Photoelectric Effect — Worked Calculations

By Aniket Bhardwaj · 17 September 2026 · Physics · Class 12

The photoelectric effect is where light stops behaving like a wave and starts behaving like a stream of particles. Conceptually it is one equation. In practice students lose marks on the arithmetic: converting nanometres, switching between electronvolts and joules, and confusing the maximum kinetic energy with the energy every electron carries. This guide works four numericals in full and shows the shortcut that makes almost all of them one-line problems.

The one equation, and what it says

h ν = φ + KE_max so KE_max = h ν − φ = (h c / λ) − φ

Read it as an energy budget. One photon arrives carrying hν. Part of that energy, the work function φ, is spent freeing the electron from the metal surface. Whatever is left becomes the electron's kinetic energy. One photon frees at most one electron — that is the whole particle-picture claim, and it is why brighter light does not produce faster electrons.

The supporting relations

Photon energy: E = h ν = h c / λ
Threshold: φ = h ν₀ = h c / λ₀ (below ν₀, or above λ₀, nothing is emitted)
Stopping potential: e V₀ = KE_max, so V₀ = KE_max / e
Electron speed: KE_max = ½ m v_max², so v_max = √(2 KE_max / m)

Constants and units

Quantity	Value used here	Note
Planck constant h	6.626 × 10⁻³⁴ J s	Since the 2019 SI redefinition h is fixed exactly at 6.62607015 × 10⁻³⁴ J s; exams round it
Speed of light c	3.00 × 10⁸ m s⁻¹	Exactly 299 792 458 m s⁻¹ by definition; 3 × 10⁸ is the exam value
Electronic charge e	1.602 × 10⁻¹⁹ C	Also the joules in one electronvolt
Electron mass m	9.109 × 10⁻³¹ kg	Needed only when a speed is asked for
Product h c	1240 eV nm	The shortcut — see below

The shortcut worth memorising: hc = 1240 eV nm

Work it out once and you never do it again:

h c = 6.626 × 10⁻³⁴ × 3.00 × 10⁸ = 1.988 × 10⁻²⁵ J m
In electronvolts: 1.988 × 10⁻²⁵ ÷ 1.602 × 10⁻¹⁹ = 1.240 × 10⁻⁶ eV m = 1240 eV nm

So the energy of a photon in electronvolts is simply 1240 ÷ (wavelength in nm). Almost every photoelectric question becomes a single division.

A word about work-function values

Different books quote slightly different work functions for the same metal, because φ depends on the surface condition of the sample, not only on the element. That is a real physical fact, not sloppy printing. Always use the value printed in your own question paper or NCERT table and do not carry a remembered number in from another book. The examples below therefore treat φ as given data.

Worked example 1 — photon energy of violet light

Find the energy of a photon of wavelength 400 nm, in eV and in joules.

Using the shortcut: E = 1240 ÷ 400 = 3.10 eV
In joules: 3.10 × 1.602 × 10⁻¹⁹ = 4.97 × 10⁻¹⁹ J

Cross-check from first principles:
E = hc/λ = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (400 × 10⁻⁹)
= 1.988 × 10⁻²⁵ ÷ 4.00 × 10⁻⁷ = 4.97 × 10⁻¹⁹ J — the same answer, so the shortcut is safe.

The frequency, if asked: ν = c/λ = 3.00 × 10⁸ ÷ 400 × 10⁻⁹ = 7.50 × 10¹⁴ Hz.

Worked example 2 — kinetic energy, stopping potential and electron speed

The 400 nm light of example 1 falls on a metal whose work function is given as 2.30 eV. Find KE_max, the stopping potential and the maximum electron speed.

KE_max = E − φ = 3.10 − 2.30 = 0.80 eV
In joules: 0.80 × 1.602 × 10⁻¹⁹ = 1.28 × 10⁻¹⁹ J

Stopping potential: V₀ = KE_max ÷ e. With KE_max already in eV this is numerically the same number: V₀ = 0.80 V. That is not a coincidence — it is the reason the electronvolt was invented.

Maximum speed:
v_max = √(2 × 1.28 × 10⁻¹⁹ ÷ 9.109 × 10⁻³¹)
= √(2.563 × 10⁻¹⁹ ÷ 9.109 × 10⁻³¹) = √(2.814 × 10¹¹)
= 5.30 × 10⁵ m s⁻¹

About 0.2% of the speed of light, so treating the electron non-relativistically was fair.

Worked example 3 — threshold wavelength and threshold frequency

For the same metal (φ = 2.30 eV), find the longest wavelength that can eject an electron.

λ₀ = 1240 ÷ φ(in eV) = 1240 ÷ 2.30 = 539 nm

Threshold frequency:
ν₀ = φ ÷ h = (2.30 × 1.602 × 10⁻¹⁹) ÷ (6.626 × 10⁻³⁴)
= 3.685 × 10⁻¹⁹ ÷ 6.626 × 10⁻³⁴ = 5.56 × 10¹⁴ Hz

Cross-check: ν₀ should equal c/λ₀ = 3.00 × 10⁸ ÷ 539 × 10⁻⁹ = 5.57 × 10¹⁴ Hz. The tiny difference is only the rounding inside the 1240 shortcut.

Worked example 4 — the question that has no emission at all

The same metal is now lit with 600 nm light from a very bright lamp. What happens?

E = 1240 ÷ 600 = 2.07 eV, and φ = 2.30 eV.
Since 2.07 eV < 2.30 eV, no electrons are emitted at all — no matter how bright the lamp is or how long it is left on.

This is the observation classical wave theory could not explain. A wave picture says energy accumulates steadily, so a bright enough lamp should eventually free an electron. It never does. Each photon must individually carry enough energy, and 600 nm photons simply do not.

What changing intensity and frequency actually does

You change	Number of electrons per second	Maximum KE of each electron	Stopping potential
Intensity up, same frequency	Increases	Unchanged	Unchanged
Frequency up, same intensity	Roughly unchanged or fewer	Increases	Increases
Frequency below ν₀	Zero	Not defined — nothing emitted	Zero

A graph of stopping potential V₀ against frequency ν is a straight line of slope h/e. Putting numbers in: 6.626 × 10⁻³⁴ ÷ 1.602 × 10⁻¹⁹ = 4.14 × 10⁻¹⁵ V s. Its intercept on the frequency axis is ν₀, and the same slope is obtained for every metal — which is exactly how Planck's constant can be measured on a school bench.

Common mistakes that cost marks

Not converting nanometres. 400 nm is 400 × 10⁻⁹ m. Using 400 m gives an answer wrong by eighteen orders of magnitude.
Mixing eV and J in one equation. Convert everything to one unit first. If φ is in eV, keep the photon energy in eV too.
Forgetting that KE_max is a maximum. Only electrons right at the surface emerge with that energy; those from deeper down lose more and come out slower. The formula gives the top of the range, not the energy of every electron.
Believing brighter light means faster electrons. Intensity controls how many electrons; frequency controls how fast. This is the single most tested idea in the chapter.
Treating the threshold as something intensity can overcome. Below ν₀ the current is exactly zero, however intense or prolonged the light.
Confusing stopping potential with an accelerating voltage. V₀ is the reverse bias that just stops the fastest electron reaching the collector.
Quoting a remembered work function. Use the value your question supplies — published values differ between sources.

Where this appears in exams

Exam	Typical use
CBSE / ICSE Class 12	Dual nature of radiation and matter: KE_max, V₀, threshold, Einstein's equation
JEE / NEET	Graph-based questions on V₀ against ν, and photon-count problems
Class 12 chemistry	The same E = hc/λ powers atomic-structure and spectral-line calculations
Later chemistry courses	Photoelectron spectroscopy measures ionisation energies by exactly this principle

Check your photon-energy arithmetic instantly. The Modern Physics calculator handles the photon energy and de Broglie wavelength relations used throughout this article, so you can confirm an E = hc/λ value before building the rest of your answer on it.

Open the Modern Physics Calculator (photon energy & de Broglie) →

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