Photoelectric effect

Physics· 10 min read

1. Observations and Failure of Wave Theory★★☆☆☆⏱ 3 min

📘 Definition

Photoelectric effect

Emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency is absorbed by the surface.

Example:

Photoemission from a zinc plate illuminated by ultraviolet radiation

Early 20th century experiments produced four key observations that could not be explained by the classical wave model of light, which assumes energy is spread evenly across a wavefront:

Photoelectrons are only emitted if incident frequency is above a metal-specific threshold frequency $f_{0}$
Emission is instantaneous, even at very low intensity
Maximum kinetic energy of photoelectrons depends only on frequency, not intensity
Increasing intensity increases the number of photoelectrons emitted per second, not their maximum kinetic energy

📐 Worked Example

A student observes instantaneous photoelectron emission when low-intensity blue light shines on a sodium surface. They then shine high-intensity low-frequency red light on the same surface, but detect no emission. Explain why no photoelectrons are released.

1
First, recall that photoelectric emission only occurs when incident frequency is above the metal's threshold frequency.
2
Red light has a frequency below sodium's threshold frequency. The classical wave model incorrectly predicts that high intensity would eventually provide enough energy for emission.
3
In the photon model, one photon interacts with one electron. Each photon of red light has energy below sodium's work function, so no electron can gain enough energy to escape, regardless of how many photons (how high the intensity) hit the surface.

2. Einstein's Photoelectric Equation★★★☆☆⏱ 4 min

📘 Definition

Work function

$Φ$

Minimum energy required to release one electron from the surface of a metal, dependent on the metal type.

📘 Definition

Threshold frequency

$f_{0}$

Minimum frequency of incident radiation that can cause photoelectric emission, related to work function by $Φ = h f_{0}$ .

Einstein explained the observations by assuming light behaves as discrete packets called photons, each with energy $E = h f$ . One photon interacts with one electron at the metal surface. The electron uses at least the work function energy to escape the metal, and any remaining energy becomes its kinetic energy. This gives the core photoelectric equation:

hf = \Phi + KE_{max}

The maximum kinetic energy can also be written in terms of stopping potential $V_{s}$ , the reverse potential needed to stop the most energetic photoelectrons: $K E_{ma x} = e V_{s}$ .

📐 Worked Example

The work function of caesium is 2.14 eV. Calculate (a) the threshold frequency, (b) the stopping potential for incident radiation of frequency $6.5 \times 1 0^{14}$ Hz. Take $h = 6.63 \times 1 0^{- 34} J s$ , $e = 1.60 \times 1 0^{- 19} C$ .

1
Convert work function from eV to joules:
2
$\Phi = 2.14 \times 1.60 \times 10^{-19} = 3.424 \times 10^{-19} J$
3
(a) Calculate threshold frequency from $f_{0} = \frac{Φ}{h}$ :
4
$f_0 = \frac{3.424 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.16 \times 10^{14} Hz$
5
(b) Substitute into Einstein's equation rearranged for stopping potential, where $e V_{s} = h f - Φ$ :
6
$hf = 6.63 \times 10^{-34} \times 6.5 \times 10^{14} = 4.31 \times 10^{-19} J$
7
$V_s = \frac{4.31 \times 10^{-19} - 3.424 \times 10^{-19}}{1.60 \times 10^{-19}} = 0.55 V$

3. Graphical Interpretation★★★☆☆⏱ 2 min

Rearranging Einstein's equation gives a linear relationship between maximum kinetic energy (or stopping potential) and incident frequency. This linear relationship confirms the photon model and allows experimental measurement of Planck's constant and work function.

Graph Feature	Physical Meaning
Gradient of $K E_{ma x}$ vs $f$	$h$ (Planck constant)
Gradient of $V_{s}$ vs $f$	$h / e$
Y-intercept of $K E_{ma x}$ vs $f$	$- Φ$
X-intercept (where $K E_{ma x} = 0$ )	Threshold frequency $f_{0}$

📐 Worked Example

A graph of $V_{s}$ (V) against $f$ (Hz) for an unknown metal has a gradient of $4.1 \times 1 0^{- 15} V / H z$ and an x-intercept at $5.0 \times 1 0^{14} H z$ . Calculate $h$ and the work function of the metal.

1
From $V_{s} = \frac{h}{e} f - \frac{Φ}{e}$ , gradient equals $\frac{h}{e}$ :
2
$h = \text{gradient} \times e = 4.1 \times 10^{-15} \times 1.60 \times 10^{-19} = 6.56 \times 10^{-34} Js$
3
The x-intercept is threshold frequency $f_{0}$ , so $Φ = h f_{0}$ :
4
$\Phi = 6.56 \times 10^{-34} \times 5.0 \times 10^{14} = 3.28 \times 10^{-19} J = 2.05 eV$

4. Common Pitfalls

Wrong move:

Forgetting to convert work function from eV to joules before calculating threshold frequency.

Why:

Planck's constant is usually given in $J s$ , so mixing energy units gives a result that is orders of magnitude wrong.

Correct move:

Always convert all energy values to joules before substitution, or use $h$ in $e V s$ for unit consistency.

Wrong move:

Claiming higher intensity of incident light increases the maximum kinetic energy of photoelectrons.

Why:

Intensity measures the number of photons per second, not the energy per photon. Maximum kinetic energy depends only on frequency.

Correct move:

State that higher intensity increases the rate of photoelectron emission, not their maximum kinetic energy.

Wrong move:

Thinking emission can occur below threshold frequency if you wait long enough at high intensity.

Why:

This is a wrong prediction from classical wave theory. The photon model requires each individual photon to have enough energy.

Correct move:

No emission can ever occur below threshold frequency, regardless of intensity or exposure time.

Wrong move:

Taking the gradient of a $V_{s}$ vs $f$ graph to be equal to $h$ directly.

Why:

Confusing the $K E_{ma x}$ vs $f$ relationship with the $V_{s}$ vs $f$ relationship.

Correct move:

Gradient equals $h$ for $K E_{ma x}$ vs $f$ , and $h / e$ for $V_{s}$ vs $f$ .

Wrong move:

Assuming all photoelectrons have kinetic energy equal to $K E_{ma x}$ .

Why:

$K E_{ma x}$ describes only electrons emitted from the metal surface. Electrons from deeper inside lose extra energy escaping.

Correct move:

Only use $K E_{ma x}$ in Einstein's equation, as it refers to the most energetic emitted electrons.

5. Quick Reference Cheatsheet

Concept	Key Formula/Relationship
Photon energy	$E = h f = \frac{h c}{λ}$
Work function - threshold frequency	$Φ = h f_{0}$
Einstein's photoelectric equation	$h f = Φ + K E_{ma x}$
Stopping potential relation	$K E_{ma x} = e V_{s}$
Gradient of $K E_{ma x}$ vs $f$	$= h$
Gradient of $V_{s}$ vs $f$	$= h / e$
Core principle	One photon interacts with one electron

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
Threshold frequency multiple choice
2022 · 2
Graph analysis for work function
2021 · 4
Explain observations vs wave theory

What's Next

The photoelectric effect provided the first definitive experimental evidence for the particle nature of light, forming the foundation of all modern quantum physics. Mastering its principles and calculations is critical for all subsequent quantum topics in the CIE 9702 syllabus. The photon model introduced here is extended to matter particles in wave-particle duality, and applied to atomic energy levels to explain line spectra, both core exam topics.