Photons and the Photoelectric Effect — AP Physics 2
1. Introduction to the Photoelectric Effect ★★☆☆☆ ⏱ 2 min
The photoelectric effect is the phenomenon where electrons are ejected from a material (typically a metal) when light of sufficient frequency is incident on its surface. Early 20th century experiments showed results that could not be explained by the classical wave model of light: the kinetic energy of ejected electrons depends only on light frequency, not intensity, and no electrons are ejected below a threshold frequency regardless of intensity.
Albert Einstein explained this in 1905 by proposing that light is made of discrete energy packets called photons, rather than a continuous wave. This topic accounts for 10-16% of your total AP Physics 2 exam score, and appears regularly in both multiple-choice and free-response questions, often testing conceptual understanding of the photon model versus classical wave theory.
2. Photon Energy and Quantization of Light ★★☆☆☆ ⏱ 4 min
Every photon of light with frequency $f$ has energy proportional to its frequency, given by:
E = hf = \frac{hc}{\lambda}
where $h = 6.626 \times 10^{-34} \text{ J·s}$ is Planck's constant, $c = 3 \times 10^8 \text{ m/s}$ is the speed of light, and $\lambda$ is the wavelength of the light. A very useful approximation for AP problems converts $hc$ to electron-volt nanometer units: $hc \approx 1240 \text{ eV·nm}$. This eliminates unit conversions when working with wavelength in nanometers, the most common unit for visible/UV light.
Intensity of light in the photon model is the number of photons incident per unit area per unit time, not the energy per photon. Higher intensity means more photons, not more energetic photons, which is the key difference from the classical wave model.
Exam tip: Always memorize $hc = 1240 \text{ eV·nm}$ for the AP exam. It cuts down calculation time for photon energy problems by 75% and eliminates unit conversion errors.
3. Einstein's Photoelectric Effect Equation ★★★☆☆ ⏱ 4 min
The minimum frequency of photon that can eject an electron is called the **threshold frequency** $f_0$, where $hf_0 = \Phi$, so $f_0 = \frac{\Phi}{h}$. The corresponding maximum wavelength that can eject an electron is threshold wavelength $\lambda_0 = \frac{hc}{\Phi}$.
By conservation of energy, the energy of the incoming photon goes into escaping the metal plus the kinetic energy of the ejected electron. For the most loosely bound electrons (which have the highest kinetic energy after ejection), this gives Einstein's photoelectric equation:
hf = \Phi + K_{max}
A key result: increasing the intensity of light (adding more photons) does not change $K_{max}$, it only increases the number of electrons ejected. Only increasing frequency (increasing energy per photon) increases $K_{max}$.
Exam tip: When asked to explain why no electrons are ejected by high-intensity light below threshold frequency, always explicitly state that intensity corresponds to number of photons, not energy per photon; each individual photon still has energy below the work function.
4. Stopping Potential and Graphical Analysis ★★★☆☆ ⏱ 4 min
In the classic photoelectric effect experiment, $K_{max}$ is measured experimentally using a reverse potential difference (called the stopping potential $V_s$) between the metal emitter and a collector plate. The stopping potential is the minimum voltage that stops the most energetic electrons from reaching the collector, so all of the maximum kinetic energy is converted to electric potential energy:
K_{max} = eV_s
Substituting into Einstein's equation gives a linear relationship between $V_s$ and incident frequency $f$, which is used to measure Planck's constant experimentally:
V_s = \left(\frac{h}{e}\right)f - \frac{\Phi}{e}
This is a straight line with slope equal to $\frac{h}{e}$, which is the same for all metals, and x-intercept equal to the threshold frequency $f_0$. The y-intercept is $-\frac{\Phi}{e}$, so you can calculate the work function directly from the graph.
Exam tip: If a question shows a $V_s$ vs $f$ graph for two different metals, the lines are always parallel (same slope = same $h/e$ for all metals); any answer option claiming different slopes for different metals is automatically wrong.
5. Concept Check (AP Style) ★★★☆☆ ⏱ 2 min
Common Pitfalls
Why: Students mix up energy flow, incorrectly thinking the electron receives both the photon energy and the work function to escape.
Why: Confuses classical wave theory predictions with the photon model, mixing up intensity (number of photons) and energy per photon.
Why: The $hc=1240 \text{ eV·nm}$ shortcut works for nanometers, but students accidentally use nanometers when calculating energy in joules.
Why: Forgets the $1/e$ factor in the linear relation between $V_s$ and $f$.
Why: Misapplies classical mass-energy relations to relativistic photons.