PhysicsCBSE Class 12 · Dual Nature of Radiation and Matter

Einstein's Photoelectric Equation Class 12: Photons and Work Function

Updated 15 October 2026 · Fact-checked

Einstein proposed that light consists of photons of energy hν. A photon gives all its energy to one electron, which uses the work function φ₀ to escape and keeps the rest as kinetic energy: K_max = hν − φ₀ = eV₀. Emission needs hν ≥ φ₀, so the threshold frequency is ν₀ = φ₀/h. This explains every photoelectric observation.

One photon gives all its energy hν to one electron: part pays the work function φ₀ to escape, and the rest becomes K_max = eV₀.

Illustration drawn for QuestionBuddy.

Understand Einstein's Photoelectric Equation

The photoelectric experiments showed that light behaves as if its energy came in lumps. In 1905 Einstein took this literally: light is a stream of particles, photons.

Photons. A photon of light of frequency ν (wavelength λ) has:

energy $E = h ν = \frac{h c}{λ}$ , where $h = 6.63 \times 1 0^{- 34}$ J s is Planck's constant (handy form: $E (eV) \approx \frac{1240}{λ ( nm )}$ );
momentum $p = \frac{h ν}{c} = \frac{h}{λ}$ ;
speed c in vacuum and zero rest mass; it is electrically neutral.

All photons of a given frequency have the same energy; a more intense beam simply has more photons per second. For a beam of power P, the number of photons per second is $N = \frac{P}{h ν}$ . In collisions with electrons, energy and momentum are conserved, but photons may be absorbed or created, so their number need not be conserved.

Work function. Electrons are held inside a metal. The minimum energy needed to free an electron from the surface is the work function $ϕ_{0}$ (a few eV; it depends on the metal and its surface).

Einstein's equation. In the photoelectric effect, one photon is absorbed by one electron. The electron spends at least $ϕ_{0}$ to escape; the most loosely bound ones leave with the maximum kinetic energy: $K_{ma x} = h ν - ϕ_{0}$ . Since $K_{ma x} = e V_{0}$ : $e V_{0} = h ν - ϕ_{0}$ , or $V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}$ . Emission needs $h ν \geq ϕ_{0}$ , giving the threshold frequency $ν_{0} = \frac{ϕ _{0}}{h}$ and threshold wavelength $λ_{0} = \frac{h c}{ϕ _{0}}$ . So the equation can also be written $K_{ma x} = h (ν - ν_{0})$ .

How it explains the observations.

Threshold frequency: if hν < φ₀, no single photon can free an electron, and electrons cannot add up energy from several photons, so there is no emission however intense the light.
K_max depends on frequency, not intensity: K_max = hν − φ₀ contains only ν; intensity changes the number of photons, hence the number of electrons (current), not their energy.
Linear V₀-ν graph: the equation is a straight line of slope h/e, the same for all metals, with intercept ν₀ on the frequency axis. Millikan's careful measurements confirmed this and gave a value of h agreeing with Planck's.
No time lag: the energy arrives as a whole photon and is absorbed at once by one electron.

Particle nature of light. The photoelectric effect shows that light, which shows interference and diffraction as a wave, also behaves as particles when it exchanges energy with matter. This dual nature leads to the idea that matter, too, can behave as waves (de Broglie).

Einstein's Photoelectric Equation: formulas and key facts

Photon energy and momentum: $E = h ν = \frac{h c}{λ}, p = \frac{h}{λ}$; E(eV) ≈ 1240/λ(nm).
Einstein's photoelectric equation: $K_{ma x} = h ν - ϕ_{0} = e V_{0}$; One photon, one electron.
Threshold frequency and wavelength: $ν_{0} = \frac{ϕ _{0}}{h}, λ_{0} = \frac{h c}{ϕ _{0}}$; No emission for ν < ν₀ (λ > λ₀).
Stopping potential: $V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}$; Slope h/e; ν-intercept ν₀.
Number of photons per second: $N = \frac{P}{h ν} = \frac{P λ}{h c}$; P = power of the beam.

How to solve Einstein's Photoelectric Equation questions

Use this method for photon and photoelectric numericals.

1Convert the given wavelength or frequency into photon energy in eV: $E \approx \frac{1240}{λ ( nm )}$ eV or $E = \frac{h ν}{e}$ eV.
2Compare with the work function: if E < φ₀, no emission.
3K_max = E − φ₀ (in eV); the stopping potential in volt has the same number.
4For speed, convert K_max to joule and use $v = \frac{2 K}{m}$ .
5For threshold values: ν₀ = φ₀/h, λ₀ = 1240/φ₀(eV) nm.
6With two sets of data, write the equation twice and subtract to eliminate φ₀ (or find h).
7For photon numbers, divide the power by the energy of one photon in joule.

Einstein's Photoelectric Equation 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.

Choosing light for a photocell

Suppose a burglar alarm uses a photocell whose emitter has a work function of 2.0 eV. The designer first tries an invisible infrared beam of wavelength 900 nm, then a 500 nm beam.

(a) Find the photon energy in each case. (b) Which beam works, and what is the maximum kinetic energy of the photoelectrons? (c) Would a stronger infrared lamp work?

Try it first, then show the solutionSolution

900 nm: $E \approx \frac{1240}{900} \approx 1.38$ eV. 500 nm: $E \approx \frac{1240}{500} = 2.48$ eV.
Only 500 nm exceeds φ₀: K_max = 2.48 − 2.0 = 0.48 eV.
No: each infrared photon has only 1.38 eV, less than φ₀; more photons do not help because one photon frees one electron.

Answer: (a) about 1.38 eV and 2.48 eV; (b) the 500 nm beam, with K_max ≈ 0.48 eV; (c) no.

Memory hook: One photon, one electron: a weak photon fails however many you send.

Photons in a beam of light

Imagine a 100 W filament lamp that gives out about 5 W as visible light. Take the average visible wavelength as 550 nm.

(a) Find the energy of one 550 nm photon. (b) Estimate the number of visible photons emitted per second. (c) Why don't we notice light arriving in separate packets?

Try it first, then show the solutionSolution

$E = \frac{h c}{λ} = \frac{6.63 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{5.5 \times 1 0 ^{- 7}} \approx 3.6 \times 1 0^{- 19}$ J.
$N = \frac{5}{3.6 \times 1 0 ^{- 19}} \approx 1.4 \times 1 0^{19}$ photons per second.
So many photons arrive every second that the light appears continuous; the graininess shows only in experiments with very faint light or single-electron effects.

Answer: (a) about 3.6 × 10⁻¹⁹ J; (b) about 1.4 × 10¹⁹ per second; (c) the numbers are enormous, so the flow looks smooth.

Memory hook: A lamp is a photon fire hose: 10¹⁹ every second.

Tricks to remember Einstein's Photoelectric Equation

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

Pay the toll, keep the change
Photon energy hν pays the work function toll φ₀; the change is K_max.
Use it when: Writing Einstein's equation.
1240 over nanometres
Photon energy in eV ≈ 1240/λ(nm).
Use it when: Fast calculations.
One photon, one electron
Energy is never pooled from several photons.
Use it when: Explaining the threshold.
Slope h over e for everyone
All metals give parallel V₀-ν lines; only the intercept (work function) changes.
Use it when: Graph questions.

Einstein's Photoelectric Equation: solved examples

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

Example 1: Kinetic energy and stopping potential · 2 marks · Very short answerBoard level

Light of wavelength 400 nm falls on a metal of work function 2.3 eV. Find (a) the photon energy, (b) the maximum kinetic energy of the photoelectrons and (c) the stopping potential.

Try it first, then show the solutionSolution

(a) $E \approx \frac{1240}{400} = 3.1$ eV. (½)
(b) $K_{ma x} = 3.1 - 2.3 = 0.8$ eV $\approx 1.28 \times 1 0^{- 19}$ J. (1)
(c) $V_{0} = 0.8$ V. (½)

Answer: (a) 3.1 eV; (b) 0.8 eV; (c) 0.8 V.

Example 2: Einstein's equation and the observations · 3 marks · Short answerBoard level

Write Einstein's photoelectric equation and use it to explain (a) the existence of a threshold frequency, (b) why the maximum kinetic energy is independent of intensity, and (c) the linear dependence of stopping potential on frequency.

Try it first, then show the solutionSolution

$K_{ma x} = h ν - ϕ_{0}$ . (a) If hν < φ₀, a single photon cannot free an electron, and energy from several photons is not pooled, so emission needs ν ≥ ν₀ = φ₀/h. (1)
(b) K_max depends only on ν (and φ₀). Greater intensity means more photons, so more electrons but each with the same maximum energy. (1)
(c) $e V_{0} = h ν - ϕ_{0}$ gives $V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}$ , a straight line of slope h/e. (1)

Answer: K_max = hν − φ₀ explains the threshold, the intensity-independence of K_max and the linear V₀-ν graph.

Example 3: Work function from data · 3 marks · Short answerHarder board level

When light of wavelength 250 nm falls on a metal, the stopping potential is 2.46 V. Find (a) the work function, (b) the threshold wavelength and (c) the stopping potential for light of wavelength 350 nm.

Try it first, then show the solutionSolution

(a) $E = \frac{1240}{250} = 4.96$ eV; $ϕ_{0} = 4.96 - 2.46 = 2.5$ eV. (1)
(b) $λ_{0} = \frac{1240}{2.5} = 496$ nm. (1)
(c) $E = \frac{1240}{350} \approx 3.54$ eV; $V_{0} \approx 3.54 - 2.5 = 1.04$ V. (1)

Answer: (a) 2.5 eV; (b) 496 nm; (c) about 1.04 V.

Example 4: Photons from a laser · 3 marks · Short answerHarder board level

A laser pointer emits 5 mW at 650 nm. Find (a) the energy and momentum of one photon and (b) the number of photons emitted per second.

Try it first, then show the solutionSolution

(a) $E = \frac{h c}{λ} = \frac{6.63 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{6.5 \times 1 0 ^{- 7}} \approx 3.06 \times 1 0^{- 19}$ J. (1)
$p = \frac{h}{λ} \approx \frac{6.63 \times 1 0 ^{- 34}}{6.5 \times 1 0 ^{- 7}} \approx 1.02 \times 1 0^{- 27}$ kg m/s. (1)
(b) $N = \frac{5 \times 1 0 ^{- 3}}{3.06 \times 1 0 ^{- 19}} \approx 1.6 \times 1 0^{16}$ per second. (1)

Answer: (a) about 3.06 × 10⁻¹⁹ J and 1.02 × 10⁻²⁷ kg m/s; (b) about 1.6 × 10¹⁶ per second.

Example 5: Comparing two frequencies · 5 marks · Long answerHOTS / competency

A metal has a work function of 2 eV.
(a) Photons of 4 eV and 6 eV fall on it in turn. Find the ratio of the maximum speeds of the photoelectrons. (2)
(b) Show that doubling the frequency of the light more than doubles K_max. (2)
(c) What frequency of light is just enough to cause emission? (1)

Try it first, then show the solutionSolution

(a) K_max = 2 eV and 4 eV. (1)
v ∝ √K, so $v_{1} : v_{2} = 2 : 4 = 1 : 2$ . (1)
(b) $K_{1} = h ν - ϕ_{0}$ ; $K_{2} = 2 h ν - ϕ_{0} = 2 K_{1} + ϕ_{0}$ . (1)
Since φ₀ > 0, K₂ > 2K₁. (1)
(c) $ν_{0} = \frac{ϕ _{0}}{h} = \frac{2 \times 1.6 \times 1 0 ^{- 19}}{6.63 \times 1 0 ^{- 34}} \approx 4.8 \times 1 0^{14}$ Hz. (1)

Answer: (a) 1 : √2; (b) K₂ = 2K₁ + φ₀ > 2K₁; (c) about 4.8 × 10¹⁴ Hz.

Example 6: Threshold and energy · 4 marks · Case-basedHarder board level

Read the passage and answer.

A metal has a threshold frequency of 5.0 × 10¹⁴ Hz. Light of frequency 7.5 × 10¹⁴ Hz falls on it. (h = 6.63 × 10⁻³⁴ J s)
(a) Find the work function in eV. (1)
(b) Find the maximum kinetic energy of the photoelectrons in eV. (1)
(c) Find the stopping potential. (1)
(d) What happens if the intensity of the light is tripled? (1)

Try it first, then show the solutionSolution

(a) $ϕ_{0} = h ν_{0} = 6.63 \times 1 0^{- 34} \times 5 \times 1 0^{14} \approx 3.3 \times 1 0^{- 19}$ J ≈ 2.07 eV. (1)
(b) $K_{ma x} = h (ν - ν_{0}) = 6.63 \times 1 0^{- 34} \times 2.5 \times 1 0^{14} \approx 1.66 \times 1 0^{- 19}$ J ≈ 1.04 eV. (1)
(c) V₀ ≈ 1.04 V. (1)
(d) Three times as many photoelectrons per second (current triples); K_max and V₀ are unchanged. (1)

Answer: (a) about 2.07 eV; (b) about 1.04 eV; (c) about 1.04 V; (d) current triples, V₀ unchanged.

Where students lose marks in Einstein's Photoelectric Equation

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

Mistake: Using wavelength in nm directly in hc/λ with h in J s.
Unit mismatch.
Do this: Either convert λ to m, or use E(eV) ≈ 1240/λ(nm).
Usually costs: The answer mark.
Mistake: Mixing eV and J in Einstein's equation.
Subtracting a work function in eV from a photon energy in J.
Wrong working
K = 3.06 × 10⁻¹⁹ − 2.3
Right working
K = 1.91 eV − 2.3 eV (no emission), or convert both to J
Do this: Put all energies in the same unit.
Usually costs: Most of the numerical.
Mistake: Writing K = hν − φ₀ for all electrons.
Forgetting 'maximum'.
Do this: Only the most loosely bound electrons leave with hν − φ₀; others have less.
Usually costs: ½ mark.
Mistake: Saying photon energy depends on intensity.
Wave-theory thinking.
Do this: E = hν depends only on frequency; intensity sets the number of photons.
Usually costs: 1 mark.

Exam tips

Einstein's equation was set in every board paper of 2023-2026 in our bank, for example 2023 (55/1/1 Q29), 2024 (55/5/1 Q22), 2025 (55/7/1 Q26) and 2026 (55/1/1 Q30).
Learn E(eV) ≈ 1240/λ(nm); it saves time and avoids power-of-ten errors.
When explaining observations, quote the equation first and then argue from it.
Give the slope (h/e) and intercepts of the V₀-ν graph when a graph question appears.
Write units on every line: eV, J, V and Hz.

Practice questions on Einstein's Photoelectric Equation

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
A graph of maximum kinetic energy of photoelectrons against frequency of incident light is a straight line whose slope equals
(a) h/e (b) h (c) e/h (d) φ₀
Hint
K_max = hν − φ₀.
Answer
(b) h.
Q2 · 1 mark · Very short answerBoard level
Define work function. How is it related to the threshold frequency?
Hint
Minimum energy to free an electron.
Answer
Work function is the minimum energy needed to free an electron from the surface of a metal; φ₀ = hν₀.
Q3 · 1 mark · Very short answerHarder board level
The threshold wavelength of a metal is 500 nm. Find its work function in eV.
Hint
1240/λ₀.
Answer
φ₀ ≈ 1240/500 = 2.48 eV.
Q4 · 1 mark · Very short answerHarder board level
Find the stopping potential when light of wavelength 200 nm falls on a metal of work function 4.5 eV.
Hint
Photon energy 6.2 eV.
Answer
E ≈ 1240/200 = 6.2 eV; V₀ = 6.2 − 4.5 = 1.7 V.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Photoelectric emission begins without any noticeable time lag.
Reason (R): The energy of a photon is delivered in a single event to a single 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
No gradual build-up.
Answer
(a).
Q6 · 2 marks · Very short answerHOTS / competency
Light of frequency 2ν₀ and then 3ν₀ falls on a metal of threshold frequency ν₀. Find the ratio of the maximum kinetic energies of the photoelectrons.
Hint
K = h(ν − ν₀).
Answer
K₁ = hν₀, K₂ = 2hν₀; ratio 1 : 2.
Q7 · 2 marks · Very short answerHOTS / competency
Find the momentum of a photon of energy 3.3 × 10⁻¹⁹ J. Why can a photon have momentum though it has no rest mass?
Hint
p = E/c.
Answer
p = E/c = 3.3 × 10⁻¹⁹/3 × 10⁸ = 1.1 × 10⁻²⁷ kg m/s. A photon always moves at c and carries energy; its momentum is p = E/c = h/λ, which does not need rest mass.

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.Einstein's photoelectric equation?Show answer
K_max = hν − φ₀.
2.Energy of a photon?Show answer
E = hν = hc/λ.
3.Momentum of a photon?Show answer
p = h/λ.
4.Threshold frequency in terms of work function?Show answer
ν₀ = φ₀/h.
5.Photon energy in eV for wavelength in nm?Show answer
About 1240/λ.
6.Number of photons per second in a beam of power P?Show answer
N = P/(hν).

Einstein's Photoelectric Equation: frequently asked questions

What is the work function?

The minimum energy needed to remove an electron from the surface of a metal. It is a few electron volts and depends on the metal and the state of its surface.

Why can't two low-energy photons together eject an electron?

In the photoelectric effect a photon is absorbed by one electron in a single event. The chance of one electron absorbing two photons at once is negligible at ordinary intensities, so each photon must have enough energy by itself.

How did Millikan verify Einstein's equation?

He measured stopping potentials for many frequencies and found a straight line V₀ against ν, with slope h/e giving a value of Planck's constant in agreement with Planck's, as Einstein's equation predicts.

Does a photon have mass?

A photon has zero rest mass. It always travels at the speed of light and carries energy hν and momentum h/λ.

What does the photoelectric effect tell us about light?

That light exchanges energy with matter in packets (photons), showing particle nature, while interference and diffraction show its wave nature. Light has a dual nature.