The photoelectric effect is when photons hit a photoactive material and eject electrons (photoelectrons). Each photon has energy hf; to remove an electron the photon must have at least the material’s work function φ (so there’s a threshold frequency f0 = φ/h). If f ≥ f0 electrons are emitted, and the maximum kinetic energy of an ejected electron is Kmax = hf − φ. Kmax depends on the light’s frequency, not on how many photons hit the surface—that’s evidence light is quantized into photons. In lab setups you shine monochromatic light on a metal cathode in a vacuum cell, measure current vs. applied potential, and find the stopping potential where current → 0; eVstop = Kmax lets you determine φ. This is directly tested by AP (Topic 15.5: threshold frequency, work function, Einstein photoelectric equation). For a clear AP-aligned review, see the Topic 15.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD). More unit review and tons of practice problems are at the unit page (https://library.fiveable.me/ap-physics-2-revised/unit-15) and practice set (https://library.fiveable.me/practice/ap-physics-2-revised).

Only certain light frequencies eject electrons because light comes in photons with energy E = hf. Each electron needs at least the material’s work function φ to escape. If hf < φ (frequency below the threshold frequency), a single photon doesn’t have enough energy, so no electrons are emitted no matter how bright the light is. Once f ≥ f_threshold, electrons can be emitted and the maximum kinetic energy is K_max = hf − φ—so higher frequency (not higher intensity) makes faster photoelectrons. This behavior (threshold frequency, quantized photons, K_max = hf − φ) is exactly what the CED asks you to know for Topic 15.5. For a quick review and examples, check the Photoelectric Effect study guide on Fiveable (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD). For broader unit review and lots of practice problems, use the Unit 15 page and Fiveable practice questions (https://library.fiveable.me/ap-physics-2-revised/unit-15; https://library.fiveable.me/practice/ap-physics-2-revised).

Threshold frequency is the minimum light frequency f_threshold needed to free an electron from a material by the photoelectric effect. If f < f_threshold, no electrons are emitted no matter how bright (intense) the light is—intensity only changes the number of photons, not their individual energy. Each photon has energy E = hf. To eject an electron you need hf ≥ φ, where φ is the work function (minimum energy to remove an electron). So f_threshold = φ/h. If hf > φ, emitted electrons can have excess kinetic energy: K_max = hf − φ (that’s the Einstein photoelectric equation on the CED). On the AP exam you won’t need actual work-function values (they’re given if needed), but you should be able to use K_max = hf − φ and relate it to stopping potential in experiments. For a quick review, check the Topic 15.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD) and try practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Number of photons vs. energy per photon: they’re two different things. The energy of each photon is E = hf (so set by frequency), and that determines whether an individual photon can overcome the material’s work function φ and how much kinetic energy an emitted electron can have (Kmax = hf − φ). The number of photons hitting the surface is basically the light’s intensity (photon flux). Increasing the photon number increases how many electrons are ejected per second (photoelectric current), but it does not change Kmax. If the photon energy is below the threshold frequency (so hf < φ) no electrons are emitted no matter how many photons hit. This distinction is exactly what the CED highlights for Topic 15.5 and is commonly tested (use Kmax = hf − φ on FRQs). For a quick review, see the Topic 15.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD) and grab practice problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

Use Kmax = hf − φ exactly like an energy budget: each photon gives energy hf; φ (the work function) is the minimum energy to free an electron; the leftover is the maximum kinetic energy of the ejected electron. Steps to solve problems 1. Identify what’s given/asked (f, λ, φ, Kmax, or stopping potential Vstop). 2. If wavelength is given, convert to frequency: f = c/λ. 3. Plug into Kmax = hf − φ. If Kmax comes out negative → no electrons are emitted (f < threshold f0 = φ/h). 4. If they ask for stopping potential, use Kmax = eVstop (so Vstop = Kmax/e). 5. Watch units: h = 6.63×10^−34 J·s, c = 3.00×10^8 m/s, e = 1.60×10^−19 C. Example: given λ and φ in eV, convert φ to joules (φ[eV]×1.60×10^−19), find f = c/λ, compute Kmax, then Vstop = Kmax/e. This matches the CED essential knowledge (15.5.A.2–3). For more worked examples and practice problems, see the Topic 15.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD) and the AP Physics 2 practice bank (https://library.fiveable.me/practice/ap-physics-2-revised).

The work function φ is the minimum energy needed to free an electron from a material—basically the energy barrier an electron must overcome to escape the metal (CED 15.5.A.3.i). In the photoelectric equation Kmax = hf − φ (CED 15.5.A.3.ii), φ sets the threshold frequency f₀: φ = h f₀. If hf < φ no electrons are emitted. Different materials have different φ because their electrons are bound differently: atomic/solid electronic structure, Fermi level, lattice arrangement, and surface chemistry (oxide layers, contamination) change how tightly electrons sit in the material. Metals with higher electron binding energy have larger φ and thus need higher-frequency light to emit electrons. On the AP exam you won’t need to memorize φ values (they’ll be given if needed). For a quick AP-aligned refresher, see the Topic 15.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD). For more review and practice, check the unit page (https://library.fiveable.me/ap-physics-2-revised/unit-15) and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Because light comes in photons, each with energy E = hf, emission needs every electron’s single-photon energy to meet or exceed the work function φ. The threshold frequency f₀ is defined by hf₀ = φ. If your light has f < f₀, each photon has hf < φ, so no single photon can free an electron—increasing brightness just gives you more low-energy photons, not more energy per photon, so no electrons are emitted. This is exactly what the CED stresses: emission requires a minimum frequency regardless of photon number (15.5.A.2.i and 15.5.A.2.ii), and Kmax = hf − φ only applies when hf > φ (15.5.A.3.ii). Note: very intense lasers can produce multi-photon effects, but that’s beyond AP scope and the typical photoelectric setup. For a quick review, see the Topic 15.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD) and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Think of the classic photoelectric setup as a vacuum tube with two metal plates (a photoactive cathode and a collecting anode) connected to a variable power supply and an ammeter. Monochromatic light shines on the cathode; if each photon has energy hf ≥ φ (the work function), electrons are ejected (photoelectrons) into the vacuum. Some of those electrons reach the anode and produce a measurable current (photoelectric current). You vary the potential difference (make the anode more negative relative to the cathode) until the current drops to zero—that voltage is the stopping (or cutoff) potential. At that point eVstop = Kmax = hf − φ, so measuring Vstop vs. frequency lets you determine φ and confirm photons are quantized. The whole experiment is done in vacuum to prevent collisions and uses monochromatic light so hf is well defined (CED 15.5.A, 15.5.A.2–3). For a quick review, see Fiveable’s photoelectric study guide (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD) and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Good question—think of photons like individual bullets: each electron needs one photon with enough energy (hf ≥ φ) to escape. The maximum kinetic energy of an emitted electron comes from a single photon's energy minus the work function: Kmax = hf − φ (CED 15.5.A.3.ii). So if you increase the number of photons (light intensity) you increase how many electrons get hit per second → larger photoelectric current, but each electron still only gets energy from the one photon that hit it. That’s why Kmax (or the stopping potential in an experiment) doesn’t change with intensity, only with frequency; raise f and hf rises, so Kmax rises. Also remember the threshold frequency: below it no electrons are emitted regardless of how many photons hit (CED 15.5.A.2 and 15.5.A.2.i). For a short review, check the Topic 15.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD) and try practice problems (https://library.fiveable.me/practice/ap-physics-2-revised) to see current vs. stopping-potential graphs.

If the light’s frequency is exactly the threshold frequency f0, each photon has just enough energy to free an electron (hf0 = φ) but no extra to give it kinetic energy. So electrons can be emitted, but their maximum kinetic energy Kmax = hf − φ = 0. Practically that means emitted electrons barely escape the surface (zero KE), the stopping potential is essentially 0 V, and you’d measure a very small or easily interrupted photoelectric current (rate still depends on photon flux). This is exactly what the CED emphasizes: emission requires f ≥ f0, and Kmax = hf − φ (Topic 15.5). For extra practice or to review how this shows light’s particle nature, check the Topic 15.5 study guide on Fiveable (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD) and try related practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

The photoelectric effect shows light acts like particles (photons) because experiments contradict a pure-wave picture. Key observations from the CED: (1) Electrons are only emitted if incident light has a minimum frequency f₀ (threshold frequency), independent of intensity. (2) Increasing intensity (more light waves per second) raises the number of emitted electrons but not their maximum kinetic energy. These match Einstein’s photon idea: each photon has energy E = hf, and one photon transfers its energy to one electron. If hf < φ (the work function), no electron is emitted regardless of brightness; if hf > φ, the maximum KE = hf − φ (Kmax = hf − φ). The stopping-potential experiment (adjust potential until current stops) measures Kmax and confirms the linear dependence on frequency, not intensity—strong evidence for quantized photons. For the AP exam, be ready to use Kmax = hf − φ and explain threshold frequency (Topic 15.5; see the Fiveable study guide for this topic: https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD). For extra practice, check unit resources and practice problems (https://library.fiveable.me/ap-physics-2-revised/unit-15 and https://library.fiveable.me/practice/ap-physics-2-revised).

You adjust the potential until no current flows to find the stopping potential—the voltage that just prevents the fastest (maximum-energy) photoelectrons from reaching the anode. At that point the electric potential energy change eVstop equals the electron’s maximum kinetic energy, so Kmax = eVstop. Using Einstein’s photoelectric equation Kmax = hf − φ, you can therefore get eVstop = hf − φ and solve for φ or check the linear relationship between Vstop and frequency. That’s why the CED’s typical setup varies the potential until current = 0: it isolates the maximum kinetic energy of emitted electrons (e is the elementary charge) and gives direct experimental evidence for photons and the work function. For more AP-style review and practice on this topic, see the Photoelectric Effect study guide (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD) and unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-15).

Photon energy (Eph) = hf. In the photoelectric effect a single photon gives its energy to one electron. If Eph is less than the work function φ (minimum energy to remove an electron) no electrons are emitted. If Eph ≥ φ, electrons are emitted and the maximum kinetic energy of an emitted electron is Kmax = hf − φ. So increasing the light’s frequency raises Kmax linearly; increasing intensity (more photons/sec) only raises the number of emitted electrons, not their individual energies. You can measure Kmax by the stopping potential Vstop: eVstop = Kmax. These are the exact AP ideas in CED 15.5.A (threshold frequency, work function, Einstein photoelectric equation). For a quick review, see the Topic 15.5 study guide (Fiveable) (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD). For more practice problems across the unit, check Fiveable practice (https://library.fiveable.me/practice/ap-physics-2-revised).

In the lab you would watch a photoactive metal emit electrons when hit with monochromatic light and measure two main things: the photoelectric current and the stopping (retarding) potential. Using a vacuum photoelectric cell, you’d vary light frequency (and intensity) and adjust the potential difference between cathode and anode until the current drops to zero—that stopping potential Vstop gives Kmax = eVstop. Key observations: (1) below a threshold frequency no electrons are emitted, (2) above threshold the maximum kinetic energy of electrons increases linearly with frequency (Kmax = hf − φ), and (3) changing intensity changes current (number of electrons) but not Kmax. From a plot of eVstop vs. f you can get Planck’s constant (slope = h) and the work function φ (intercept). This matches the AP CED goals (15.5.A, stopping potential, work function)—review the Topic 15.5 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD) and practice more problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

Check the metal’s work function φ (minimum energy to free an electron). Convert the given light frequency f to photon energy using E = hf (h = Planck’s constant). If hf ≥ φ (equivalently f ≥ f_threshold where φ = h f_threshold), electrons can be emitted; if hf < φ, no photoelectric emission occurs no matter the intensity. The maximum kinetic energy of emitted electrons is K_max = hf − φ, and you can measure K_max experimentally from the stopping potential V_s via eV_s = K_max. On the AP exam you may be given φ or f_threshold; use K_max = hf − φ and remember emission depends on frequency (not photon number), which is key evidence for photons (CED 15.5.A.2–3). For a quick review and practice problems, see the topic study guide (https://library.fiveable.me/ap-physics-2-revised/unit-7/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD) and unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-15).