TECHNIQUE: Photoelectric Graph-Reading

The graph-reading craft of this article is the everyday language of light detectors. Every photomultiplier tube, photodiode and image sensor is characterised by an $I$–$V$ curve with a saturation plateau, and engineers still read a device's "cut-off" and "saturation" straight off that plot exactly as you now do. Photocathode materials are chosen by their work function — a low-$\varphi$ metal like caesium extends the threshold $f_0$ into the visible, which is why caesium and its alloys line the photocathodes of night-vision tubes and scintillation detectors. The $V_0$–$f$ slope that Millikan used to weigh Planck's constant is now run in reverse: knowing $h$, a measured stopping potential reads out a photon's frequency, the principle behind photoelectron spectroscopy, which maps the electronic structure of materials by measuring the kinetic energy of electrons that light kicks out. And the reflex you are building — name the axes, read the slope, watch how the curve moves when you turn a knob — is the core skill of every experimental physicist facing an unfamiliar plot.

The bookkeeping is the physics here, so start from the one law and watch the three graphs fall out of it. When light of frequency $f$ strikes a metal of work function $\varphi$, the most energetic escaping electron carries

where: $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ is Planck's constant; $\varphi$ is the minimum energy to free an electron (measured in joules here, or in electronvolts as $\varphi/e$); $f_0$ is the threshold frequency, the lowest frequency that ejects any electron at all. Below $f_0$ the photon simply cannot pay the escape cost and no current flows, however bright the light. This is the equation of a straight line in $f$ — and that single fact is why the exam draws it as a line. To measure $KE_{\max}$ you make the collector plate slightly negative and turn it down until even the fastest electron is turned back. The voltage that just stops the current is the stopping potential $V_0$, and because the retarding field does work $eV_0$ on each electron,

where: $e = 1.60\times10^{-19}\ \mathrm{C}$ is the electron charge; $V_0$ is the stopping (cut-off) potential; the quantity $h/e = 4.14\times10^{-15}\ \mathrm{V\,s}$ is a pure combination of constants. Read equation (2) as $y = mx + c$: plotting $V_0$ against $f$ gives a straight line of slope $h/e$, exactly the same for every metal, shifted only by its intercept. That universality — one slope for all photocathodes — is the single most tested fact in this whole family. The third graph, the photocurrent against collector voltage, is not a line but a rising-then-flat curve, and it carries two independent pieces of information at once. Slide the collector positive and every ejected electron is swept up: the current rises and then flattens at the saturation current, whose height is set by how many electrons are emitted per second — i.e. by the light's intensity. Slide the collector negative and the current falls to zero at $-V_0$, whose value is set by how energetic the fastest electron is — i.e. by the frequency. Symbolically,

where: $I_{\text{sat}}$ is the plateau current; the two arrows point at the two knobs. Brighten the light and the plateau rises but the cut-off voltage does not budge; redden or blue-shift the light and the cut-off moves but the plateau (at fixed power-per-photon-count) responds through the photon count. Keep those two levers separate and the $I$–$V$ graph never traps you.

Equation (2) is the workhorse of the exam. Because $V_0 = (h/e)f - \varphi/e$, the plot of stopping potential against frequency is a straight line, and the three things a paper can ask about a line are its slope and its two intercepts. Read them off directly:

where: the slope is a ratio of two universal constants and so is identical for every metal — caesium, sodium, tungsten all give the same $4.14\times10^{-15}\ \mathrm{V\,s}$. Only the intercepts move: a metal with a larger work function has a threshold $f_0$ further to the right and a $y$-intercept further down. Two metals therefore give two parallel lines. The physical reading is Millikan's 1916 masterpiece: measure the slope of that line and you have read Planck's constant straight out of a photoelectric experiment, $h = e \times \text{slope}$, without ever counting a photon. That is the single most examined inference in the chapter, so drill the three questions the line can pose. "What is the slope?" — $h/e$, always. "Where does it cross the frequency axis?" — at $f_0 = \varphi/h$, the threshold. "What is the value at $f = 0$?" — the (extrapolated) $y$-intercept $-\varphi/e$, from which $\varphi = e \times |y\text{-intercept}|$. A line for sodium ($\varphi = 2.28\ \mathrm{eV}$) crosses the axis at $f_0 = 5.50\times10^{14}\ \mathrm{Hz}$ and would extrapolate back to $-2.28\ \mathrm{V}$; a line for caesium ($\varphi = 2.14\ \mathrm{eV}$) crosses at $f_0 = 5.16\times10^{14}\ \mathrm{Hz}$ and back-extrapolates to $-2.14\ \mathrm{V}$ — parallel, shifted, same slope.

The same data plotted as energy rather than voltage gives the second line. Since $KE_{\max} = hf - \varphi$, the plot of maximum kinetic energy against frequency is a straight line with

where: the slope is now Planck's constant itself, not $h/e$, because the vertical axis is energy (joules or electronvolts), not volts. This is the trap the examiner loves: the $V_0$–$f$ line and the $KE_{\max}$–$f$ line look identical — same $x$-intercept $f_0$, same "rise from a threshold" shape — but their slopes differ by a factor of $e$. Ask yourself what the vertical axis measures: volts $\Rightarrow$ slope $h/e$; energy $\Rightarrow$ slope $h$. Get the axis right and you get the constant right. Note also that if the vertical axis is $KE_{\max}$ in electronvolts, the numerical slope equals $h/e = 4.14\times10^{-15}$ (because dividing joules by $e$ converts to eV) — so "eV per hertz" and "volts per hertz" share the same number, $4.14\times10^{-15}$, while "joules per hertz" is $6.63\times10^{-34}$. Watch the unit on the axis, not just the letter.

The photocurrent curve is the one that rewards careful reading because it moves in two independent ways. Fix the frequency and brighten the light: more photons per second eject more electrons per second, so the saturation current rises in proportion to intensity — but each electron's maximum energy is unchanged, so the curve still dies at exactly the same stopping potential. Now fix the intensity and raise the frequency: each electron leaves faster, so the cut-off $-V_0$ moves further negative, while the saturation plateau responds only through how many photons the fixed power now delivers. The two readings never cross-contaminate:

• Height of the plateau — set by intensity (photon arrival rate). Double the intensity, double $I_{\text{sat}}$.
• Cut-off voltage $-V_0$ — set by frequency alone. It is completely independent of how bright the light is.

The classic MCQ shows two curves at the same frequency and different intensity: they share the same $-V_0$ and differ only in plateau height — and the wrong option is always the one that also shifts the cut-off. A second classic shows two frequencies at the same intensity: now the cut-offs differ. Decide first which knob the question turned, and the curve's behaviour is fixed.

Everything above collapses into one table you can carry into the hall. Identify the axes, then read.

Drill 1 · Board level — reading the V₀–f line

A stopping-potential-versus-frequency graph for a metal is a straight line of slope $4.14\times10^{-15}\ \mathrm{V\,s}$ that crosses the frequency axis at $f_0 = 5.50\times10^{14}\ \mathrm{Hz}$. Find (a) Planck's constant, (b) the work function in eV, and (c) the stopping potential at $f = 9.0\times10^{14}\ \mathrm{Hz}$.

Step 1. Planck's constant from the slope. The slope of a $V_0$–$f$ line is $h/e$, so $h = e\times\text{slope} = (1.60\times10^{-19})(4.14\times10^{-15}) = \boxed{6.62\times10^{-34}\ \mathrm{J\,s}}$ (the accepted value is $6.63\times10^{-34}$; the gap is rounding).

Step 2. Work function from the threshold. $\varphi = h f_0$, and in electronvolts $\varphi = (h/e) f_0 = (4.14\times10^{-15})(5.50\times10^{14}) = \boxed{2.28\ \mathrm{eV}}$ — this is sodium.

Step 3. Stopping potential at the new frequency, from equation (2): $V_0 = (h/e)(f - f_0) = (4.14\times10^{-15})(9.0\times10^{14} - 5.50\times10^{14}) = (4.14\times10^{-15})(3.5\times10^{14}) = \boxed{1.45\ \mathrm{V}}$.

Step 4. Sense-check: the frequency sits above threshold, so $V_0 > 0$ as it must, and at $f = f_0$ the formula gives $V_0 = 0$ — the line's $x$-intercept, consistent.

Drill 2 · Board level — reading the I–V characteristic

In a photoelectric tube, monochromatic light of fixed frequency produces a saturation photocurrent of $2.0\ \mu\mathrm{A}$ and a stopping potential of $1.45\ \mathrm{V}$. The intensity of the light is now doubled, the frequency unchanged. State the new saturation current and the new stopping potential.

Step 1. Identify which knob moved: intensity doubled, frequency fixed. Intensity owns the plateau; frequency owns the cut-off.

Step 2. Saturation current is proportional to intensity, so doubling the intensity doubles it: $I_{\text{sat}}' = 2\times2.0\ \mu\mathrm{A} = \boxed{4.0\ \mu\mathrm{A}}$.

Step 3. Stopping potential depends only on frequency, which did not change, so $V_0' = \boxed{1.45\ \mathrm{V}}$ — unchanged.

Step 4. The trap option would raise both numbers; brighter light never changes the maximum electron energy, so the cut-off voltage cannot move. Only the plateau responds to brightness.

Drill 3 · Competitive level — two-point slope gives h and φ

Two points are read off a metal's $V_0$–$f$ line: $(6.0\times10^{14}\ \mathrm{Hz},\ 0.207\ \mathrm{V})$ and $(10.0\times10^{14}\ \mathrm{Hz},\ 1.863\ \mathrm{V})$. Determine Planck's constant, the threshold frequency, and the work function.

Step 1. Two-point slope: $\text{slope} = \dfrac{1.863 - 0.207}{(10.0 - 6.0)\times10^{14}} = \dfrac{1.656}{4.0\times10^{14}} = 4.14\times10^{-15}\ \mathrm{V\,s}$.

Step 2. Planck's constant: $h = e\times\text{slope} = (1.60\times10^{-19})(4.14\times10^{-15}) = \boxed{6.62\times10^{-34}\ \mathrm{J\,s}}$.

Step 3. Threshold from $V_0 = 0$: $f_0 = f_1 - \dfrac{V_{0,1}}{\text{slope}} = 6.0\times10^{14} - \dfrac{0.207}{4.14\times10^{-15}} = 6.0\times10^{14} - 0.50\times10^{14} = \boxed{5.50\times10^{14}\ \mathrm{Hz}}$.

Step 4. Work function: $\varphi = (h/e)f_0 = (4.14\times10^{-15})(5.50\times10^{14}) = \boxed{2.28\ \mathrm{eV}}$ (sodium). Cross-check with the other point: $V_0 = (4.14\times10^{-15})(10.0\times10^{14} - 5.50\times10^{14}) = (4.14\times10^{-15})(4.5\times10^{14}) = 1.863\ \mathrm{V}$ — matches the given datum exactly. Timing note: about 60 seconds. The whole problem is one two-point slope and one substitution — do not attempt to fit a line by eye or average several slopes. Pick the two cleanest points, and remember $h = e\times\text{slope}$, never the slope itself.

Drill 4 · Competitive level — which graph gives h directly

A student plots maximum kinetic energy (in joules) against frequency for caesium ($\varphi = 2.14\ \mathrm{eV}$, $f_0 = 5.16\times10^{14}\ \mathrm{Hz}$). What is the slope of the line, and what is $KE_{\max}$ at $f = 8.0\times10^{14}\ \mathrm{Hz}$? Contrast the slope with that of the $V_0$–$f$ plot.

Step 1. For a $KE_{\max}$–$f$ line (energy on the vertical axis) the slope is Planck's constant itself: $\text{slope} = h = \boxed{6.63\times10^{-34}\ \mathrm{J\,s}}$ — a factor $e$ larger than the $V_0$–$f$ slope of $4.14\times10^{-15}\ \mathrm{V\,s}$.

Step 2. Kinetic energy at the given frequency: $KE_{\max} = h(f - f_0) = (6.63\times10^{-34})(8.0\times10^{14} - 5.16\times10^{14}) = (6.63\times10^{-34})(2.84\times10^{14}) = 1.88\times10^{-19}\ \mathrm{J}$.

Step 3. Convert to electronvolts: $KE_{\max} = \dfrac{1.88\times10^{-19}}{1.60\times10^{-19}} = \boxed{1.18\ \mathrm{eV}}$, so the stopping potential here would be $V_0 = 1.18\ \mathrm{V}$.

Step 4. The contrast is the whole point: the energy-axis slope is $h$; the volt-axis slope is $h/e$. Same line shape, same $f_0$, slopes differing by $e$. If an MCQ asks "which plot has slope equal to Planck's constant?", the answer is $KE_{\max}$ (in joules) versus $f$, not $V_0$ versus $f$. Timing note: about 40 seconds. The reflex is to name the vertical axis first — energy or volts — and only then write the slope. Getting that one word right settles whether the answer is $h$ or $h/e$.

Confusing the two slopes. The $V_0$–$f$ line has slope $h/e = 4.14\times10^{-15}\ \mathrm{V\,s}$; the $KE_{\max}$–$f$ line has slope $h = 6.63\times10^{-34}\ \mathrm{J\,s}$. They look identical. Read the vertical-axis unit — volts or joules — before writing a slope. CBSE/NEET/JEE all set this deliberately.

Reporting the slope as $h$ when the axis is volts. If you read $V_0$–$f$ and quote $6.63\times10^{-34}$, you have skipped the $\times e$. The slope you read is $h/e$; multiply by $e$ to get $h$.

Letting intensity move the stopping potential. On the $I$–$V$ graph, brighter light raises the saturation plateau and nothing else. The cut-off $-V_0$ is fixed by frequency. The trap option changes both.

Letting frequency alone fix the saturation height. Raising the frequency shifts the cut-off but does not, by itself, raise the plateau — the plateau tracks the photon arrival rate (intensity), not the photon energy.

Forgetting that below $f_0$ there is no current at all. No amount of intensity ejects an electron below threshold; the $I$–$V$ curve is flat at zero, and the $V_0$–$f$ point simply does not exist there. Do not extrapolate a stopping potential below $f_0$ as if the metal still emitted.

Misreading a $KE_{\max}$-in-eV axis. If kinetic energy is plotted in electronvolts, the numerical slope is $4.14\times10^{-15}$ (eV per Hz) — the same digits as the volt slope — because dividing by $e$ converts joules to eV. It is not $6.63\times10^{-34}$; that number belongs only to a joule axis.

Board scope. This is a technique module aimed at the multiple-choice papers. Reading the $V_0$–$f$ slope, the $KE_{\max}$–$f$ slope, and the $I$–$V$ curve is exam-critical for JEE-Main, JEE-Advanced, NEET and AP Physics 2, where a graph question is nearly guaranteed and speed buys marks. You will recognise these same three photoelectric graphs from your CBSE, ICSE or A-Level (9702) chapter and can apply the identical reading tricks, but the "TEC" tag marks them as especially exam-critical only on the JEE, NEET and AP-2 papers this module targets. IGCSE (0625) carries no photoelectric content at all, and IB Standard Level does not carry the photoelectric equation (it is HL-only), so these graphs are not on those two syllabi.

Texts. NCERT Physics Class 12, Ch. 11 ("Dual Nature of Radiation and Matter"), the stopping-potential and $I$–$V$ graphs · H. C. Verma, Concepts of Physics Vol. 2, "Photoelectric Effect and Wave–Particle Duality" · D. C. Pandey / Cengage photoelectric problem sets for graph-based MCQ practice · any JEE/NEET previous-year compilation for the $V_0$–$f$ slope family.

Provenance. Technique module built on the concept chapters; the numbers are standard exam anchors. Recompute: slope $h/e = 4.14\times10^{-15}\ \mathrm{V\,s}$; slope $h = 6.63\times10^{-34}\ \mathrm{J\,s}$; $h = e\times\text{slope} = 6.62\times10^{-34}\ \mathrm{J\,s}$ from a read slope; $f_0 = \varphi/h$ for Cs ($2.14\ \mathrm{eV}\to5.16\times10^{14}\ \mathrm{Hz}$) and Na ($2.28\ \mathrm{eV}\to5.50\times10^{14}\ \mathrm{Hz}$); $y$-intercepts $-\varphi/e$ and $-\varphi$; two-point slope $= 4.14\times10^{-15}$ from $(6.0\times10^{14},0.207)$ and $(10.0\times10^{14},1.863)$; $V_0(9.0\times10^{14},\mathrm{Na}) = 1.45\ \mathrm{V}$; $KE_{\max}(8.0\times10^{14},\mathrm{Cs}) = 1.88\times10^{-19}\ \mathrm{J} = 1.18\ \mathrm{eV}$; saturation-current ratio $2$ on doubling intensity.

Prerequisites. The Photoelectric Effect & Its Laws (the four laws these graphs display) and Einstein's Photoelectric Equation (the line $KE_{\max} = hf - \varphi$ every plot is built on). Leads to. the university seam The Photoelectric Effect and the Photon, where the same graphs are re-derived with the full photon picture.

Go deeper (UG). The photoelectric graphs, the Fowler–DuBridge current model, and the temperature dependence of the near-threshold curve are developed in the university article The Photoelectric Effect and the Photon. Enrichment only — this article needs nothing beyond Einstein's equation and the two experimental facts about intensity and frequency.