_0$.
• Forgetting that the stopping potential $V_s$ is defined by the fastest electrons only — slower photoelectrons are turned back by smaller retarding voltages, but $V_s$ itself is fixed by the maximum-energy cutoff.
• Mixing up photon momentum $p = h/\lambda$ with photon energy $E = h\nu$ — related but distinct quantities; momentum is naturally expressed via wavelength, energy via frequency (though $E = hc/\lambda$ also works).

Ask the tutor: Want to practice extracting $h$ and $W_0$ from a different pair of stopping-potential readings, or see why intensity and frequency play such different roles here?