Photoelectric Effect

Einstein's photoelectric equation

When light hits a metal surface, each photon can knock out at most one electron, and only if the photon carries enough energy to first overcome the metal's work function $\phi$ , the minimum energy binding an electron to the surface. Whatever energy is left over becomes the electron's kinetic energy:

h\nu = \phi + KE_{max}

where $h\nu$ is the photon's energy ( $h$ is Planck's constant, $6.626\times10^{-34}$ J·s). Below the threshold frequency $\nu_0 = \phi/h$ , no electrons are emitted at all, no matter how intense the light is, a single low-energy photon still can't individually clear the work function.

The stopping potential $V_0$ is the reverse voltage that just stops the fastest electrons, so it measures $KE_{max}$ directly: $eV_0 = KE_{max}$ . In electron-volts, the two are the same number.

What depends on what

The particle picture predicts how each quantity responds to a change in the light:

Effect of changing the light, for a fixed metal

Change	$KE_{max}$ and stopping potential	Photocurrent
Higher intensity, same frequency	No change	Increases (more photons, more electrons)
Higher frequency, same intensity	Increases	Set by photons per second, not by frequency itself: at equal intensity, higher-frequency light delivers fewer photons, so the saturation current can be lower
Frequency below $\nu_0$	No emission	Zero, at any intensity

Emission also starts with no measurable delay, even in dim light, because each electron is freed by a single photon rather than by energy slowly building up.

Worked example

Sodium has a work function of 2.3 eV. Light of wavelength 400 nm shines on it. Find the maximum kinetic energy of the emitted electrons, the stopping potential, and the threshold wavelength above which no electrons are emitted.

Solution: Working in electron-volts and nanometres, photon energy is $E = hc/\lambda$ , and $hc \approx 1240\text{ eV}\cdot\text{nm}$ is a convenient constant to keep on hand:

E = \dfrac{1240}{400} = 3.1\text{ eV}

Subtracting the work function gives the maximum kinetic energy:

KE_{max} = E - \phi = 3.1 - 2.3 = 0.8\text{ eV}

So the stopping potential is $V_0 = 0.8$ V. The threshold wavelength is where $KE_{max}=0$ : the photon energy exactly equals the work function:

\lambda_0 = \dfrac{hc}{\phi} = \dfrac{1240}{2.3} \approx 539\text{ nm}

Any wavelength longer than this (lower photon energy) produces no photoelectrons at all from sodium, regardless of the light's intensity.

Worked example: work function from a stopping potential

Light of wavelength 400 nm falls on a metal and the stopping potential is found to be 1.0 V. Find the metal's work function and threshold wavelength.

Solution: The photon energy is again $\dfrac{1240}{400} = 3.1$ eV, and $KE_{max} = eV_0 = 1.0$ eV:

\phi = 3.1 - 1.0 = 2.1\text{ eV}, \qquad \lambda_0 = \dfrac{1240}{2.1} \approx 590\text{ nm}

Common mistakes

Getting the threshold direction backwards. No emission happens above the threshold wavelength (below the threshold frequency). Longer wavelength means less energy per photon.
Expecting brighter light to give faster electrons. Intensity changes how many electrons come out, not how fast the fastest ones are.
Mixing joules and electron-volts. $hc = 1240$ eV·nm only works with $\lambda$ in nm and energies in eV. With $h = 6.626\times10^{-34}$ J·s, use metres and joules throughout.
Treating $KE_{max}$ as every electron's energy. It's the maximum. Electrons from deeper in the metal lose some energy on the way out and leave slower.

Where this comes up in exams

The photoelectric effect, Hertz and Lenard's observations and Einstein's photoelectric equation are listed in the Dual Nature of Matter and Radiation units of the JEE Main 2026 and NEET (UG) 2026 syllabi. The JEE Advanced 2026 syllabus lists the photoelectric effect.

Checked against JEE (Main) 2026 syllabus (NTA), NEET (UG) 2026 syllabus (NMC, published by NTA) and JEE (Advanced) 2026 syllabus. Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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