Experimental study of the photoelectric effect (intensity potential frequency effects)

Picture a small evacuated glass tube. Inside it sits a metal plate wired to the negative terminal of a battery, facing a second plate a short distance away, wired through a sensitive ammeter and a variable voltage supply that can be dialled positive or negative. Shine light of a chosen colour and brightness onto the first plate and, if the light is energetic enough, the ammeter twitches — a tiny current is flowing across empty space, carried by electrons that light itself has knocked loose. This is the photoelectric effect, and by the early twentieth century the phenomenon itself was not in doubt. What mattered far more, and what this section is about, was the patient dial-turning that followed: change the brightness of the light and watch the current; change the retarding voltage and watch the current fall; change the colour (frequency) and watch a completely different quantity, the stopping voltage, move instead. Those three controlled experiments produced three clean, reproducible trends — and every one of them turned out to be a trap for the classical wave theory of light, which predicted the wrong behaviour on two of the three counts. Getting the trends exactly right, including which quantity changes and which stays frozen, is worth more marks in this chapter than almost anything else, because it is exactly the kind of "state the graph, state the reasoning" question CBSE loves. This article walks through the apparatus, draws and reads the two signature graphs — photocurrent versus collector potential, and stopping potential versus frequency — and shows how a straight line on the second graph, sketched by nineteen-year-old data points, quietly hands you the value of Planck's constant.

Key takeaways

The photoelectric apparatus has an emitter plate (light falls here) and a collector plate, with a variable voltage between them and an ammeter measuring photocurrent.
Increasing intensity (frequency fixed) increases the saturation current in direct proportion, but leaves the stopping potential $V_0$ unchanged.
Increasing the retarding (negative) potential on the collector reduces the photocurrent smoothly to zero at $V_0$ , because photoelectrons are emitted with a spread of kinetic energies up to a maximum $KE_{max}=eV_0$ .
Increasing frequency (intensity fixed) leaves the saturation current unchanged but increases the stopping potential; below a threshold frequency $\nu_0$ there is no emission at all, whatever the intensity.
A graph of $V_0$ against $\nu$ is a straight line that misses the origin: its slope is $h/e$ (the same for every metal) and its frequency-intercept is the threshold frequency $\nu_0$ of that particular metal.
These three trends — current tracks intensity, energy tracks frequency, and there is a hard frequency threshold with no time lag — are precisely the facts a wave picture of light cannot reproduce, which is why Einstein's photon equation $eV_0=h\nu-\phi_0$ was needed.
Key idea: intensity controls how many photoelectrons are ejected per second (hence the saturation current), while frequency alone controls the energy of each photoelectron (hence the stopping potential), and the straight-line V0-versus-nu graph turns that qualitative fact into a measurement of h.

The core insight

In the photoelectric experiment, changing the light's intensity changes only the number of photoelectrons emitted per second (the saturation current), while changing its frequency changes only the energy of each photoelectron (the stopping potential), and the resulting straight-line graph of stopping potential against frequency has slope $h/e$ and a frequency-intercept equal to the metal's threshold frequency.

Where it fits in the syllabus

The CBSE 2025-26 unit "Dual Nature of Radiation and Matter" explicitly lists "Experimental study of the photoelectric effect (qualitative: effect of intensity, frequency, stopping potential)" as prescribed content, alongside Einstein's photoelectric equation $eV_0=h\nu-\phi_0$ . That combination is exactly what this section covers: the syllabus wants the trends described in words and read off standard graphs, not derived from first principles, but it does expect you to connect those trends to Einstein's equation and to be comfortable using the equation on two data points, as in the worked examples below. Photocell engineering, and any device-level detail beyond a single qualitative mention, sits outside the prescribed scope and is not the concern of this article.

The physical picture

The apparatus is simple to draw and worth sketching from memory: an evacuated tube, an emitter plate illuminated by monochromatic light of adjustable frequency and intensity, a collector plate, a battery and potentiometer arrangement that lets the collector's potential relative to the emitter be swept from strongly positive through zero to strongly negative, and a microammeter reading the resulting photocurrent. Three separate experiments are then done, each holding two variables fixed and changing the third. With frequency and intensity fixed, sweeping the collector potential from a large positive value down through zero and into negative values traces out the first signature curve. At large positive potential every photoelectron emitted per second is swept across and collected, and the current sits flat at its saturation value $I_s$ — making the collector more positive still cannot raise the current further, because there are no more electrons to collect. As the potential is reduced and then made negative (retarding), the current falls, smoothly rather than abruptly, because photoelectrons leave the emitter with a whole range of kinetic energies from zero up to some maximum, and only those with enough energy can still climb the opposing potential hill. The current reaches exactly zero at the stopping potential $V_0$ , the retarding voltage that turns back even the fastest photoelectrons.

eV_0 = KE_{max} \qquad (1)

where: $e$ is the electron's charge magnitude ( $1.6\times10^{-19}\ \text{C}$ ), $V_0$ is the stopping potential in volts, and $KE_{max}$ is the maximum kinetic energy any photoelectron carries as it leaves the emitter. Physically, equation (1) says the stopping potential is simply a convenient, easily measured stand-in for the one number — maximum photoelectron energy — that the experiment is really trying to pin down.

Figure

Photocurrent plotted against collector potential for two intensities of the same colour light, showing both curves reaching the same stopping potential on the negative axis but the brighter light reaching a higher saturation current on the positive axis

Repeating this whole sweep at a second, higher intensity of the same colour produces a second curve of the same shape, crossing the potential axis at the very same point $-V_0$ , but flattening out at a proportionally higher saturation current, because doubling the number of photons striking the plate per second doubles the number of photoelectrons ejected per second while leaving the energy of each one untouched. Repeating the sweep instead at a higher frequency (intensity unchanged) does the opposite: the saturation current is unchanged, but the curve now crosses zero at a larger stopping potential, because each individual photoelectron now carries more energy.

eV_0 = h\nu - \phi_0 \qquad (2)

where: $h$ is Planck's constant ( $6.63\times10^{-34}\ \text{J s}$ ), $\nu$ is the frequency of the incident light, and $\phi_0=h\nu_0$ is the metal's work function, the minimum energy needed to free an electron from its surface, expressed through the metal's own threshold frequency $\nu_0$ . Rearranged as $V_0=(h/e)\nu-\phi_0/e$ , equation (2) is the equation of a straight line in $V_0$ and $\nu$ , and that straight line is exactly what the third experiment traces out.

Figure

Stopping potential plotted against frequency of incident light for two different metals, both graphs being straight lines with the same slope but different frequency intercepts, neither line passing through the origin

Every metal gives a different straight line, but the two graphs above always agree on one thing: the slope is identical, because $h/e$ is a fixed constant of nature, not a property of the metal. Only the intercept — the threshold frequency $\nu_0$ , hence the work function $\phi_0$ — differs from one surface to another. That single geometric fact, that all such lines are parallel, is one of the strongest pieces of evidence for the photon picture of light, since a wave theory gives no reason at all for a universal slope across every element in the periodic table.

Derivation AReading Planck's constant and the work function off the stopping-potential line

Goal: a photoelectric experiment on an unknown metal gives only two measured pairs of stopping potential and frequency; the aim is to show, using nothing beyond straight-line algebra, how those two points alone yield an experimental estimate of the slope $h/e$ , the metal's work function $\phi_0$ , and its threshold frequency $\nu_0$ , without ever measuring $h$ or $\phi_0$ directly.

Step 1. Start from the rearranged Einstein equation as the equation of a line, $V_0=\left(\dfrac{h}{e}\right)\nu-\dfrac{\phi_0}{e}$ , and identify its two constants: the slope is $m=h/e$ and the intercept (the value of $V_0$ at $\nu=0$ , found by extending the line backward) is $c=-\phi_0/e$ .

Step 2. Given two measured points $(\nu_1,V_{01})$ and $(\nu_2,V_{02})$ on this line, compute the slope directly from the data using the ordinary two-point slope formula, $m=\dfrac{V_{02}-V_{01}}{\nu_2-\nu_1}$ , exactly as for any straight line through two known points.

Step 3. Since the theory says $m=h/e$ , multiply the measured slope by $e$ to obtain an experimental estimate of Planck's constant, $h_{\text{expt}}=m\,e$ . This is, historically, exactly how Millikan first measured $h$ from photoelectric data and found it to agree with the value Planck had already extracted from black-body radiation — powerful cross-confirmation of the quantum idea from two completely different experiments.

Step 4. Substitute either data point back into $V_0=m\nu-\phi_0/e$ and solve for the work function in electron-volts, $\dfrac{\phi_0}{e}=m\nu_1-V_{01}$ , then find the threshold frequency as the point where the line crosses $V_0=0$ , namely $\nu_0=\dfrac{\phi_0/e}{m}$ . This threshold frequency is the smallest frequency of light that can eject any photoelectron at all from that metal, however intense the beam.

Worked examples

Board level - saturation current and stopping potential under changing intensity

Problem: In a photoelectric experiment using light of frequency $\nu_1$ , the stopping potential is found to be $0.5\ \text{V}$ . If the intensity of this same light is doubled while its frequency is kept unchanged, what happens to (a) the saturation current and (b) the stopping potential?

Step 1. Doubling the intensity, at fixed frequency, means twice as many photons of the same individual energy $h\nu_1$ strike the emitter plate per second, so twice as many photoelectrons are ejected per second.

Step 2. The saturation current is, by definition, proportional to the number of photoelectrons collected per second, so it doubles: if it was $I_s$ before, it becomes $2I_s$ after.

Step 3. The stopping potential is fixed by $KE_{max}=h\nu_1-\phi_0$ , which depends only on the photon energy and the work function — neither of which changed when the light got brighter. The stopping potential therefore stays at exactly $0.5\ \text{V}$ , unchanged, while the saturation current doubles.

Competitive level - extracting h and the work function from two data points

Problem: For a certain metal, the stopping potential is measured as $0.3\ \text{V}$ for light of frequency $6\times10^{14}\ \text{Hz}$ and $1.0\ \text{V}$ for light of frequency $8\times10^{14}\ \text{Hz}$ . Using this data, estimate (a) the experimental value of $h/e$ and of $h$ , (b) the metal's work function in eV, (c) its threshold frequency, and (d) predict the stopping potential at $10\times10^{14}\ \text{Hz}$ .

Step 1. Slope from the two points: $m=\dfrac{1.0-0.3}{(8-6)\times10^{14}}=\dfrac{0.7}{2\times10^{14}}=3.5\times10^{-15}\ \text{V s}$ . This is the experimental estimate of $h/e$ .

Step 2. Multiplying by $e=1.6\times10^{-19}\ \text{C}$ gives $h_{\text{expt}}=3.5\times10^{-15}\times1.6\times10^{-19}=5.6\times10^{-34}\ \text{J s}$ — the right order of magnitude compared with the accepted $6.63\times10^{-34}\ \text{J s}$ , the small gap being exactly what you would expect from fitting a line through only two data points instead of many.

Step 3. Work function: using the point $(6\times10^{14},0.3)$ , $\dfrac{\phi_0}{e}=m\nu_1-V_{01}=3.5\times10^{-15}\times6\times10^{14}-0.3=2.1-0.3=1.8\ \text{V}$ , so $\phi_0\approx1.8\ \text{eV}$ . Checking against the second point: $3.5\times10^{-15}\times8\times10^{14}-1.8=2.8-1.8=1.0\ \text{V}$ , which matches the given data exactly, confirming the line fit.

Step 4. Threshold frequency: $\nu_0=\dfrac{\phi_0/e}{m}=\dfrac{1.8}{3.5\times10^{-15}}\approx5.1\times10^{14}\ \text{Hz}$ . Predicted stopping potential at $\nu=10\times10^{14}\ \text{Hz}$ : $V_0=m\nu-\phi_0/e=3.5\times10^{-15}\times10\times10^{14}-1.8=3.5-1.8=\mathbf{1.7\ \text{V}}$ , consistent with simply extending the straight-line trend. Timing note: In an exam, do the slope calculation first and keep it as an exact fraction ( $0.7/2\times10^{14}$ ) before converting to decimal, so rounding error doesn't creep into every later step; this two-point method is a favourite in competitive papers precisely because it tests whether you can move fluently between a physical graph and plain coordinate geometry.

Common traps

Believing that increasing intensity increases the stopping potential — it does not; intensity changes only the number of photoelectrons (and hence the saturation current), never their maximum kinetic energy.
Believing that increasing frequency increases the saturation current — it does not, at fixed photon flux; frequency changes the energy of each photoelectron, not directly how many are emitted per second.
Misreading the $V_0$ vs $\nu$ graph as passing through the origin — it never does; it always intersects the frequency axis at the metal's own threshold frequency $\nu_0$ , and its intercept on the $V_0$ axis has no direct physical meaning on its own.
Assuming the photocurrent drops to zero the instant any retarding potential is applied — it falls off gradually and reaches zero only at $V_0$ , since photoelectrons leave with a spread of kinetic energies up to $KE_{max}$ , not a single fixed value.
Forgetting that the slope of the $V_0$ vs $\nu$ line is the same constant $h/e$ for every metal — only the intercept (work function/threshold frequency) is metal-dependent; a question asking you to compare two metals is testing this parallel-lines fact.

You should be able to…

Sketch the photocurrent-versus-collector-potential graph and correctly label the saturation current and the stopping potential on it.
State precisely what changes and what stays the same when intensity is increased at fixed frequency, and explain why in terms of photon number versus photon energy.
State precisely what changes and what stays the same when frequency is increased at fixed intensity, and explain the threshold-frequency cutoff.
Sketch the stopping-potential-versus-frequency graph, correctly showing it as a straight line that misses the origin, with slope $h/e$ and frequency intercept $\nu_0$ .
Use two data points on a $V_0$ vs $\nu$ graph to compute an experimental estimate of $h$ , the work function, and the threshold frequency.
Explain, in one or two sentences, why these three trends together could not be explained by the classical wave theory of light.

Ask the tutor: want to practise sketching both graphs together and reading off h and the work function from a fresh pair of stopping-potential data points? Ask the AI tutor to run through one with you.

Why doesn't a very bright beam of low-frequency light eject any photoelectrons at all, no matter how long you shine it

Because in the photon picture each individual photon carries a fixed energy $h\nu$ , and a single electron is freed by absorbing a single photon in one shot. If $h\nu$ is below the work function $\phi_0$ , no single photon carries enough energy to free an electron, and making the beam brighter only adds more photons of that same insufficient energy — it cannot make any one photon stronger. A classical wave picture, where energy simply accumulates over time, predicted emission would eventually happen at any frequency given enough intensity and waiting time; experiment showed a hard frequency cutoff instead, with no time lag once above it.

If saturation current only depends on intensity, why do the two intensity curves in the graph reach zero at exactly the same point

Because the stopping potential is set purely by the maximum kinetic energy any photoelectron carries, and that maximum energy is fixed by the photon energy $h\nu$ and the work function $\phi_0$ — quantities that have nothing to do with how many photons per second are arriving. Brightness changes the population of photoelectrons (more of them at every energy up to the same ceiling) but never raises that ceiling itself, so however the current curve is scaled vertically, it always comes back to zero at the same $V_0$ .

Is the stopping potential graph exactly a straight line in real experiments, or is that an idealisation

Real photoelectric data does fit a straight line extremely well once experimental noise is averaged out over several frequencies, which is exactly why this experiment became a trusted way to measure $h$ independently of Planck's original black-body work. Millikan spent years refining the apparatus (freshly-cut metal surfaces in high vacuum) specifically to get a clean straight line, because early data was scattered enough that some physicists doubted Einstein's equation until the linearity was firmly established.

Do I need to memorise the numeric value of h and e for this topic, or just the trends

For this particular qualitative-experiment topic, the syllabus mainly wants the three trends and the shapes of the two graphs described correctly in words. That said, being able to plug in $h=6.63\times10^{-34}\ \text{J s}$ and $e=1.6\times10^{-19}\ \text{C}$ to read a slope, as in the competitive-level example here, is a very common extension question, so it is worth having both constants ready rather than looking them up mid-exam.