Photoelectric Effect
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Direct answer
When light of frequency nu falls on a metal of work function phi, electrons are emitted only if nu exceeds the threshold nu0 = phi/h, with maximum kinetic energy K_max = h nu − phi (Einstein's equation). The stopping potential satisfies e V_s = K_max. Intensity controls only the number of electrons per second — the photocurrent — and never their maximum kinetic energy: the two facts that overturned the wave picture of light.
What you must remember
- Einstein's equation: K_max = h nu − phi = h(nu − nu0); no emission below nu0 = phi/h, equivalently beyond threshold wavelength lambda0 = h c/phi.
- Stopping potential V_s = (h nu − phi)/e; the V_s-versus-nu graph is a straight line of slope h/e (identical for every metal) with frequency intercept nu0 and voltage intercept −phi/e.
- Photocurrent above threshold is proportional to intensity; K_max is independent of intensity, depending only on frequency and the metal.
- Emission is instantaneous — no lag between illumination and emission, as expected when a single photon hands all its energy to one electron.
- Saturation current flows when all emitted electrons are collected; the stopping potential is the retarding voltage that cuts the current to zero.
- Photon: E = h nu = h c/lambda, approximately 1240 eV-nm divided by wavelength in nm; momentum p = h/lambda.
- Work function is the minimum escape energy from the surface; deeper electrons emerge slower, which is why K_max defines the fast edge of the emitted spectrum.
Common confusion
The perennial confusion is intensity versus frequency. Doubling intensity doubles the photoelectrons per second (larger saturation current) but leaves the stopping potential untouched; raising frequency alone increases K_max even at feeble intensity, and no intensity, however large, ejects electrons below threshold — dim ultraviolet succeeds where intense red fails. Students also mix the graphs: the current–voltage curve rises with intensity, while the V_s–nu line shifts only with the metal, keeping slope h/e.
Exam-focused takeaway
JEE Main tests the Einstein equation directly — K_max or V_s from frequency and work function, wavelength-to-energy conversion, threshold quantities — as numerical-value questions worth taking slowly. JEE Advanced layers it: identifying metals from graph intercepts, emitter and collector of different metals handled by energy conservation, photon momentum linked to de Broglie wavelength, and power-to-photocurrent estimates. Keep the two causal chains separate — intensity sets number, frequency sets energy — and the traps dissolve.
Frequently asked questions
What does intensity control?
Only the rate of photoelectron emission, hence the photocurrent; maximum kinetic energy and stopping potential are untouched by it.
Why does intense red light fail to eject electrons?
Each red photon carries less than the work function, and emission is one photon to one electron; accumulating many sub-threshold photons does not help at ordinary intensities.
What is the stopping potential?
The retarding voltage that just stops the fastest electrons, so e V_s = K_max; its variation with frequency yields h/e from the slope and phi from the intercept.
Why is emission instantaneous?
A photon transfers its entire quantum in one absorption event, so the electron gains the full h nu at once instead of waiting to accumulate wave energy.
What are threshold frequency and wavelength?
nu0 = phi/h and lambda0 = h c/phi — the sharp cutoff of a given metal; light below nu0 ejects nothing, above it ejects with K_max growing linearly with frequency.
How do V_s–nu graphs differ across metals?
They are parallel (common slope h/e), but the metal with the larger work function has its threshold further right and its voltage intercept more negative.