# Photoelectric Effect Graphs > Photoelectric graphs for JEE Physics: saturation current vs intensity, stopping potential lines and the universal slope h/e of the V0–ν plot. - Canonical URL: https://prepelephant.com/topics/jee/physics/photoelectric-graphs - Exam / course: JEE · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Photoelectric Effect Graphs", PrepElephant, https://prepelephant.com/topics/jee/physics/photoelectric-graphs ## 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. ## Three graphs, one metal Take a metal with work function φ = 2.0 eV. Threshold frequency: ν0 = φ/h = 2.0/(4.14 × 10⁻¹⁵) ≈ 4.83 × 10¹⁴ Hz — below this, nothing happens at any intensity. Illuminate at ν = 6 × 10¹⁴ Hz: photon energy hν = 2.48 eV, so K_max = 0.48 eV and V0 = 0.48 V. On the V0-ν graph this metal is a line through (4.83 × 10¹⁴, 0) with slope 4.14 × 10⁻¹⁵. Now double the light intensity: saturation current doubles, V0 holds at 0.48 V; raise the frequency to 7 × 10¹⁴ Hz instead: V0 climbs to 0.90 V while the saturation current stays put. The composite graph question draws current-voltage curves for two intensities at one frequency — same V0 intercept, different plateau heights — or for two frequencies at one intensity — same plateau height, different intercepts — or two metals — parallel V0-ν lines with different thresholds. Each version is answered by asking which knob changed: intensity moves the plateau, frequency moves the intercept, the metal moves both intercepts of the V0-ν line while the slope h/e never moves. A second metal with φ = 3.0 eV has ν0 = 7.25 × 10¹⁴ Hz; light at 6 × 10¹⁴ Hz ejects from the first metal and not from the second. ## 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.