Photoelectric Effect Graphs
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Direct answer
Set up a photocell, vary the voltage and frequency, and three straight-line graphs encode the whole photoelectric effect. Photocurrent against voltage rises to a saturation plateau whose height grows linearly with light intensity and not at all with frequency; the curve crosses the voltage axis at the stopping potential V0 where eV0 = hν − φ. Plot stopping potential against frequency and you get a straight line for every metal: slope h/e = 4.14 × 10⁻¹⁵ V s — the same universal value for all metals, the fact Millikan used to measure Planck's constant — with x-intercept at the threshold frequency ν0 = φ/h and y-intercept −φ/e. Doubling intensity doubles the saturation current and leaves V0 unchanged; raising frequency raises V0 and leaves saturation current untouched — every graph question is built on keeping that separation clean.
What you must remember
- Current-voltage curve: rises from zero (at −V0) to a saturation plateau; saturation current I_s ∝ intensity, and the plateau appears because every emitted electron is already being collected.
- Stopping potential: eV0 = hν − φ = K_max; only frequency moves V0, only intensity moves I_s.
- V0 versus ν: straight line of slope h/e = 4.14 × 10⁻¹⁵ V s (identical for every metal), x-intercept ν0 = φ/h (threshold), y-intercept −φ/e; steeper thinking is wrong — the slope never changes with the metal, only the intercepts shift.
- K_max versus ν: line of slope h starting at ν0; below threshold the current is zero no matter how intense the light.
- Intensity dependences: at fixed frequency above threshold, photocurrent ∝ intensity while K_max stays constant — brighter light means more electrons, not faster ones.
- Instantaneity: emission begins within about 10⁻⁹ s of illumination regardless of intensity, which no wave picture explains gracefully.
- Millikan's use: the universal slope h/e let him determine h from the graph itself, confirming Einstein's 1905 photon equation.
Graph traps examiners set
The number-one trap is the slope of the V0-ν line: it is h/e for every metal without exception, and options offering a metal-dependent slope (or e/h, or h) harvest the careless. Second, the y-intercept is negative, −φ/e, because V0 is zero at threshold; students who plot K_max versus ν instead get a positive φ/h intercept, and mixing the two graphs is the designed error. Third, saturation current against intensity is linear at fixed frequency but the curve against voltage is not — the plateau's existence is why. Fourth, sub-threshold behaviour: below ν0 the current is identically zero. Main tests single-graph reading; Advanced combines graphs — the V0-ν lines for two metals and which emits at a given frequency, or h and φ read from supplied axis numbers in one pass.
Frequently asked questions
What does the slope of the stopping potential versus frequency graph represent?
Planck's constant divided by e, numerically 4.14 × 10⁻¹⁵ V s, and it is identical for every metal — the universality that let Millikan measure h.
Why does the saturation current not change with frequency?
Because at saturation every emitted electron is collected already; frequency controls each electron's energy, and only intensity controls how many electrons there are.
How does doubling light intensity affect the photocurrent graph?
The saturation plateau doubles in height while the stopping potential stays exactly where it was — brighter light gives more electrons at the same maximum energy.
What are the intercepts of the V0 versus ν line?
The x-intercept is the threshold frequency ν0 = φ/h and the y-intercept is −φ/e, both metal-specific even though the slope h/e is universal.
Can photoemission occur below the threshold frequency if the light is very intense?
No — emission requires a single photon to carry at least the work function; intensity only changes the number of photons, never each photon's energy.