Einstein's Photoelectric Equation

Key idea: Apply Einstein’s photoelectric equation Kmax = hf − Φ and the stopping potential relation Kmax = eVs, including graph interpretation (A Level Physics).

GCE A-Level H2 Physics 2027

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Learning objectives

Use photon energy and momentum and analyse the photoelectric effect.
Apply de Broglie wavelength and wave-particle evidence.
Interpret wavefunctions, probability density and superposition.
Apply uncertainty and infinite-square-well energy quantisation.
Analyse atomic energy levels and emission or absorption spectra.

The current 9478 outcome names threshold frequency and photon energy E = hf, but does not explicitly name stopping potential or Einstein’s photoelectric equation. Use this page as supporting method work where your school or an older paper includes it.

1. Definitions (Must Know)

A. Work function, Φ

The work function, Φ, is the minimum energy needed to remove an electron from the metal surface.

B. Einstein’s photoelectric equation

For the most energetic photoelectrons:

Kₘₐₓ = hf-Φ

C. Stopping potential, Vₛ

The stopping potential, Vₛ, is the minimum retarding potential difference needed to reduce the photoelectric current to zero.

It links to maximum kinetic energy:

Kₘₐₓ = eVₛ

2. Key Ideas (What Earns Marks)

Threshold frequency: Kₘₐₓ = 0 ⇒ hf₀ = Φ ⇒ f₀ = Φ/h
Combine with stopping potential: eVₛ = hf-Φ ⇒ Vₛ = (h/e)f-Φ/e
Vₛ vs f is a straight line:
- gradient: h/e
- intercept: -Φ/e

hf and Φ are energies (J). eVₛ is also energy (C·V = J).

3. Detailed Explanations

Monochromatic light ejects electrons from an emitter in an evacuated tube. The electrons travel towards a collector connected to a microammeter. An adjustable retarding supply makes the emitter positive and collector negative; the stopping potential is reached when the photocurrent becomes zero. — Reverse the electrode polarity to oppose the photoelectrons, then increase the retarding potential until even the fastest electrons cannot reach the collector.

How Vₛ is found (I–V curve)

The stopping potential is found by increasing a retarding potential until the photocurrent just falls to zero.

Qualitative photoelectric current–voltage curves showing saturation current, stopping potential, and how intensity vs frequency affects the graph. — Intensity changes the saturation current (how many electrons per second). Frequency changes the stopping potential (how energetic the fastest electrons are).

Values and uncertainty for Photoelectric I–V curves (qualitative)
Series	Anode potential, V (V)	Photocurrent (scaled)
Same f, higher intensity	-2.5	0
Same f, higher intensity	-2	0
Same f, higher intensity	-1.2	0
Same f, higher intensity	-0.8	0.25
Same f, higher intensity	-0.4	0.62
Same f, higher intensity	0	0.9
Same f, higher intensity	0.4	1
Same f, higher intensity	0.8	1
Same f, higher intensity	1.2	1
Same f, lower intensity	-2.5	0
Same f, lower intensity	-2	0
Same f, lower intensity	-1.2	0
Same f, lower intensity	-0.8	0.12
Same f, lower intensity	-0.4	0.31
Same f, lower intensity	0	0.45
Same f, lower intensity	0.4	0.5
Same f, lower intensity	0.8	0.5
Same f, lower intensity	1.2	0.5
Higher f (more energetic): larger Vs	-2.5	0
Higher f (more energetic): larger Vs	-1.8	0
Higher f (more energetic): larger Vs	-1.4	0.18
Higher f (more energetic): larger Vs	-1	0.36
Higher f (more energetic): larger Vs	-0.6	0.46
Higher f (more energetic): larger Vs	0	0.5
Higher f (more energetic): larger Vs	0.6	0.5
Higher f (more energetic): larger Vs	1.2	0.5

Values and uncertainty for Photoelectric I–V curves (qualitative)
Series	Anode potential, V (V)	Photocurrent (scaled)
Same f, higher intensity	-2.5	0
Same f, higher intensity	-2	0
Same f, higher intensity	-1.2	0
Same f, higher intensity	-0.8	0.25
Same f, higher intensity	-0.4	0.62
Same f, higher intensity	0	0.9
Same f, higher intensity	0.4	1
Same f, higher intensity	0.8	1
Same f, higher intensity	1.2	1
Same f, lower intensity	-2.5	0
Same f, lower intensity	-2	0
Same f, lower intensity	-1.2	0
Same f, lower intensity	-0.8	0.12
Same f, lower intensity	-0.4	0.31
Same f, lower intensity	0	0.45
Same f, lower intensity	0.4	0.5
Same f, lower intensity	0.8	0.5
Same f, lower intensity	1.2	0.5
Higher f (more energetic): larger Vs	-2.5	0
Higher f (more energetic): larger Vs	-1.8	0
Higher f (more energetic): larger Vs	-1.4	0.18
Higher f (more energetic): larger Vs	-1	0.36
Higher f (more energetic): larger Vs	-0.6	0.46
Higher f (more energetic): larger Vs	0	0.5
Higher f (more energetic): larger Vs	0.6	0.5
Higher f (more energetic): larger Vs	1.2	0.5

A. Why only the maximum matters

Electrons come from different depths in the metal and can lose energy in collisions before escaping.

So emitted electrons have a range of kinetic energies. The equation uses Kₘₐₓ for the most energetic electrons (often those that escape with minimal loss).

B. From Kₘₐₓ to stopping potential

Stopping potential is defined so that even the most energetic electrons are just stopped:

Kₘₐₓ = eVₛ

Combine with Einstein’s equation:

eVₛ = hf-Φ

4. Common Mistakes

Treating Φ as a “force” (it is an energy).
Using Vₛ when the question is about a different potential in the circuit.
Using f₀ formula but forgetting to use the same units (Hz, J, J·s).

5. Exam Tips

If asked for a “threshold wavelength”, use f = c/λ: f₀ = Φ/h ⇒ λ₀ = c/f₀ = hc/Φ
If given a Vₛ vs f graph, read:
- gradient → h/e,
- x-intercept → f₀.

6. Worked Examples

Problem

Light of frequency

8.0 × 10¹⁴ Hz illuminates a metal with work function

2.4 × 10⁻¹⁹ J. Using

h = 6.63 × 10⁻³⁴ J s, find

Kₘₐₓ.

Study the worked solution

Check the energy balance
Method
One photon supplies energy hf; the work function is the minimum removal energy.
Reason
Only the remaining energy can become maximum electron kinetic energy.
Working
Kₘₐₓ = hf-Φ
Calculate photon energy
Method
Multiply Planck’s constant by frequency.
Reason
J s multiplied by s⁻¹ gives joules.
Working
hf = (6.63 × 10⁻³⁴)(8.0 × 10¹⁴) = 5.304 × 10⁻¹⁹ J
Subtract the work function
Method
Use unrounded energy before the final rounding.
Reason
The emitted electron retains the surplus.
Working
Kₘₐₓ = 5.304 × 10⁻¹⁹-2.4 × 10⁻¹⁹ = 2.9 × 10⁻¹⁹ J

Problem

Using the preceding unrounded

Kₘₐₓ result and

e = 1.60 × 10⁻¹⁹ C, find the stopping potential.

Try this before viewing the solution

Relation, substitution and value

Hints

Hint 1: match energy to electric work

At the stopping condition,

Kₘₐₓ = eVₛ.

View solution step by step

Select relation
Method
Set maximum kinetic energy equal to electric stopping work.
Reason
The fastest photoelectrons are just prevented from reaching the collector.
Working
Kₘₐₓ = eVₛ
Calculate
Method
Divide by the elementary charge.
Reason
One joule per coulomb is one volt.
Working
Vₛ = (2.9 × 10⁻¹⁹)/(1.60 × 10⁻¹⁹) = 1.8 V

Learner claim

A metal has

Φ = 3.0 × 10⁻¹⁹ J. A learner says threshold frequency depends on light intensity because stronger light supplies more energy. Diagnose the claim and calculate

f₀.

Try this before viewing the solution

Condition at threshold

Kmax = 0 and hf0 = PhiIntensity must be highPhi = 0

View solution step by step

State threshold condition
Method
At threshold, the most energetic emitted electron has zero kinetic energy.
Reason
One photon just supplies the removal energy.
Working
0 = hf₀-Φ
Separate intensity
Method
Intensity changes photon arrival rate, not energy per photon.
Reason
Photon energy is hf, controlled by frequency.
Working
E_γ = hf
Calculate
Method
Divide the work function by h.
Reason
The quotient has units s⁻¹.
Working
f₀ = (3.0 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 4.5 × 10¹⁴ Hz

Examination question

For one metal,

Vₛ = 0.40 V at

5.0 × 10¹⁴ Hz and

Vₛ = 1.20 V at

7.0 × 10¹⁴ Hz. From

Vₛ = (h/e)f-Φ/e, find

h and

Φ. Use

e = 1.60 × 10⁻¹⁹ C. [5 marks]

Try this before viewing the solution

Five-mark graph-data calculation

View solution step by step

Find gradient
1 mark
Method
Use changes in stopping potential and frequency.
Reason
The two points define the line’s gradient.
Working
m = (1.20-0.40)/((7.0-5.0) × 10¹⁴) = 4.0 × 10⁻¹⁵ V s
Interpret gradient
1 mark
Method
Set m = h/e.
Reason
This follows from the coefficient of f.
Working
h = me
Calculate h
1 mark
Method
Multiply by the elementary charge.
Reason
V s times C gives J s.
Working
h = (4.0 × 10⁻¹⁵)(1.60 × 10⁻¹⁹) = 6.4 × 10⁻³⁴ J s
Find work function
2 marks
Method
Substitute either point into Φ = hf-eVₛ.
Reason
The intercept energy is the photon energy not retained as kinetic energy.
Working
Φ = (6.4 × 10⁻³⁴)(5.0 × 10¹⁴)-(1.60 × 10⁻¹⁹)(0.40) = 2.56 × 10⁻¹⁹ J

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Self-mark gradient, interpretation, Planck constant and the two-part work-function method.

Mark points

1 mark: Calculates the stopping-potential gradient.
1 mark: Identifies gradient as h over e.
1 mark: Obtains 6.4 times 10 to the minus 34 joule seconds.
2 marks: Selects the energy relation and obtains 2.56 times 10 to the minus 19 joules.

Independent transfer

A metal has work function

2.5 eV. Using

hc = 1240 eV·nm, find its threshold wavelength and state what happens for longer-wavelength light.

Try this before viewing the solution

Wavelength calculation and boundary interpretation

Hints

Hint 1: use the threshold equality

Orient with

hc/λ₀ = Φ and remember that longer wavelength means lower photon energy.

View solution step by step

Rearrange
Method
Use the photon-energy form at threshold.
Reason
E = hc/λ equals the work function when Kₘₐₓ = 0.
Working
λ₀ = hc/Φ
Calculate
Method
Keep eV and nm units paired.
Reason
The supplied hc avoids an unnecessary joule conversion.
Working
λ₀ = 1240/2.5 = 496 nm ≈ 5.0 × 10² nm
Interpret boundary
Method
Longer wavelengths do not eject electrons in this model.
Reason
Their lower-frequency photons have E < Φ.
Working
λ > λ₀ ⇒ f < f₀

Next step

Now explain independently why stopping potential plotted against frequency is linear.

7. Mind Stretchers

Mind stretcher 1: Explain the straight-line graphExtension

Explain why a plot of Vₛ against f is a straight line.

Show Answer

From eVₛ = hf-Φ: Vₛ = (h/e)f-Φ/e

This is of the form y = mx + c, so the graph is a straight line with constant gradient h/e.

Mind stretcher 2: Meaning of the interceptsExtension

In Vₛ = (h/e)f-Φ/e, what do the x-intercept and y-intercept represent physically?

Show Answer

The x-intercept occurs when Vₛ = 0, so f = f₀ = Φ/h (the threshold frequency).

The y-intercept is -Φ/e, a negative voltage value; it encodes the work function (how much energy per electron must be supplied before any kinetic energy remains).

8. Optional (Enrichment)

A. Threshold wavelength

Sometimes the threshold is quoted as a wavelength λ₀: λ₀ = hc/Φ

Use this only if the question explicitly gives wavelength instead of frequency.

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H2 Physics
Edition: GCE A-Level H2 Physics 2027

Einstein's Photoelectric Equation

Continue where you stopped

Learning objectives

1. Definitions (Must Know)

A. Work function, Φ

B. Einstein’s photoelectric equation

C. Stopping potential, Vₛ

2. Key Ideas (What Earns Marks)

3. Detailed Explanations

How Vₛ is found (I–V curve)

A. Why only the maximum matters

B. From Kₘₐₓ to stopping potential

4. Common Mistakes

5. Exam Tips

6. Worked Examples

Find Kₘₐₓ

Check the energy balance

Calculate photon energy

Subtract the work function

Find stopping potential

Select relation

Calculate

Find threshold frequency

State threshold condition

Separate intensity

Calculate

Find h and Φ from two data points

Find gradient

Interpret gradient

Calculate h

Find work function

Threshold wavelength from work function (eV form)

Rearrange

Calculate

Interpret boundary

7. Mind Stretchers

Mind stretcher 1: Explain the straight-line graphExtension

Mind stretcher 2: Meaning of the interceptsExtension

8. Optional (Enrichment)

A. Threshold wavelength