PhysicsISC Class 12 · Dual Nature of Radiation and Matter

Photoelectric Effect and Einstein's Photoelectric Equation

Q: Why don't all photoelectrons have the same energy?

Electrons below the surface lose some energy in collisions before escaping; only those at the surface leave with the maximum, h\nu - W_0.

Q: Can very bright red light eject electrons from zinc?

No. Each red photon carries less energy than zinc's work function, and an electron cannot add up energy from several photons in the ordinary photoelectric effect.

Q: Does a photon have mass?

It has zero rest mass but carries energy hν and momentum \frac{h}{\lambda}.

Q: What does the ISC 2027 syllabus need here?

Hertz and Lenard's observations, quantisation of radiation, Einstein's equation E_{max} = h\nu - W_0, threshold frequency, work function, why only the particle nature of light explains the effect, and photon momentum p = \frac{h}{\lambda}.

Q: How was this asked recently?

ISC 2025 Q8 (2 marks) asked definitions of threshold frequency and work function; ISC 2024 Q16(ii) (2 marks) compared metals of work functions 8, 2 and 4 eV under UV light.

Updated 15 October 2026 · Fact-checked

Einstein's photoelectric equation is $E_{max} = h\nu - W_0$: a photon of energy hν gives all its energy to one electron, which uses at least the work function $W_0$ to escape, leaving at most $h\nu - W_0$ as kinetic energy. Emission occurs only if ν is at least the threshold frequency $\nu_0 = \frac{W_0}{h}$.

One photon gives all its energy hν to one electron: W₀ pays for escape and the rest is Kmax, so emission needs ν at least ν₀.

Illustration drawn for QuestionBuddy.

Understand Photoelectric Effect and Einstein's Equation

The photoelectric effect. When light of high enough frequency falls on a clean metal surface, electrons (photoelectrons) are emitted. Hertz noticed that ultraviolet light helped sparks jump a gap; Hallwachs and Lenard showed that the light ejects negative charges, later identified as electrons.

Experimental facts (Lenard and others).
1. For each metal there is a threshold frequency $ν_{0}$ . Below it no electrons are emitted, however intense the light.
2. Above $ν_{0}$ , the maximum kinetic energy of photoelectrons increases linearly with frequency and is independent of intensity.
3. The number of photoelectrons per second (the photocurrent) is proportional to the intensity.
4. Emission is practically instantaneous (within about $1 0^{- 9}$ s), even for very faint light.

Why wave theory fails. In the wave picture energy is spread continuously over the wavefront, so (a) bright light of any frequency should eventually eject electrons, (b) brighter light should give faster electrons, and (c) faint light should need a measurable time for an electron to collect enough energy. All three contradict the facts.

Einstein's explanation. Extending Planck's idea, Einstein proposed that light consists of photons, each of energy
$E = h ν = \frac{h c}{λ},$
with $h = 6.63 \times 1 0^{- 34}$ J s. A photon is absorbed whole by a single electron. Part of its energy, at least the work function $W_{0}$ (minimum energy to remove an electron from the surface), is used to escape; the rest appears as kinetic energy. The fastest electrons (from the surface) have
$E_{ma x} = \frac{1}{2} m v_{ma x}^{2} = h ν - W_{0} .$
This explains every fact:

Emission needs $h ν \geq W_{0}$ , i.e. $ν \geq ν_{0} = \frac{W _{0}}{h}$ (threshold frequency) or $λ \leq λ_{0} = \frac{h c}{W _{0}}$ (threshold wavelength).
$E_{ma x}$ depends on ν, not on intensity; more intense light means more photons, so more electrons but not faster ones.
One photon acts at once, so there is no time lag.

Useful numbers. $h c \approx 1240$ eV nm, so $E (eV) = \frac{1240}{λ ( nm )}$ . 1 eV = $1.6 \times 1 0^{- 19}$ J.

Photon momentum and number. A photon has zero rest mass but carries momentum $p = \frac{E}{c} = \frac{h ν}{c} = \frac{h}{λ}$ . A beam of power P and photon energy E delivers $N = \frac{P}{E}$ photons per second.

Wave-particle duality. Interference and diffraction show light's wave nature; the photoelectric effect can be explained only by its particle (quantum) nature.

Photoelectric Effect and Einstein's Equation: formulas and key facts

Photon energy: $E = h ν = \frac{h c}{λ}$; hc ≈ 1240 eV nm.
Einstein's equation: $E_{ma x} = \frac{1}{2} m v_{ma x}^{2} = h ν - W_{0}$; Energy conservation for one photon.
Threshold frequency: $ν_{0} = \frac{W _{0}}{h}$; Minimum frequency for emission.
Threshold wavelength: $λ_{0} = \frac{h c}{W _{0}}$; Maximum wavelength.
Alternative form: $E_{ma x} = h (ν - ν_{0})$; Using ν₀.
Photon momentum: $p = \frac{h}{λ} = \frac{E}{c}$; kg m/s.
Photons per second: $N = \frac{P}{h ν}$; P in watts.

How to solve Photoelectric Effect and Einstein's Equation questions

Work in electronvolts whenever the work function is given in eV.

1Photon energy: $E = \frac{1240}{λ ( nm )}$ eV, or $h ν$ in joules ÷ $1.6 \times 1 0^{- 19}$ .
2Compare E with $W_{0}$ : if $E < W_{0}$ , no emission, whatever the intensity.
3If emission occurs: $E_{ma x} = E - W_{0}$ (eV); convert to joules for speed: $v = \frac{2 E _{ma x}}{m}$ .
4Threshold: $ν_{0} = \frac{W _{0}}{h}$ (with $W_{0}$ in J), $λ_{0} = \frac{1240}{W _{0}}$ nm.
5Photocurrent: photons per second $\frac{P}{E}$ × fraction that eject electrons × e.
6For two wavelengths on one metal, write two equations and subtract to eliminate $W_{0}$ .

Photoelectric Effect and Einstein's 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.

Automatic street lights

Suppose a street-light switch uses a photocell whose emitter has work function 2.0 eV, and daylight contains plenty of green light of 550 nm.

Will daylight trigger the cell, and why does the lamp switch on at dusk?

Try it first, then show the solutionSolution

$E = \frac{1240}{550} \approx 2.25$ eV > 2.0 eV, so photoelectrons are emitted and a current flows by day.
At dusk fewer photons arrive, the photocurrent drops, and the circuit switches the lamp on.

Answer: Yes; the current depends on how many photons arrive, so it falls at dusk.

Memory hook: Frequency decides "whether"; intensity decides "how many".

Spacecraft charging in sunlight

Imagine an aluminium spacecraft (work function about 4.2 eV) bathed in solar ultraviolet of 150 nm.

Find the maximum kinetic energy of the photoelectrons, and explain why the craft becomes positively charged.

Try it first, then show the solutionSolution

$E = \frac{1240}{150} \approx 8.27$ eV; $E_{ma x} = 8.27 - 4.2 \approx 4.07$ eV.
Electrons leave the surface, so the craft loses negative charge and becomes positive.
Once its potential reaches about +4 V, even the fastest electrons are pulled back and the charging stops.

Answer: ≈ 4.1 eV; the craft charges up to roughly +4 V.

Memory hook: Lose electrons, gain plus.

Tricks to remember Photoelectric Effect and Einstein's Equation

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

Twelve-forty
$E (eV) = \frac{1240}{λ ( nm )}$ .
Use it when: Fast photon energies.
Pay to leave, keep the change
Photon energy = exit fee ( $W_{0}$ ) + change ( $E_{ma x}$ ).
Use it when: Einstein's equation.
Colour for speed, brightness for number
ν controls $E_{ma x}$ ; intensity controls current.
Use it when: Graph and theory questions.
Threshold: minimum ν, maximum λ
$ν_{0}$ is the least frequency; $λ_{0}$ the greatest wavelength.
Use it when: Definitions.

Photoelectric Effect and Einstein's Equation: solved examples

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

Example 1: Condition for emission · 1 mark · MCQBoard level

Photoelectrons are emitted from a metal of work function $W_{0}$ by light of frequency ν only if:
(a) $h ν < W_{0}$
(b) $h ν \geq W_{0}$
(c) the light is very intense
(d) the light falls for a long time

Try it first, then show the solutionSolution

One photon must supply at least $W_{0}$ to one electron.

Answer: (b).

Example 2: A UV photon · 2 marks · Short answerBoard level

Find the energy (in eV) and the momentum of a photon of ultraviolet light of wavelength 310 nm.

Try it first, then show the solutionSolution

$E = \frac{1240}{310} = 4.0$ eV. (1)
$p = \frac{h}{λ} = \frac{6.63 \times 1 0 ^{- 34}}{310 \times 1 0 ^{- 9}} \approx 2.14 \times 1 0^{- 27}$ kg m/s. (1)

Answer: 4.0 eV; ≈ $2.14 \times 1 0^{- 27}$ kg m/s.

Example 3: Einstein's explanation · 3 marks · Long answerHarder board level

Write Einstein's photoelectric equation. Use it to explain (a) the existence of a threshold frequency, (b) why the maximum kinetic energy does not depend on intensity, and (c) why there is no time lag.

Try it first, then show the solutionSolution

$E_{ma x} = h ν - W_{0}$ . (a) $E_{ma x} \geq 0$ requires $h ν \geq W_{0}$ , i.e. $ν \geq ν_{0} = \frac{W _{0}}{h}$ . (1)
(b) Each electron absorbs one photon of energy hν; intensity changes only the number of photons, not hν, so $E_{ma x}$ is unchanged. (1)
(c) Energy arrives in a single packet, absorbed at once by one electron, so emission is immediate. (1)

Answer: $E_{ma x} = h ν - W_{0}$ , with the three explanations above.

Example 4: Speed of the fastest photoelectrons · 3 marks · Short answerHarder board level

Light of wavelength 248 nm falls on sodium (work function 2.3 eV). Find the maximum kinetic energy and speed of the photoelectrons, and the threshold wavelength of sodium. ( $m_{e} = 9.1 \times 1 0^{- 31}$ kg)

Try it first, then show the solutionSolution

$E = \frac{1240}{248} = 5.0$ eV; $E_{ma x} = 5.0 - 2.3 = 2.7$ eV. (1)
$v = \frac{2 \times 2.7 \times 1.6 \times 1 0 ^{- 19}}{9.1 \times 1 0 ^{- 31}} \approx 9.74 \times 1 0^{5}$ m/s. (1)
$λ_{0} = \frac{1240}{2.3} \approx 539$ nm. (1)

Answer: 2.7 eV; ≈ $9.7 \times 1 0^{5}$ m/s; ≈ 539 nm.

Example 5: From beam power to photocurrent · 3 marks · Short answerHOTS / competency

A 1.5 mW beam of 248 nm light falls on a photocathode, and 2% of the photons eject electrons. Find the number of photons per second and the saturation photocurrent.

Try it first, then show the solutionSolution

Photon energy $= 5.0$ eV $= 8.0 \times 1 0^{- 19}$ J. (1)
$N = \frac{1.5 \times 1 0 ^{- 3}}{8.0 \times 1 0 ^{- 19}} \approx 1.88 \times 1 0^{15}$ photons per second. (1)
Electrons per second $= 0.02 N = 3.75 \times 1 0^{13}$ ; $I = 3.75 \times 1 0^{13} \times 1.6 \times 1 0^{- 19} = 6.0$ µA. (1)

Answer: ≈ $1.88 \times 1 0^{15}$ s⁻¹; 6.0 µA.

Example 6: Three metals, one lamp · 4 marks · Case-basedHOTS / competency

Suppose UV light of 220 nm falls on caesium (W = 2.1 eV), zinc (W = 4.3 eV) and platinum (W = 5.6 eV).
(i) Find the photon energy. (1 mark)
(ii) Find $E_{ma x}$ for each metal. (1 mark)
(iii) Which metal has the lowest threshold frequency? Find it. (1 mark)
(iv) What changes if the lamp's intensity is doubled? (1 mark)

Try it first, then show the solutionSolution

(i) $E = \frac{1240}{220} \approx 5.64$ eV.
(ii) Cs: ≈ 3.54 eV; Zn: ≈ 1.34 eV; Pt: only a few hundredths of an eV (it is just above threshold).
(iii) Caesium: $ν_{0} = \frac{2.1 \times 1.6 \times 1 0 ^{- 19}}{6.63 \times 1 0 ^{- 34}} \approx 5.07 \times 1 0^{14}$ Hz.
(iv) Twice as many photons, so about twice the photocurrent; $E_{ma x}$ for each metal is unchanged.

Answer: (i) ≈ 5.64 eV; (ii) ≈ 3.54, 1.34 eV, almost zero; (iii) Cs, ≈ $5.07 \times 1 0^{14}$ Hz; (iv) more electrons, same $E_{ma x}$ .

Where students lose marks in Photoelectric Effect and Einstein's Equation

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

Mistake: Missing "minimum" in definitions
The ISC 2025 examiners (Q8) saw definitions of threshold frequency and work function without the key word minimum, or with "maximum".
Do this: Threshold frequency: minimum frequency for emission. Work function: minimum energy to remove an electron from the surface.
Usually costs: 1 mark each.
Mistake: Writing "photons are emitted"
Also 2025 Q8: candidates wrote photons where electrons are emitted.
Do this: Photons are absorbed; electrons are emitted.
Usually costs: 1 mark.
Mistake: Linking intensity with stopping potential or $E_{ma x}$
In 2024 Q16(i) candidates said the stopping potential changes with intensity.
Do this: Intensity changes only the number of electrons.
Usually costs: 1 mark.
Mistake: Answering without a reason
In 2024 Q16(ii) the right metal was named without using $E_{ma x} = h ν - W$ .
Do this: Quote the equation: largest $W$ gives smallest $E_{ma x}$ ; smallest W gives lowest $ν_{0}$ .
Usually costs: 1 mark.

Exam tips

Use hc = 1240 eV nm only when the work function is in eV; otherwise work in joules.
Write units of energy clearly (eV or J); do not mix them in one line.
In theory questions, contrast each fact with what wave theory predicts.
Give the word "photoelectrons" and "photon" correctly.

Practice questions on Photoelectric Effect and Einstein's 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 metal has a work function of 3.1 eV. Its threshold wavelength is about:
(a) 200 nm
(b) 400 nm
(c) 600 nm
(d) 800 nm
Hint
$\frac{1240}{W _{0}}$ .
Answer
(b).
Q2 · 2 marks · Short answerBoard level
Write the relation between work function and threshold frequency, and find the threshold frequency of a metal with $W_{0}$ = 4.14 eV.
Hint
$W_{0} = h ν_{0}$ .
Answer
$W_{0} = h ν_{0}$ ; $ν_{0} \approx 1.0 \times 1 0^{15}$ Hz.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Doubling the intensity of light of fixed frequency does not change the maximum kinetic energy of photoelectrons.
Reason (R): Intensity changes the number of photons, not the energy of each photon.
(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
One photon per electron.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
Light of 200 nm falls on aluminium (work function 4.2 eV). Find the maximum kinetic energy of the photoelectrons in eV and J.
Hint
1240/200.
Answer
2.0 eV = $3.2 \times 1 0^{- 19}$ J.
Q5 · 3 marks · Short answerHarder board level
State the four experimental facts of the photoelectric effect and say which one shows most clearly that light energy is quantised.
Hint
Threshold, $E_{ma x}$ , current, time.
Answer
Threshold frequency exists; $E_{ma x}$ rises with ν, not intensity; current ∝ intensity; no time lag. The threshold (and no time lag) shows energy arrives in packets.
Q6 · 3 marks · Short answerHOTS / competency
When light of wavelength λ falls on a metal, $E_{ma x}$ = 1.0 eV; with λ/2 it is 4.0 eV. Find the work function and λ.
Hint
Let photon energy at λ be E: E − W = 1, 2E − W = 4.
Answer
E = 3.0 eV, $W_{0}$ = 2.0 eV; λ = $\frac{1240}{3}$ ≈ 413 nm.
Q7 · 2 marks · Short answerHOTS / competency
A 2 W lamp emits light of 600 nm. How many photons does it emit per second?
Hint
Photon energy ≈ $3.3 \times 1 0^{- 19}$ J.
Answer
≈ $6.0 \times 1 0^{18}$ per second.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a burglar alarm uses an infrared LED of 940 nm aimed at a photocell whose emitter has work function 1.9 eV.
(i) Find the photon energy of the LED. (1 mark)
(ii) Will the alarm's photocell respond? (1 mark)
(iii) What is the longest wavelength that would work? (1 mark)
(iv) Would a brighter 940 nm LED help? Explain. (1 mark)
Hint
1240/940.
Answer
(i) ≈ 1.32 eV. (ii) No, 1.32 eV < 1.9 eV. (iii) ≈ 653 nm. (iv) No: more photons, but each still has too little energy.

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
$E_{ma x} = h ν - W_{0}$ .
2.Threshold frequency?Show answer
The minimum frequency of incident light that can eject electrons from a metal.
3.Photon energy at 1240 nm?Show answer
1 eV.
4.Effect of intensity on photocurrent?Show answer
Photocurrent is proportional to intensity.
5.Momentum of a photon?Show answer
$p = \frac{h}{λ}$ .

Photoelectric Effect and Einstein's Equation: frequently asked questions

Why don't all photoelectrons have the same energy?

Electrons below the surface lose some energy in collisions before escaping; only those at the surface leave with the maximum, h\nu - W_0.

Can very bright red light eject electrons from zinc?

No. Each red photon carries less energy than zinc's work function, and an electron cannot add up energy from several photons in the ordinary photoelectric effect.

Does a photon have mass?

It has zero rest mass but carries energy hν and momentum \frac{h}{\lambda}.

What does the ISC 2027 syllabus need here?

Hertz and Lenard's observations, quantisation of radiation, Einstein's equation E_{max} = h\nu - W_0, threshold frequency, work function, why only the particle nature of light explains the effect, and photon momentum p = \frac{h}{\lambda}.

How was this asked recently?

ISC 2025 Q8 (2 marks) asked definitions of threshold frequency and work function; ISC 2024 Q16(ii) (2 marks) compared metals of work functions 8, 2 and 4 eV under UV light.