Photoelectric effect

Point an ultraviolet lamp at a clean sheet of zinc and something strange happens: electrons fly off the surface instantly, as if the metal had been struck. Turn the lamp brighter and more electrons leave, but each one leaves with exactly the same top speed as before. Now switch to red light — however dazzlingly bright you make it, however long you wait, not a single electron escapes. Classical physics, which pictured light as a spread-out wave carrying energy continuously, had no way to explain either fact. A wave should be able to shake an electron loose eventually if you shine it long enough and hard enough; it should never care about colour, only about brightness. The metal surface, it turned out, did not agree with the wave. The way out of this puzzle, found by Albert Einstein in 1905, was to say that light itself is not spread out like a wave when it interacts with matter — it arrives in discrete parcels of energy called photons, each carrying $E = h\nu$ , fixed entirely by its frequency. One photon meets one electron in a one-to-one transaction: it is absorbed whole or not at all, and the entire photon energy goes to that single electron. If the photon does not carry enough energy to free the electron from the metal, brightness cannot help — millions of underpowered photons still cannot combine their energy on one electron. This is the photoelectric effect, and it is the experiment that forced light to reveal a particle side, launching the wave–particle dual nature that gives this NCERT chapter its name. This article builds Einstein's photoelectric equation from that one clean idea, reproduces the stopping-potential numbers you have already met, and shows how a two-point graphical trick — plotting stopping potential against frequency — lets an experimenter measure Planck's constant $h$ itself, without ever touching a Planck-radiation curve.

Key takeaways

Light striking a metal surface can knock out electrons — called photoelectrons — but only if the light's frequency is at or above a metal-specific threshold frequency $\nu_0$ ; below it, no emission occurs at any intensity.
Einstein's photoelectric equation ties it together algebraically: $eV_0 = h\nu - \phi_0$ , where $\phi_0 = h\nu_0$ is the metal's work function, the minimum energy needed to free an electron from the surface.
Stopping potential $V_0$ is the retarding voltage that just stops even the fastest photoelectrons, giving a direct experimental handle on their maximum kinetic energy: $eV_0 = KE_{max}$ .
Intensity controls the number, frequency controls the energy — brighter light above threshold gives more photoelectrons (bigger saturation current) at the same maximum kinetic energy; higher frequency gives faster photoelectrons regardless of intensity.
A graph of $V_0$ against $\nu$ is a straight line whose slope is $h/e$ (the same for every metal) and whose intercepts hand over $\phi_0$ and $\nu_0$ directly — the classic way $h$ is measured from this effect.
No time lag — photoemission is instantaneous (within about $10^{-9}$ s of illumination), which is itself a fact only the photon picture explains cleanly; a spread-out wave would need measurable time to accumulate enough energy on one electron.
Key idea: Einstein's photon picture treats each quantum of light as a single indivisible energy packet $h\nu$ handed to one electron, so $eV_0 = h\nu - \phi_0$ explains why frequency (not intensity) decides whether and how energetically photoemission happens.

The core insight

Photoelectric emission is a one-photon-one-electron event obeying $eV_0 = h\nu - \phi_0$ , so a photon below the metal's threshold energy $\phi_0 = h\nu_0$ can never eject an electron no matter how intense the beam, while every photon above threshold hands its excess energy $h\nu - \phi_0$ straight to one electron's kinetic energy.

Where it fits in the syllabus

This is the theoretical centrepiece of the NCERT Class 12 chapter "Dual Nature of Radiation and Matter" (Unit IX), sitting right after the qualitative Hertz–Lenard observations and the experimental study of stopping potential you have already met, and right before the wave-nature-of-matter (de Broglie) half of the same chapter. Everything here is on the current CBSE 2025-26 prescription: the photoelectric equation $eV_0 = h\nu - \phi_0$ , work function $\phi_0 = h\nu_0$ , and the qualitative dependence of stopping potential and saturation current on frequency and intensity. Kept honestly out of this section — because the current unit text does not carry them as testable anchors — are a quantitative treatment of the Davisson–Germer electron-diffraction experiment (background colour only), X-ray production/Bremsstrahlung, and the Compton effect, all of which belong to competitive-exam or university physics rather than this board syllabus.

The physical picture

Treat the incoming light as a stream of photons, each an indivisible packet of energy $E = h\nu$ that depends only on frequency, never on how bright the beam looks. When one such photon strikes the metal surface and is absorbed by a single electron near the surface, that electron receives the entire photon energy in one instant. Some of that energy, a fixed amount $\phi_0$ characteristic of the metal, is spent freeing the electron from the attractive forces holding it inside the metal; whatever energy is left over becomes the electron's kinetic energy as it leaves. The electrons that escape with the least resistance — those sitting exactly at the surface — carry away the maximum possible share, giving the single clean relation

KE_{max} = h\nu - \phi_0

where: $h = 6.63\times10^{-34}\ \text{J s}$ is Planck's constant, $\nu$ is the frequency of the incident light (Hz), and $\phi_0 = h\nu_0$ is the work function of the metal (J, or conventionally quoted in eV), fixed by the metal's identity and surface condition. The physical reading is a strict energy budget: photon energy in, work-function toll paid first, kinetic energy left over — and if $h\nu$ is smaller than $\phi_0$ , the toll cannot be paid, so nothing leaves at all, however many such photons arrive per second. Because the maximum kinetic energy of the escaping electrons can be measured experimentally by finding the retarding stopping potential $V_0$ that just cancels it ( $eV_0 = KE_{max}$ ), the same relation is usually written in its measurable, experiment-facing form — Einstein's photoelectric equation:

eV_0 = h\nu - \phi_0

where: $e = 1.6\times10^{-19}\ \text{C}$ is the electronic charge and $V_0$ is the stopping potential (V). Every quantity on the right is fixed once the metal and the light frequency are fixed, which is exactly why $V_0$ depends only on frequency, never on intensity — a fact that flatly contradicted the wave picture and is the single most tested qualitative statement in this chapter.

Figure

A graph of stopping potential on the vertical axis against light frequency on the horizontal axis for two different metals, each a straight line with the same slope h over e, one line shifted to the right of the other because it has a higher threshold frequency and larger work function, both lines crossing the frequency axis at their own threshold frequency

Derivation AEinstein's photoelectric equation from energy conservation

The goal is to turn the single physical assumption that light delivers its energy to matter in one-photon-one-electron packets into the testable algebraic relation $eV_0 = h\nu - \phi_0$ , using nothing beyond the statement of energy conservation for one photon absorption event — no calculus and no need to derive the quantization $E = h\nu$ itself, which is quoted here as Einstein's postulate (a result that itself rests on Planck's earlier work on black-body radiation, outside this chapter's scope).

Step 1. State the photon energy. A photon of frequency $\nu$ carries a fixed energy $E = h\nu$ , entirely independent of how many other photons are present in the beam — this is the postulate that intensity (photon number per second) and photon energy are separate, unrelated quantities.

Step 2. Apply conservation of energy to a single absorption event. When one photon is absorbed by one electron at the metal surface, all of the photon's energy goes to that electron; none is shared between electrons and none is left in the electromagnetic field. The absorbed energy $h\nu$ must be accounted for by exactly two things: the energy $\phi_0$ needed to overcome the attractive forces binding the electron to the metal (its work function), and whatever kinetic energy the electron carries away, $KE$ . So $h\nu = \phi_0 + KE$ .

Step 3. Identify the maximum case. Electrons freed from just below the surface lose no additional energy escaping and so carry the largest possible share of the leftover energy; electrons freed from deeper inside lose some energy in collisions on the way out and emerge slower. The maximum kinetic energy is therefore $KE_{max} = h\nu - \phi_0$ — this is the useful, testable form, since it is the fastest electrons that experiment can isolate.

Step 4. Convert to the measurable stopping-potential form. The stopping potential $V_0$ is defined as the retarding voltage that exactly cancels $KE_{max}$ , so $eV_0 = KE_{max}$ by definition of potential difference. Substituting Step 3 gives Einstein's photoelectric equation, $eV_0 = h\nu - \phi_0$ , and since $\phi_0 = h\nu_0$ this can equally be written $eV_0 = h(\nu - \nu_0)$ — a straight line in $V_0$ versus $\nu$ , slope $h/e$ , crossing the $\nu$ -axis exactly at the threshold frequency. Result: $eV_0 = h\nu - \phi_0 = h(\nu-\nu_0)$ ; the slope $h/e = 4.14\times10^{-15}\ \text{V s}$ is a universal constant, the same for every metal, which is what makes the two-point graphical method below work.

Worked examples

Board level - stopping potential and work function for cesium light source

Ultraviolet-adjacent light of frequency $\nu = 8.06\times10^{14}\ \text{Hz}$ falls on a cesium surface whose work function is $\phi_0 = 2.14\ \text{eV}$ . Find the stopping potential and the maximum kinetic energy of the photoelectrons.

Step 1. Convert the photon energy to electron volts, since the work function is given in eV. $h\nu = (6.63\times10^{-34})(8.06\times10^{14}) = 5.34\times10^{-19}\ \text{J}$ . Dividing by $e = 1.6\times10^{-19}\ \text{C}$ converts this to $5.34\times10^{-19}/1.6\times10^{-19} = 3.34\ \text{eV}$ .

Step 2. Apply Einstein's equation directly in eV, where it reads simply $eV_0(\text{in eV}) = h\nu(\text{in eV}) - \phi_0(\text{in eV})$ : $eV_0 = 3.34 - 2.14 = 1.20\ \text{eV}$ . Since $1\ \text{eV}$ corresponds to a stopping potential of exactly $1\ \text{V}$ , this reads off immediately as $V_0 = 1.20\ \text{V}$ .

Step 3. Convert to joules for a complete answer: $KE_{max} = eV_0 = (1.6\times10^{-19}\ \text{C})(1.20\ \text{V}) = 1.92\times10^{-19}\ \text{J}$ . Answer: stopping potential $V_0 = 1.2\ \text{V}$ , and $KE_{max} = 1.92\times10^{-19}\ \text{J} = 1.2\ \text{eV}$ — the same maximum kinetic energy met in the introductory stopping-potential experiment, now derived from the incident frequency and the metal's work function rather than measured directly.

Competitive level - measuring Planck's constant from two stopping-potential readings

The same cesium surface is illuminated in turn by two frequencies. At $\nu_1 = 8.06\times10^{14}\ \text{Hz}$ the stopping potential is measured as $V_{01} = 1.20\ \text{V}$ (as above); at a second, higher frequency $\nu_2 = 1.20\times10^{15}\ \text{Hz}$ the stopping potential is measured as $V_{02} = 2.83\ \text{V}$ . Use only these two readings to determine Planck's constant $h$ and the work function $\phi_0$ , and check them against the accepted values.

Step 1. Write Einstein's equation at both frequencies and subtract to eliminate $\phi_0$ : $e(V_{02}-V_{01}) = h(\nu_2-\nu_1)$ , so the slope of the $V_0$ -versus- $\nu$ line is $\dfrac{h}{e} = \dfrac{V_{02}-V_{01}}{\nu_2-\nu_1}$ .

Step 2. Substitute the numbers: $\dfrac{h}{e} = \dfrac{2.83 - 1.20}{(1.20\times10^{15}) - (8.06\times10^{14})} = \dfrac{1.63}{3.94\times10^{14}} = 4.14\times10^{-15}\ \text{V s}$ .

Step 3. Recover $h$ by multiplying by $e$ : $h = (4.14\times10^{-15})(1.6\times10^{-19}) = 6.62\times10^{-34}\ \text{J s}$ — matching the accepted value $6.63\times10^{-34}\ \text{J s}$ to three significant figures, which is the entire point of the two-point method: it measures $h$ using only a voltmeter and a frequency-selectable light source, no black-body furnace required.

Step 4. Recover $\phi_0$ by substituting back into either original equation, using point 1: $\phi_0 = h\nu_1 - eV_{01} = (6.62\times10^{-34})(8.06\times10^{14}) - (1.6\times10^{-19})(1.20) = 5.34\times10^{-19} - 1.92\times10^{-19} = 3.41\times10^{-19}\ \text{J} = 2.14\ \text{eV}$ . Answer: $h \approx 6.62\times10^{-34}\ \text{J s}$ , $\phi_0 = 2.14\ \text{eV}$ , both consistent with the cesium surface assumed throughout. Timing note: in a competitive-exam setting, do not recompute $h\nu_1$ and $h\nu_2$ separately in joules and then subtract — go straight for the slope $\dfrac{\Delta V_0}{\Delta \nu} = \dfrac{h}{e}$ first. This one-line shortcut turns a three-minute two-equation solve into a thirty-second subtraction and division, and it is the standard JEE pattern whenever two stopping-potential readings are given together.

Common traps

Writing $eV_0 = h\nu - \phi_0$ but then plugging in intensity anywhere in the formula — intensity does not appear in Einstein's equation at all; it only changes the number of photoelectrons (and hence the saturation current), never their maximum kinetic energy or the stopping potential.
Assuming a brighter sub-threshold beam will eventually cause emission if you wait long enough — it will not, at any intensity or any duration, because no single photon in that beam ever carries enough energy to pay the work-function toll; energy from different photons cannot combine on one electron.
Forgetting that $KE_{max} = h\nu - \phi_0$ describes only the fastest photoelectrons — those freed right at the surface — while the full population emerges with a spread of kinetic energies from $0$ up to this maximum, because electrons from deeper layers lose energy escaping.
Mixing up which axis is which on the $V_0$ -versus- $\nu$ graph, or forgetting that its slope is the universal constant $h/e$ for every metal — only the intercepts (work function and threshold frequency) change from metal to metal, never the slope.
Reporting stopping potential without checking that it is calculated relative to the correct sign convention — $V_0$ is a retarding potential and the equation $eV_0 = KE_{max}$ already accounts for that by construction; do not introduce an extra negative sign.

Common trapFrequency, not intensity, is what decides whether photoemission happens at all

The single biggest mark-losing habit in this topic is reasoning about the photoelectric effect the way you would reason about heating something with a lamp — assuming more light energy, delivered for longer, must eventually do the job. That intuition is built for continuous wave energy and it is exactly what Einstein's photon picture overturns. Each photon is an indivisible unit: either it alone carries enough energy to free an electron ( $h\nu \ge \phi_0$ ) or it does not, and no accumulation across photons or across time is possible in this one-photon-one-electron process. This is why increasing intensity below threshold changes nothing (zero current stays zero, forever), while increasing intensity above threshold increases only the number of photoelectrons per second, never their maximum energy — and it is why raising frequency above threshold increases $KE_{max}$ and hence $V_0$ , but changes the saturation current only if it also changes how efficiently photons are absorbed, not by the photon-energy argument itself.

You should be able to…

State Einstein's photoelectric equation $eV_0 = h\nu - \phi_0$ in both its energy form ( $KE_{max}=h\nu-\phi_0$ ) and its stopping-potential form, and explain what each symbol represents physically.
Explain, using the one-photon-one-electron idea, why emission depends on frequency but not intensity, and why intensity instead controls the number of photoelectrons (saturation current).
Convert between joules and electron volts confidently when combining a work function and a photon energy in the same calculation.
Use two stopping-potential readings at two frequencies to extract the slope $h/e$ and hence both Planck's constant and the metal's work function.
Sketch and read a $V_0$ -versus- $\nu$ graph: identify the universal slope $h/e$ , the frequency-axis intercept $\nu_0$ , and how the line shifts for a metal with a larger work function.
Distinguish clearly between the threshold-frequency condition (an on/off gate set by frequency) and the maximum-kinetic-energy relation (a continuous quantity that grows linearly with frequency once above threshold).

Ask the tutor: want to see how these same photon-energy ideas connect forward to the de Broglie wavelength of matter, the other half of this chapter's dual nature story — ask the AI tutor to walk through it.

Why does a single photon give all its energy to just one electron instead of spreading it around?

This is exactly the postulate that makes the photon picture different from the wave picture, and it is confirmed by experiment rather than derived from anything simpler at this level. In the wave picture, energy is spread continuously across the wavefront and could in principle be absorbed gradually by many electrons at once, accumulating over time. Observation shows the opposite: emission is essentially instantaneous and either happens with the full photon energy or does not happen at all. Treating the photon as an indivisible particle-like packet, absorbed whole by a single electron, is simply the model that matches what the stopping-potential and threshold-frequency experiments actually show.

If work function is an energy, why is it sometimes written as a frequency, like h times nu-zero?

Both forms describe the same physical quantity, just packaged differently. The work function $\phi_0$ is fundamentally an energy — the minimum energy needed to remove an electron from the metal surface — and it can always be reported directly in joules or electron volts. Writing $\phi_0 = h\nu_0$ is simply converting that energy into "the frequency of a photon that would carry exactly this much energy," which is a convenient way to compare it directly against the frequency of the light you are shining, since emission starts exactly when $\nu$ reaches $\nu_0$ .

Does the photoelectric effect prove light is only a particle and never a wave?

No, and this chapter's title, dual nature, is the giveaway. The photoelectric effect shows that light exchanges energy with matter in discrete packets, which only the particle picture explains. But plenty of other phenomena — interference, diffraction, polarisation — are explained only by treating light as a wave, and cannot be explained by a simple particle picture. The resolution is that light has both aspects, and which one shows up depends on the experiment: wave behaviour governs how light propagates and spreads, particle behaviour governs how it is absorbed or emitted in an interaction with matter.

My textbook problem gives work function in electron volts but frequency in hertz — how do I avoid unit mistakes?

The safest habit is to convert everything to one consistent unit before substituting into Einstein's equation. Either convert $h\nu$ from joules into electron volts by dividing by $e = 1.6\times10^{-19}$ (as in the board-level example above), so you can subtract directly from a work function already in eV, or convert the work function from eV into joules by multiplying by $1.6\times10^{-19}$ and keep everything in SI units throughout. Mixing joules and electron volts in the same subtraction is the single most common arithmetic slip in this topic, so pick one system at the start of the problem and stay in it until the final answer, converting only at the very end if the question asks for a different unit.