PhysicsCBSE Class 12 · Dual Nature of Radiation and Matter

Photoelectric Effect Class 12: Experimental Study and Graphs

Updated 15 October 2026 · Fact-checked

In the photoelectric effect, light of high enough frequency ejects electrons from a metal surface. Experiments show that photocurrent is proportional to intensity, the stopping potential (and maximum kinetic energy) rises linearly with frequency but does not depend on intensity, no emission occurs below a threshold frequency, and emission is instantaneous. The wave theory cannot explain these results.

UV light enters through the quartz window, knocks photoelectrons out of plate C, and the microammeter reads the current when A is made positive.

Illustration drawn for QuestionBuddy.

Understand Photoelectric Effect: Experimental Study

Shine ultraviolet light on a freshly cleaned zinc plate connected to a charged electroscope and, if the electroscope is negatively charged, it slowly loses its charge. Light is knocking electrons out of the metal. This is the photoelectric effect, and its details forced physicists to rethink the nature of light.

Early observations. Hertz (1887) noticed that sparks jumped more easily across a gap when ultraviolet light fell on the cathode. Hallwachs and Lenard showed that ultraviolet light makes a metal plate emit negatively charged particles (electrons, called photoelectrons), and that no electrons are emitted, however bright the light, if its frequency is below a certain value, the threshold frequency $ν_{0}$ , which depends on the metal. Alkali metals such as sodium and caesium respond even to visible light; zinc, copper and others need ultraviolet.

Experimental set-up. An evacuated glass tube contains a photosensitive plate C (emitter) and a collector plate A. Light of chosen frequency and intensity falls on C through a quartz window (quartz lets UV through). A battery and potential divider can make A positive or negative relative to C, and a microammeter reads the photocurrent.

1. Effect of intensity. With frequency fixed (above $ν_{0}$ ) and A positive, the photocurrent is directly proportional to the intensity of light: more light, more electrons emitted per second.

2. Effect of potential. As A is made more positive, the current rises until all emitted electrons are collected: the saturation current, which is proportional to intensity. Making A negative (retarding) reduces the current, because only electrons with enough kinetic energy reach A. At a particular negative potential, the stopping potential $V_{0}$ , even the fastest electrons are turned back and the current becomes zero. So $K_{ma x} = e V_{0}$ . For a given frequency, $V_{0}$ is the same for all intensities: the I-V curves for different intensities have different saturation currents but meet the axis at the same $- V_{0}$ .

3. Effect of frequency. With the same intensity at different frequencies, the saturation current stays the same but $V_{0}$ increases with frequency. A graph of $V_{0}$ against ν is a straight line that cuts the frequency axis at the threshold frequency $ν_{0}$ . Lines for different metals are parallel (same slope) but have different intercepts.

4. No time lag. Emission starts almost instantly (within about $1 0^{- 9}$ s) once light of frequency above $ν_{0}$ falls on the metal, even if the light is very dim.

Why the wave theory fails. In the wave picture, energy is spread smoothly over the wavefront, so (a) brighter light should give faster electrons, but $K_{ma x}$ does not depend on intensity; (b) any frequency should work if the light is intense enough or shone long enough, but there is a sharp threshold; (c) a dim beam would need a long time to deliver enough energy to one electron, yet emission is instantaneous. Einstein explained all of these with photons (next topic).

Photoelectric Effect: Experimental Study: formulas and key facts

Stopping potential and maximum kinetic energy: $K_{ma x} = \frac{1}{2} m v_{ma x}^{2} = e V_{0}$; V₀ independent of intensity; increases with frequency.
Photocurrent and intensity: $I_{s a t} \propto intensity$; Frequency fixed and above threshold.
Number of electrons per second: $n = \frac{I}{e}$; From the photocurrent.
Stopping potential versus frequency: $V_{0} = \frac{h}{e} (ν - ν_{0})$; Straight line: slope h/e, intercept ν₀ on the frequency axis.
Threshold condition: $ν < ν_{0} \Rightarrow no emission$; Regardless of intensity.

How to solve Photoelectric Effect: Experimental Study questions

Use this method for experiment-based and graph questions.

1Read what is held constant: intensity, frequency or the metal.
2If intensity changes at fixed frequency: saturation current changes in proportion; stopping potential does not.
3If frequency changes at fixed intensity (above threshold): stopping potential changes; saturation current stays the same.
4From a stopping potential, find $K_{ma x} = e V_{0}$ (in eV numerically equal to V₀ in volt) and $v_{ma x} = \frac{2 e V _{0}}{m}$ .
5From a V₀-ν graph, slope = h/e, frequency intercept = ν₀, and the extended V-axis intercept = −φ₀/e.
6From a photocurrent, number of electrons per second = I/e.
7For 'why wave theory fails', give the three observations: threshold, KE independent of intensity, no time lag.

Photoelectric Effect: Experimental Study 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.

Threshold frequency in a light sensor

Suppose a light sensor uses a metal whose threshold wavelength is 600 nm (it emits electrons only for shorter wavelengths). An engineer tests it with a very bright red lamp (700 nm) and with a dim green lamp (530 nm).

(a) Find the threshold frequency. (b) Which lamp makes the sensor respond? (c) Explain why brightness alone does not matter.

Try it first, then show the solutionSolution

$ν_{0} = \frac{c}{λ _{0}} = \frac{3 \times 1 0 ^{8}}{6 \times 1 0 ^{- 7}} = 5 \times 1 0^{14}$ Hz.
Red: $ν = \frac{3 \times 1 0 ^{8}}{7 \times 1 0 ^{- 7}} \approx 4.3 \times 1 0^{14}$ Hz < ν₀: no response. Green (530 nm) has ν ≈ 5.7 × 10¹⁴ Hz > ν₀: the sensor responds, even though the lamp is dim.
Emission depends on frequency exceeding the threshold; increasing the intensity of light below threshold produces no electrons at all.

Answer: (a) 5 × 10¹⁴ Hz; (b) the dim green lamp; (c) only frequency above ν₀ can eject electrons.

Memory hook: Colour decides whether; brightness decides how many.

Photocurrent tracks intensity

Imagine a photographer's light meter based on a photocell. In sunshine it gives a saturation current of 3.0 μA; when a cloud passes, the current drops to 1.2 μA.

(a) By what fraction has the light intensity fallen? (b) How many electrons per second are emitted in sunshine? (c) Does the maximum speed of the photoelectrons change when the cloud passes? Explain.

Try it first, then show the solutionSolution

Saturation current ∝ intensity: $\frac{1.2}{3.0} = 0.4$ , so the intensity is now 40% of before (a fall of 60%).
$n = \frac{3 \times 1 0 ^{- 6}}{1.6 \times 1 0 ^{- 19}} \approx 1.9 \times 1 0^{13}$ per second.
No: if the light's spectrum is unchanged, the stopping potential (and K_max) does not depend on intensity.

Answer: (a) to 40% of its value; (b) about 1.9 × 10¹³ electrons/s; (c) no change in maximum speed.

Memory hook: Cloud cuts the number of electrons, not their top speed.

Tricks to remember Photoelectric Effect: Experimental Study

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

Intensity → current, frequency → energy
More intense light: more electrons. Higher frequency: faster electrons.
Use it when: Most graph and reasoning questions.
Stop sign is the same
I-V curves for different intensities all stop at the same −V₀.
Use it when: Drawing intensity graphs.
Same roof, different stops
Same intensity, different frequencies: same saturation 'roof', different stopping points.
Use it when: Drawing frequency graphs.
Three failures of waves: T-K-T
Threshold, Kinetic energy independent of intensity, Time lag absent.
Use it when: Why wave theory fails.

Photoelectric Effect: Experimental Study: solved examples

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

Example 1: From stopping potential to speed · 2 marks · Very short answerBoard level

The stopping potential in a photoelectric experiment is 2 V. Find the maximum kinetic energy and maximum speed of the photoelectrons.

Try it first, then show the solutionSolution

$K_{ma x} = e V_{0} = 2$ eV $= 3.2 \times 1 0^{- 19}$ J. (1)
$v_{ma x} = \frac{2 K}{m} = \frac{6.4 \times 1 0 ^{- 19}}{9.1 \times 1 0 ^{- 31}} \approx 8.4 \times 1 0^{5}$ m/s. (1)

Answer: 2 eV (3.2 × 10⁻¹⁹ J) and about 8.4 × 10⁵ m/s.

Example 2: Graphs of photocurrent · 3 marks · Short answerBoard level

Draw (describe) graphs of photocurrent against collector potential (a) for two intensities at the same frequency, and (b) for two frequencies at the same intensity. Explain the features of each.

Try it first, then show the solutionSolution

(a) Two curves meeting the V-axis at the same negative potential −V₀ but saturating at different currents; the higher intensity gives the higher saturation current, since more photons eject more electrons. (1½)
(b) Two curves with the same saturation current but different stopping potentials; the higher frequency gives the larger V₀, since K_max increases with frequency. (1½)

Answer: (a) same V₀, saturation current ∝ intensity; (b) same saturation current, larger V₀ for higher frequency.

Example 3: Reading a V₀-ν graph · 3 marks · Short answerHarder board level

In an experiment, the stopping potential is 0.497 V for light of frequency 6.0 × 10¹⁴ Hz and 1.739 V for 9.0 × 10¹⁴ Hz. Find (a) the slope of the V₀-ν line and the value of h, (b) the threshold frequency, and (c) the work function of the metal in eV.

Try it first, then show the solutionSolution

(a) Slope $= \frac{1.739 - 0.497}{3 \times 1 0 ^{14}} \approx 4.14 \times 1 0^{- 15}$ V s; $h = e \times$ slope $\approx 6.62 \times 1 0^{- 34}$ J s. (1)
(b) $ν_{0} = 6 \times 1 0^{14} - \frac{0.497}{4.14 \times 1 0 ^{- 15}} \approx 4.8 \times 1 0^{14}$ Hz. (1)
(c) $ϕ_{0} = h ν_{0}$ , i.e. in eV, $4.14 \times 1 0^{- 15} \times 4.8 \times 1 0^{14} \approx 1.99$ eV. (1)

Answer: (a) about 4.14 × 10⁻¹⁵ V s and 6.62 × 10⁻³⁴ J s; (b) about 4.8 × 10¹⁴ Hz; (c) about 2.0 eV.

Example 4: Predicting the effect of changes · 3 marks · Short answerHarder board level

For a photocell, state what happens to (i) the saturation current and (ii) the stopping potential when (a) the intensity is doubled at fixed frequency, (b) the frequency is increased at fixed intensity, and (c) light below the threshold frequency is made ten times more intense.

Try it first, then show the solutionSolution

(a) Saturation current doubles; stopping potential unchanged. (1)
(b) Saturation current unchanged (same intensity); stopping potential increases. (1)
(c) No emission at all: both remain zero. (1)

Answer: (a) I_sat doubles, V₀ same; (b) I_sat same, V₀ rises; (c) no current.

Example 5: The time-lag problem for wave theory · 5 marks · Long answerHOTS / competency

(a) State three observations of the photoelectric effect that the wave theory of light cannot explain. (3)
(b) Suppose a metal has work function 2 eV and each atom presents an area of about $1 0^{- 20}$ m² to a wave of intensity $1 0^{- 2}$ W/m². Estimate how long, on the wave picture, an electron would take to gather enough energy to escape, and compare with experiment. (2)

Try it first, then show the solutionSolution

(a) (i) No emission below the threshold frequency, however intense the light. (1)
(ii) Maximum kinetic energy depends on frequency, not on intensity. (1)
(iii) Emission is instantaneous (within about 10⁻⁹ s), even for dim light. (1)
(b) Power received per atom = $1 0^{- 2} \times 1 0^{- 20} = 1 0^{- 22}$ W; energy needed = 2 eV = 3.2 × 10⁻¹⁹ J; time $= \frac{3.2 \times 1 0 ^{- 19}}{1 0 ^{- 22}} = 3200$ s, about 53 minutes. (1)
Experiment shows emission within about 10⁻⁹ s, so the energy must arrive in concentrated packets (photons), not spread over the wavefront. (1)

Answer: (a) threshold frequency, K_max independent of intensity, no time lag; (b) about 3200 s on the wave picture, against less than 10⁻⁹ s observed.

Example 6: A photocell in a counter · 4 marks · Case-basedHarder board level

Read the passage and answer.

A photocell is used to count objects on a conveyor belt; each object interrupts a beam of light falling on the cell. With the beam on, the photocurrent is 2.0 μA.
(a) How many electrons per second leave the emitter? (1)
(b) Why must the frequency of the beam be above the threshold of the emitter? (1)
(c) What happens to the current when an object blocks the beam? (1)
(d) If a brighter lamp of the same colour is used, how do the current and the stopping potential change? (1)

Try it first, then show the solutionSolution

(a) $n = \frac{2 \times 1 0 ^{- 6}}{1.6 \times 1 0 ^{- 19}} = 1.25 \times 1 0^{13}$ per second. (1)
(b) Below the threshold no electrons are emitted at any intensity. (1)
(c) It falls to (nearly) zero, which the counter registers as one object. (1)
(d) The current increases in proportion to intensity; the stopping potential stays the same. (1)

Answer: (a) 1.25 × 10¹³/s; (b) otherwise no emission; (c) it drops to zero; (d) current rises, V₀ unchanged.

Where students lose marks in Photoelectric Effect: Experimental Study

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

Mistake: Saying higher intensity increases the stopping potential.
Mixing up number and energy of electrons.
Do this: Intensity changes the number of electrons (current), not their maximum energy.
Usually costs: 1 mark.
Mistake: Drawing different stopping potentials for different intensities.
Graph slip.
Do this: For the same frequency, all curves meet the voltage axis at the same −V₀.
Usually costs: 1 mark of the graph.
Mistake: Forgetting units when converting eV to J.
Treating eV as J.
Wrong working
K = 2 J
Right working
K = 2 eV = 3.2 × 10⁻¹⁹ J
Do this: Multiply by 1.6 × 10⁻¹⁹ to get joule.
Usually costs: The answer mark.
Mistake: Saying low-frequency light will eventually eject electrons if left on long enough.
Wave-theory thinking.
Do this: Below threshold there is no emission at all, however long or bright the light.
Usually costs: 1 mark.

Exam tips

Practise the three standard graphs: I versus intensity, I versus V (two intensities and two frequencies), and V₀ versus ν.
Label axes and mark −V₀, saturation currents and ν₀; unlabelled graphs lose marks.
Experimental features of the photoelectric effect were set in 2023 (55/1/1 Q18), 2024 (55/1/1 Q13), 2025 (55/1/1 Q26) and 2026 (55/5/1 Q16).
When asked why the wave theory fails, give three points with one line each.
Use eV for energies where possible; it keeps numbers simple.

Practice questions on Photoelectric Effect: Experimental Study

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
The stopping potential in a photoelectric experiment depends on
(a) the intensity of light only (b) the frequency of light and the metal (c) the collector area (d) the intensity and the metal
Hint
V₀ = (h/e)(ν − ν₀).
Answer
(b) the frequency of light and the metal.
Q2 · 2 marks · Very short answerBoard level
Define (a) threshold frequency and (b) stopping potential.
Hint
Minimum frequency; minimum retarding potential.
Answer
(a) The minimum frequency of incident light below which no photoelectrons are emitted from a given metal, however intense the light. (b) The minimum negative (retarding) potential of the collector at which the photocurrent becomes zero.
Q3 · 1 mark · Very short answerHarder board level
The stopping potential for a metal is 3 V. Find the maximum kinetic energy of the photoelectrons in joule.
Hint
K = eV₀.
Answer
K_max = 1.6 × 10⁻¹⁹ × 3 = 4.8 × 10⁻¹⁹ J.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The photoelectric current increases with the intensity of incident light (above threshold frequency).
Reason (R): A more intense beam has more photons per second, each of which can eject an electron.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Photon picture.
Answer
(a).
Q5 · 1 mark · Very short answerHarder board level
A photocurrent of 4.8 μA flows. How many photoelectrons leave the emitter each second?
Hint
n = I/e.
Answer
n = 4.8 × 10⁻⁶/1.6 × 10⁻¹⁹ = 3 × 10¹³ per second.
Q6 · 2 marks · Very short answerHOTS / competency
The V₀-ν graphs for two metals P and Q are parallel straight lines; Q's line cuts the frequency axis at a higher frequency. Why are the lines parallel, and which metal has the larger work function?
Hint
Slope and intercept.
Answer
The slope is h/e, a universal constant, so the lines are parallel. Q has the higher threshold frequency, so its work function hν₀ is larger.
Q7 · 2 marks · Very short answerHOTS / competency
In the I-V graph for a photocell, why does the current not become zero as soon as the collector is made slightly negative?
Hint
Electrons leave with a range of energies.
Answer
Photoelectrons are emitted with a range of kinetic energies up to K_max. A small retarding potential turns back only the slower ones; the faster ones still reach the collector. The current becomes zero only when the retarding potential reaches V₀, which stops even the fastest electrons.

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.What happens to photocurrent when intensity doubles (frequency fixed)?Show answer
It doubles.
2.Does stopping potential depend on intensity?Show answer
No.
3.Relation between stopping potential and K_max?Show answer
K_max = eV₀.
4.Slope of the V₀-ν graph?Show answer
h/e.
5.What is the threshold frequency?Show answer
The minimum frequency for photoelectric emission from a given metal.
6.Time lag in photoelectric emission?Show answer
Practically none (about 10⁻⁹ s or less).

Photoelectric Effect: Experimental Study: frequently asked questions

What is saturation current?

The maximum photocurrent, reached when the collector is positive enough to collect every emitted electron. It is proportional to the intensity of light.

Why is the stopping potential independent of intensity?

Each electron gets its energy from a single photon, whose energy depends only on frequency. A more intense beam has more photons, ejecting more electrons, but none of them faster.

Why does the wave theory fail to explain the photoelectric effect?

It predicts that brighter light should give faster electrons, that any frequency should work if intense enough, and that dim light should cause a delay. Experiments show the opposite on all three counts.

Why is the tube evacuated and the window made of quartz?

Evacuation stops electrons from colliding with gas molecules on their way to the collector. Quartz lets ultraviolet light through, which ordinary glass absorbs.

What is the significance of the V₀-ν graph?

Its straight-line form supports Einstein's equation; its slope gives h/e and its intercept on the frequency axis gives the threshold frequency, and hence the work function.