Photoelectric Effect (Einstein)
The photoelectric effect describes the emission of electrons from a metal surface when it is illuminated with light.
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Formula
E_{kin} = h \cdot f - W_AVariables & units – Photoelectric Effect (Einstein)
| Symbol | Meaning | Unit |
|---|---|---|
| E_kin | Kinetic energy of the emitted electron | J or eV |
| h | Planck constant | J·s |
| f | Frequency of the incident light | Hz |
| W_A | Work function (material-specific) | J or eV |
Derivation & background – Photoelectric Effect (Einstein)
In 1905, Albert Einstein explained the photoelectric effect observed by Hertz through the quantum hypothesis (photons with E = hf). For this he received the Nobel Prize in 1921, not for the theory of relativity.
Exam blueprint
Validity range
Applies to the external photoelectric effect when photon energy exceeds the material work function.
Derivation steps
One photon gives energy hf to one electron; the work function is paid first, the remainder becomes kinetic.
- 1Photon energy: E = h·f.
- 2Energy conservation: h·f = W_A + E_kin.
Rearrangements
Threshold frequency
Below this frequency no photoelectrons are emitted, regardless of light intensity.
Task variant
What does higher intensity change above the threshold frequency?
More photons emit more electrons; maximum kinetic energy rises only with f.
Common mistakes
Attributing electron energy to intensity instead of frequency.
Energy per photon is h·f; intensity counts photons per time.
Exam context
- Exams often ask for threshold frequency, stopping voltage and linear E(f) graphs.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Light quanta
Connects photon energy, matter waves and quantum models.
Worked example
Sodium (W_A = 2.28 eV), illuminated with UV light of f = 8×10¹⁴ Hz: E_kin = (6.626×10⁻³⁴ × 8×10¹⁴) / 1.6×10⁻¹⁹ − 2.28 ≈ 1.03 eV.
Applications
Solar cells, photodetectors, night-vision devices, digital cameras (CCD)
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Scientific sources
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Frequently asked questions about Photoelectric Effect (Einstein)
How do you calculate the kinetic energy of photoelectrons?+
Subtract the work function from the photon energy: E_kin = h·f − W_A. The photon energy is h·f with Planck constant h = 6.626×10⁻³⁴ J·s and the light frequency f in hertz. The work function W_A is the minimum energy to release an electron from the metal, often given in electronvolts. Example sodium with W_A = 2.28 eV and f = 8×10¹⁴ Hz: E_kin = (6.626×10⁻³⁴·8×10¹⁴)/1.6×10⁻¹⁹ − 2.28 ≈ 1.03 eV. Watch for consistent units; work either entirely in joules or entirely in electronvolts. If h·f is smaller than W_A, no electrons are emitted at all.
Why does higher light intensity not change the electron energy?+
Because the energy per photon depends solely on the frequency, E = h·f, and an electron always absorbs only a single photon. Higher intensity means more photons per time, but each individual photon has the same energy. Therefore more intense light above the threshold frequency releases more electrons, increasing the photocurrent, but does not change the maximum kinetic energy of the individual electrons. That depends only on f. Exactly this finding contradicted the classical wave theory of light and led Einstein in 1905 to the photon hypothesis, for which he later received the Nobel Prize.
What is the threshold frequency and how do you calculate it?+
The threshold frequency f_g is the smallest light frequency at which electrons are emitted at all. Exactly at it the photon energy just suffices to cover the work function, so the kinetic energy is zero. From h·f_g = W_A it follows that f_g = W_A/h. Below this frequency no photoelectrons are emitted, no matter how intense or how long you irradiate. For sodium with W_A = 2.28 eV = 3.65×10⁻¹⁹ J you get f_g = 3.65×10⁻¹⁹/6.626×10⁻³⁴ ≈ 5.5×10¹⁴ Hz, which lies in the green range of visible light. Convert W_A from electronvolts to joules before inserting.
What does the stopping voltage mean in the photoelectric effect?+
The stopping voltage U is the voltage you must apply to just barely stop even the fastest photoelectrons. At the stopping point the electrical work e·U equals the maximum kinetic energy: e·U = E_kin = h·f − W_A. From this you can calculate U = (h·f − W_A)/e. Measuring U for different frequencies gives a straight line U(f) with slope h/e; from it Planck constant is determined experimentally. The intercept yields the threshold frequency. The stopping voltage depends only on the frequency, not on the intensity of the light.
When does the photoelectric effect not occur at all?+
The external photoelectric effect occurs only when the photon energy h·f is at least as large as the work function W_A of the material. If the frequency is below the threshold frequency f_g = W_A/h, the energy of a single photon is not enough to release an electron, and nothing happens, regardless of the light intensity or the exposure time. Red light with low frequency therefore releases no electrons in many metals, while more energetic UV light does. Only above the threshold frequency does the effect begin abruptly, which strikingly demonstrates the quantum nature of light.
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How do you calculate with Photoelectric Effect (Einstein)?
Here is how to work through a typical Photoelectric Effect (Einstein) (Ekin = hf − WA) task step by step:
- 1
Task
What does higher intensity change above the threshold frequency?
Solution path
More photons emit more electrons; maximum kinetic energy rises only with f.