Photoelectric Effect Calculator

Calculate maximum kinetic energy of ejected electrons from photon frequency and work function.

Visible light: 4–7.5×10¹⁴ Hz
Cu: 4.5 eV, Na: 2.3 eV, Cs: 2.0 eV
Please check your inputs and try again.

Photon Energy, Work Function, and Electron Emission

The photoelectric effect is controlled by the energy of individual photons, not by the total light power alone. A photon of frequency f carries energy E = hf. To escape a material's surface, an electron must receive at least the work function φ, which is a material-dependent binding energy. Any photon energy above that minimum can appear as the maximum kinetic energy of an emitted electron.

This creates a threshold frequency f0 = φ/h. Light below f0 cannot eject electrons in the basic one-photon photoelectric model, even if the light is made more intense. Increasing intensity at a fixed frequency above threshold increases the number of incident photons and therefore can increase photoelectric current, but it does not increase the maximum kinetic energy. Raising frequency does increase maximum electron kinetic energy.

Kmax = hf − φ     f0 = φ/h
SymbolMeaningWhy it appears / units
hPlanck constant4.135667696 × 10−15 eV·s when energies are in eV
fPhoton frequencyHz; determines photon energy
φWork functioneV; minimum energy needed to remove an electron from the surface
KmaxMaximum electron kinetic energyeV; excess photon energy after overcoming φ

If hf equals φ exactly, the most energetic emitted electrons leave with essentially zero kinetic energy. When Kmax is expressed in electronvolts, its numerical value also equals the stopping potential in volts for a single electron, because eV is defined from the energy gained by one elementary charge through one volt.

Worked Examples

Example 1: UV on copper: f=1×10¹⁵ Hz, φ=4.5 eV
E=4.14 eV, φ=4.5 eV
Result: No emission — photon too weak
Need higher frequency UV for copper
Example 2: Far-UV on sodium: f=2×10¹⁵ Hz, φ=2.3 eV
E=8.27 eV, φ=2.3 eV
Result: KE = 5.97 eV
High energy electrons ejected
Example 3: Ultraviolet light on sodium
f = 8.0 × 1014 Hz, φ = 2.3 eV → hf = 3.31 eV; Kmax = 3.31 − 2.30
Result: Kmax ≈ 1.01 eV
The corresponding stopping potential is about 1.01 V because the electron kinetic energy is 1.01 eV.
Example 4: 300 nm light and a 2.0 eV work function
f = c/λ ≈ 9.99 × 1014 Hz; hf ≈ 4.13 eV; Kmax = 4.13 − 2.00
Result: Kmax ≈ 2.13 eV
Wavelength must first be converted to frequency when the working equation is written as E = hf.

Common Mistakes

⚠️
Expecting brighter sub-threshold light to eject electrons

In the basic photoelectric effect, intensity cannot compensate for photon energy below the work function. The frequency must exceed threshold.

⚠️
Mixing joules and electronvolts

Use a Planck constant in units consistent with the work function. If φ is in eV, h = 4.1357 × 10−15 eV·s is convenient.

⚠️
Reporting negative kinetic energy

If hf − φ is negative, the correct physical conclusion is no photoelectron emission in the one-photon model, not a negative electron kinetic energy.

⚠️
Confusing intensity with photon energy

Frequency sets energy per photon. Intensity describes energy delivered per area per time and mainly changes the photon arrival rate at fixed frequency.

Frequently Asked Questions

What is the threshold frequency?
The minimum frequency where f₀ = φ/h. Below this, no electrons are emitted regardless of light intensity.
Why doesn't intensity matter?
Einstein showed each photon interacts with one electron. More intensity = more photons = more electrons, but not higher KE. Only frequency determines KE.
What happens exactly at the threshold frequency?
At f = f0, the photon energy hf equals the work function φ. In the idealized model, electrons at the most favorable surface states can escape with maximum kinetic energy approaching zero. Above threshold, the excess energy appears as electron kinetic energy.
How is stopping potential related to photoelectron kinetic energy?
A reverse voltage can be applied until even the fastest photoelectrons no longer reach the collector. Then eVs = Kmax. If Kmax is stated in electronvolts, the stopping-potential magnitude has the same numerical value in volts.
Why did the photoelectric effect support the photon model of light?
Experiments showed a sharp threshold frequency and an immediate electron response, while maximum electron kinetic energy increased with frequency rather than intensity. Those observations are naturally explained when light transfers energy in discrete packets E = hf instead of only as a continuously distributed classical wave.
Does every emitted electron have Kmax?
No. Kmax is the maximum measured kinetic energy. Electrons can originate from different states or lose energy before leaving the surface, so actual emitted electrons commonly have a distribution of energies below the maximum value predicted by Einstein's photoelectric equation.

Formula Explorer connections

Interpretation: This relationship connects light propagation, geometry, wavelength, refraction, interference or image formation. Assumption: Use a consistent sign convention and units. Paraxial rays, thin elements, coherent light, vacuum wavelength or ideal optical components may be assumed.

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