Photoelectric Effect — Einstein's Equation & Applications

Shine light on a metal and electrons can fly off, but only if each photon carries enough energy. The photoelectric effect turns that statement into a calculation, and almost every introductory problem reduces to one energy-balance equation plus a threshold check.

The Formula and Its Symbols

Einstein's photoelectric equation is

Here $h$ is Planck's constant, $f$ is the light frequency, $\phi$ is the work function of the material, and $K_{max}$ is the maximum kinetic energy of the emitted electrons. Emission is possible only when the threshold condition holds:

That inequality carries the central idea: brighter light is not enough if each photon is still below threshold.

Why the Equation Holds

The equation is an energy balance for a single photon meeting a single electron. A photon delivers a fixed packet of energy $hf$. To escape the surface, the electron must first pay the work function $\phi$, the minimum energy needed to break free. Whatever energy is left over becomes kinetic energy, and the most energetic electrons carry the full remainder $hf - \phi$. That is exactly the equation.

This also explains why frequency dominates over brightness. For a fixed material, higher frequency means larger energy per photon, $hf$, so each photon can pay $\phi$ and leave more behind. Intensity is different: it sets how many photons arrive per second, not how much energy each one carries. The same reasoning gives the threshold frequency

If $f < f_0$, no single photon can pay the work function and no electrons leave; if $f \ge f_0$, emission becomes possible. This is why the effect was historically decisive: it showed that "more light" is not the right idea, energy per photon is.

Worked Example Using Einstein's Equation

Suppose a metal has work function $\phi = 2.3\,\mathrm{eV}$ and the incident light has photon energy $3.0\,\mathrm{eV}$. Always start with the threshold check:

So electrons can be emitted. Now apply the equation:

The most energetic emitted electrons have kinetic energy $0.7\,\mathrm{eV}$. If the problem also asks for the stopping potential $V_s$, use $eV_s = K_{max}$ under the stopping-potential condition, where the retarding voltage just halts the fastest electrons. In electronvolt units that gives $V_s = 0.7\,\mathrm{V}$ here.

Try It Yourself

Keep the same work function, $\phi = 2.3\,\mathrm{eV}$, but lower the photon energy to $2.1\,\mathrm{eV}$. The useful question is not "what is $K_{max}$?" but "does emission happen at all?"

Answer check: compare $2.1\,\mathrm{eV}$ with $2.3\,\mathrm{eV}$. Since $2.1 < 2.3$, the threshold condition $hf \ge \phi$ fails, so no photoelectrons are produced. Plugging into the equation would give a meaningless negative $K_{max}$, which is the signal that you skipped the threshold check.

Calculation Pitfalls

• Applying the equation before checking the threshold. $K_{max} = hf - \phi$ is only meaningful once $hf \ge \phi$. A negative result means no emission, not a real energy.
• Thinking brighter light gives faster electrons. For fixed material and frequency above threshold, more intensity means more electrons, not a larger $K_{max}$.
• Confusing photon energy with total beam energy. The threshold depends on energy per photon. An intense beam can still fail if each photon is below threshold.
• Forgetting that $K_{max}$ is a maximum. Emitted electrons span a range of energies; the equation gives the largest one in the distribution.

Where the Photoelectric Effect Is Used

It powers devices that convert incoming photons into emitted electrons: vacuum phototubes and photomultiplier tubes. It also underlies photoelectron spectroscopy, where measured electron energies reveal electronic structure and binding energies. Historically, it is one of the clearest early pieces of evidence for the quantum picture of light. Use this model whenever light strikes a surface and ejects electrons; if the problem is about refraction, interference, or ordinary circuits, reach for a different model instead.