PhysicsISC Class 12 · Dual Nature of Radiation and Matter

Stopping Potential and the Stopping Potential-Frequency Graph

Q: Why does some current flow at zero collector voltage?

Photoelectrons leave the emitter with kinetic energy, so some reach the collector even without an accelerating field.

Q: Why does the current saturate?

Once the collector is positive enough, every emitted electron is collected; the current can rise further only if more electrons are emitted (more intense light).

Q: Why is the $V_0$-ν graph a straight line?

Because eV_0 = h\nu - W_0 is linear in ν, with h and W_0 constant for a given metal.

Q: What does the ISC 2027 syllabus need here?

Experimental facts of the photoelectric effect and the determination of Planck's constant from the graph of stopping potential against frequency, using Einstein's equation.

Q: How was this asked recently?

ISC 2025 Q17 (3 marks) asked the labelled V_s-f graph and how to get h from it; ISC 2024 Q16 (3 marks) asked the effect of intensity on stopping potential and a comparison of three metals.

Updated 15 October 2026 · Fact-checked

Stopping potential $V_0$ is the least negative collector potential that stops even the fastest photoelectrons, so $eV_0 = E_{max} = h\nu - W_0$. Hence $V_0 = \frac{h}{e}\nu - \frac{W_0}{e}$: a straight line of slope $\frac{h}{e}$ that cuts the frequency axis at the threshold frequency $\nu_0$. Planck's constant is $h = e \times$ slope.

Notice the line cuts the frequency axis at ν₀, its slope is h/e, and the intercept on the V₀ axis is −W₀/e.

Illustration drawn for QuestionBuddy.

Understand Stopping Potential and the Frequency Graph

The photocell experiment. Light falls on an emitter (cathode) C in an evacuated tube; photoelectrons travel to a collector (anode) A. A variable, reversible voltage V between A and C and a microammeter measure the photocurrent.

Photocurrent against collector potential.

With A positive, the current rises and then levels off at the saturation current, when every emitted electron is collected.
With V = 0 some current still flows, because electrons leave with kinetic energy.
Making A negative repels electrons; the current falls. At $V = - V_{0}$ , even the fastest electrons are turned back and the current is zero. $V_{0}$ is the stopping potential (cut-off potential), and

$e V_{0} = E_{ma x} = \frac{1}{2} m v_{ma x}^{2} .$

Effect of intensity (fixed frequency). More photons, more electrons: the saturation current rises in proportion to intensity, but the curves meet the voltage axis at the same $- V_{0}$ . Stopping potential is independent of intensity (ISC 2024 Q16(i)).

Effect of frequency (fixed intensity). Higher frequency gives faster electrons, so $V_{0}$ is larger (the curve cuts the axis further to the left). The saturation current stays about the same for the same photon flux.

**The $V_{0}$ -ν graph (asked in ISC 2025 Q17).** From Einstein's equation $e V_{0} = h ν - W_{0}$ :
$V_{0} = (\frac{h}{e}) ν - \frac{W _{0}}{e} .$
This has the form y = mx + c, so:

the graph is a straight line with positive slope $\frac{h}{e}$ ;
it cuts the frequency axis ( $V_{0} = 0$ ) at the threshold frequency $ν_{0} = \frac{W _{0}}{h}$ ; below $ν_{0}$ there is no emission, so the line is drawn only from $ν_{0}$ upwards;
extended backwards, it cuts the $V_{0}$ axis at $- \frac{W _{0}}{e}$ , so the work function in eV is the magnitude of this intercept.

Determining Planck's constant. Measure $V_{0}$ for several frequencies, plot $V_{0}$ against ν, and find the slope:
$h = e \times slope = e \frac{Δ V _{0}}{Δ ν} .$
The slope is the same for every metal (it is a universal constant $\frac{h}{e} \approx 4.14 \times 1 0^{- 15}$ V s); different metals give parallel lines with different intercepts. Millikan's careful measurements this way confirmed Einstein's equation.

Stopping Potential and the Frequency Graph: formulas and key facts

Stopping potential: $e V_{0} = E_{ma x} = \frac{1}{2} m v_{ma x}^{2}$; V₀ in volts gives E in eV.
Einstein in terms of V₀: $e V_{0} = h ν - W_{0}$; Straight-line relation.
Graph equation: $V_{0} = \frac{h}{e} ν - \frac{W _{0}}{e}$; y = mx + c.
Planck's constant: $h = e \times slope$; Slope ≈ 4.14 × 10⁻¹⁵ V s.
Threshold frequency: $ν_{0} = \frac{W _{0}}{h}$; ν-intercept.
Two readings: $h = \frac{e ( V _{2} - V _{1} )}{ν _{2} - ν _{1}}$; Eliminates W₀.

How to solve Stopping Potential and the Frequency Graph questions

Treat $V_{0}$ in volts as $E_{ma x}$ in electronvolts; then everything is Einstein's equation.

1Single reading: $V_{0} = \frac{h ν - W _{0}}{e}$ ; with energies in eV, $V_{0}$ (V) = E (eV) − $W_{0}$ (eV).
2Two readings: subtract to get $h = \frac{e ( V _{2} - V _{1} )}{ν _{2} - ν _{1}}$ .
3Then find $W_{0} = h ν_{1} - e V_{1}$ and $ν_{0} = \frac{W _{0}}{h}$ .
4From a graph: slope → h; ν-intercept → $ν_{0}$ ; magnitude of $V_{0}$ -intercept → $W_{0}$ in eV.
5For I-V graph questions, say what happens to the saturation current and to $V_{0}$ separately.

Stopping Potential and the Frequency Graph in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Identifying a lamp's colour

Suppose a photocell with an emitter of work function 1.8 eV is lit by an unlabelled lamp, and the photocurrent stops when the collector is 0.6 V negative.

Find the wavelength of the light.

Try it first, then show the solutionSolution

Photon energy $= W_{0} + e V_{0} = 1.8 + 0.6 = 2.4$ eV.
$λ = \frac{1240}{2.4} \approx 517$ nm.

Answer: ≈ 517 nm: green light.

Memory hook: Stopping volts plus work function = photon eV.

UV sensor on a cloudy day

Imagine a UV sensor based on a photocell. On a cloudy day the UV intensity halves but its spectrum is the same.

What happens to the saturation current and to the stopping potential?

Try it first, then show the solutionSolution

Half as many photons arrive per second, so half as many electrons: the saturation current halves.
Each photon has the same energy, so $E_{ma x}$ and the stopping potential are unchanged.

Answer: Current halves; stopping potential stays the same.

Memory hook: Clouds cut the count, not the kick.

Tricks to remember Stopping Potential and the Frequency Graph

Short memory aids. Say each one aloud once, then use it on a question the same day.

Slope is h over e
"Slope times charge gives Planck".
Use it when: Determining h.
Parallel lines, different starts
Every metal: same slope, different $ν_{0}$ .
Use it when: Two-metal graphs.
Intensity lifts the top, frequency moves the foot
I-V graph: intensity raises saturation; frequency moves the cut-off to more negative V.
Use it when: Sketching I-V graphs.
Volts equal electronvolts
A 1.5 V stopping potential means $E_{ma x}$ = 1.5 eV.
Use it when: Quick conversions.

Stopping Potential and the Frequency Graph: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Slope of the graph · 1 mark · MCQBoard level

The slope of the graph of stopping potential against frequency of incident light is:
(a) h
(b) $\frac{h}{e}$
(c) $\frac{e}{h}$
(d) $W_{0}$

Try it first, then show the solutionSolution

$V_{0} = \frac{h}{e} ν - \frac{W _{0}}{e}$ , so the slope is $\frac{h}{e}$ .

Answer: (b).

Example 2: Stopping potential from wavelength · 2 marks · Short answerBoard level

Light of wavelength 310 nm falls on a metal of work function 2.5 eV. Find the stopping potential.

Try it first, then show the solutionSolution

$E = \frac{1240}{310} = 4.0$ eV. (1)
$e V_{0} = 4.0 - 2.5 = 1.5$ eV, so $V_{0} = 1.5$ V. (1)

Answer: 1.5 V.

Example 3: I-V graphs · 3 marks · Long answerHarder board level

Sketch photocurrent against collector potential (a) for two intensities $I_{1} < I_{2}$ at the same frequency, and (b) for two frequencies $ν_{1} < ν_{2}$ at the same intensity. Explain each feature.

Try it first, then show the solutionSolution

(a) Two curves meeting the V-axis at the same $- V_{0}$ ; the $I_{2}$ curve saturates higher. (1)
More intense light has more photons, so more electrons (higher saturation current), but each photon's energy is the same, so $V_{0}$ is unchanged. (1)
(b) Two curves with about the same saturation current; the $ν_{2}$ curve cuts the axis at a more negative voltage, since $e V_{0} = h ν - W_{0}$ grows with ν. (1)

Answer: Intensity changes the saturation current; frequency changes the stopping potential.

Example 4: Planck's constant from two readings · 3 marks · Short answerHarder board level

A photocell gives stopping potentials of 1.2 V at $7.5 \times 1 0^{14}$ Hz and 2.4 V at $10.4 \times 1 0^{14}$ Hz. Find Planck's constant, the threshold frequency and the work function in eV.

Try it first, then show the solutionSolution

Slope $= \frac{2.4 - 1.2}{( 10.4 - 7.5 ) \times 1 0 ^{14}} \approx 4.14 \times 1 0^{- 15}$ V s; $h = 1.6 \times 1 0^{- 19} \times 4.14 \times 1 0^{- 15} \approx 6.62 \times 1 0^{- 34}$ J s. (1)
$ν_{0} = 7.5 \times 1 0^{14} - \frac{1.2}{4.14 \times 1 0 ^{- 15}} \approx 4.6 \times 1 0^{14}$ Hz. (1)
$W_{0} = \frac{h ν _{0}}{e} \approx 1.90$ eV. (1)

Answer: ≈ $6.62 \times 1 0^{- 34}$ J s; ≈ $4.6 \times 1 0^{14}$ Hz; ≈ 1.9 eV.

Example 5: Two metals on one graph · 3 marks · Short answerHOTS / competency

Metals A ( $W_{0}$ = 2.0 eV) and B ( $W_{0}$ = 3.0 eV) are lit by photons of 5.0 eV. (a) Find the stopping potential for each. (b) Find each threshold frequency. (c) Describe their $V_{0}$ -ν graphs.

Try it first, then show the solutionSolution

(a) A: 3.0 V; B: 2.0 V. (1)
(b) $ν_{0} = \frac{W _{0}}{h}$ : A ≈ $4.83 \times 1 0^{14}$ Hz; B ≈ $7.24 \times 1 0^{14}$ Hz. (1)
(c) Two parallel straight lines (same slope $\frac{h}{e}$ ); B's line is shifted to higher frequency, cutting the ν-axis further right. (1)

Answer: 3.0 V and 2.0 V; ≈ $4.83 \times 1 0^{14}$ and $7.24 \times 1 0^{14}$ Hz; parallel lines.

Example 6: Turning the collector negative · 4 marks · Case-basedHOTS / competency

Suppose light of 400 nm falls on an emitter of work function 2.0 eV in a photocell.
(i) Find the stopping potential. (1 mark)
(ii) The collector is made 0.6 V negative. Is there still a current? Why? (1 mark)
(iii) What is the current when the collector is 1.5 V negative? (1 mark)
(iv) The intensity is doubled at 400 nm. What happens to the saturation current and to $V_{0}$ ? (1 mark)

Try it first, then show the solutionSolution

(i) $E = \frac{1240}{400} = 3.1$ eV; $V_{0} = 3.1 - 2.0 = 1.1$ V.
(ii) Yes, a smaller current: electrons with kinetic energy above 0.6 eV still reach the collector.
(iii) Zero: 1.5 V exceeds $V_{0}$ , so even the fastest electrons are turned back.
(iv) Saturation current doubles; $V_{0}$ stays 1.1 V.

Answer: (i) 1.1 V; (ii) yes, reduced; (iii) zero; (iv) current doubles, $V_{0}$ unchanged.

Where students lose marks in Stopping Potential and the Frequency Graph

The slips examiners see most often, and what to write instead.

Mistake: Saying the slope is h
The ISC 2025 examiners (Q17) noted candidates who took the slope itself as Planck's constant.
Wrong working
h = slope ≈ $4.1 \times 1 0^{- 15}$ .
Right working
h = e × slope = $1.6 \times 1 0^{- 19} \times 4.14 \times 1 0^{- 15} \approx 6.6 \times 1 0^{- 34}$ J s.
Do this: Write $V_{0} = \frac{h}{e} ν - \frac{W _{0}}{e}$ first.
Usually costs: 1 mark.
Mistake: Curved or unlabelled graph
Also 2025 Q17: curves drawn instead of a straight line; axes missing or wrongly labelled.
Do this: Straight line, $V_{0}$ on y-axis, ν on x-axis, cutting the ν-axis at $ν_{0}$ .
Usually costs: 1 mark.
Mistake: Stopping potential changing with intensity
In 2024 Q16(i) many said it increases or decreases with intensity.
Do this: $V_{0}$ depends only on ν and the metal.
Usually costs: 1 mark.
Mistake: Line drawn through the origin
Students forget the threshold.
Do this: The line starts at $ν_{0}$ , not at zero.
Usually costs: 1 mark.

Exam tips

Write the straight-line form $V_{0} = \frac{h}{e} ν - \frac{W _{0}}{e}$ before discussing slope and intercepts.
Draw I-V curves with saturation and the cut-off clearly marked.
In numericals, keep $V_{0}$ in volts and energies in eV to avoid conversions.
Give h with units J s and three significant figures.

Practice questions on Stopping Potential and the Frequency Graph

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In the graph of stopping potential against frequency, the point where the line meets the frequency axis gives:
(a) Planck's constant
(b) the threshold frequency
(c) the work function in joules
(d) the saturation current
Hint
$V_{0} = 0$ there.
Answer
(b).
Q2 · 1 mark · Short answerBoard level
Photoelectrons are just stopped by a potential of 2.4 V. Find their maximum kinetic energy in eV and in joules.
Hint
$e V_{0}$ .
Answer
2.4 eV = $3.84 \times 1 0^{- 19}$ J.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The saturation photocurrent increases when the intensity of light is increased at fixed frequency.
Reason (R): A more intense beam delivers more photons per second and so ejects more electrons.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Count of electrons.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
A metal has work function 2.2 eV, and the stopping potential is 1.3 V. Find the wavelength of the light.
Hint
E = 3.5 eV.
Answer
≈ 354 nm.
Q5 · 2 marks · Short answerHarder board level
The extended $V_{0}$ -ν line for a metal cuts the $V_{0}$ axis at −1.8 V. Find the work function and the threshold frequency.
Hint
Intercept = $- \frac{W _{0}}{e}$ .
Answer
1.8 eV; ≈ $4.34 \times 1 0^{14}$ Hz.
Q6 · 3 marks · Short answerHOTS / competency
When the frequency of light on a metal is doubled, the stopping potential rises from 0.8 V to 3.0 V. Find the work function and the original photon energy.
Hint
E − W = 0.8; 2E − W = 3.0.
Answer
E = 2.2 eV; $W_{0}$ = 1.4 eV.
Q7 · 2 marks · Short answerHOTS / competency
Explain why the $V_{0}$ -ν graphs for different metals are parallel, and what is different about them.
Hint
Slope and intercept.
Answer
The slope $\frac{h}{e}$ depends only on universal constants, so it is the same for all metals; the intercepts differ because each metal has its own work function and threshold frequency.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student records stopping potentials of 0.27, 0.68, 1.09 and 1.51 V at frequencies of 5.5, 6.5, 7.5 and $8.5 \times 1 0^{14}$ Hz.
(i) Describe the graph. (1 mark)
(ii) Find its slope. (1 mark)
(iii) Find Planck's constant. (1 mark)
(iv) Find the threshold frequency and work function. (1 mark)
Hint
Use the first and last points.
Answer
(i) A straight line with positive slope. (ii) $\frac{1.51 - 0.27}{3 \times 1 0 ^{14}} \approx 4.13 \times 1 0^{- 15}$ V s. (iii) h ≈ $6.61 \times 1 0^{- 34}$ J s. (iv) $ν_{0} \approx 4.85 \times 1 0^{14}$ Hz; $W_{0}$ ≈ 2.0 eV.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define stopping potential.Show answer
The minimum negative potential of the collector at which the photocurrent becomes zero.
2.Slope of $V_{0}$ -ν graph?Show answer
$\frac{h}{e}$ .
3.Does $V_{0}$ depend on intensity?Show answer
No.
4. $E_{ma x}$ = 1.7 eV: $V_{0}$ ?Show answer
1.7 V.
5.What does the ν-intercept give?Show answer
Threshold frequency.

Stopping Potential and the Frequency Graph: frequently asked questions

Why does some current flow at zero collector voltage?

Photoelectrons leave the emitter with kinetic energy, so some reach the collector even without an accelerating field.

Why does the current saturate?

Once the collector is positive enough, every emitted electron is collected; the current can rise further only if more electrons are emitted (more intense light).

Why is the $V_0$-ν graph a straight line?

Because eV_0 = h\nu - W_0 is linear in ν, with h and W_0 constant for a given metal.

What does the ISC 2027 syllabus need here?

Experimental facts of the photoelectric effect and the determination of Planck's constant from the graph of stopping potential against frequency, using Einstein's equation.

How was this asked recently?

ISC 2025 Q17 (3 marks) asked the labelled V_s-f graph and how to get h from it; ISC 2024 Q16 (3 marks) asked the effect of intensity on stopping potential and a comparison of three metals.