Photoelectric Effect Calculations: Work Function and Kinetic Energy

Einstein’s explanation of the photoelectric effect comes down to one equation. With it, you can find the minimum frequency of light that ejects electrons from a metal, how fast the fastest electrons travel, and even the value of Planck’s constant from a graph. This article sets out the equation and works through the standard types of calculation, in both joules and electronvolts.

The photoelectric equation

hf = φ + KE_max

• hf = energy of one incoming photon (h = 6.626 × 10⁻³⁴ J s)
• φ = work function of the metal: the minimum energy needed to free an electron from the surface
• KE_max = maximum kinetic energy of an emitted electron (some electrons lose extra energy on the way out, so this is the maximum)

Related equations:

• Threshold frequency: f₀ = φ ÷ h (the photon energy exactly equals φ, so KE_max = 0)
• Threshold wavelength: λ₀ = hc ÷ φ (the longest wavelength that works)
• Kinetic energy and speed: KE = ½mv², with electron mass m = 9.109 × 10⁻³¹ kg
• Stopping potential: eV_s = KE_max (the voltage needed to stop the fastest electrons), with e = 1.602 × 10⁻¹⁹ C
• Units: 1 eV = 1.602 × 10⁻¹⁹ J; hc = 1.988 × 10⁻²⁵ J m; E (eV) ≈ 1240 ÷ λ (nm)

Worked example 1: threshold frequency and wavelength

The work function of sodium is 2.28 eV. Find the threshold frequency and threshold wavelength.

Convert φ to joules: 2.28 × 1.602 × 10⁻¹⁹ = 3.65 × 10⁻¹⁹ J

f₀ = φ ÷ h = 3.65 × 10⁻¹⁹ ÷ 6.626 × 10⁻³⁴ = 5.51 × 10¹⁴ Hz

λ₀ = c ÷ f₀ = 3.00 × 10⁸ ÷ 5.51 × 10¹⁴ = 5.44 × 10⁻⁷ m = 544 nm (green)

Light with a wavelength longer than 544 nm (yellow, orange, red) won’t eject electrons from sodium, however bright it is.

Worked example 2: maximum kinetic energy

Ultraviolet light of wavelength 250 nm falls on sodium (φ = 3.65 × 10⁻¹⁹ J). Calculate KE_max.

Photon energy: E = hc ÷ λ = 1.988 × 10⁻²⁵ ÷ 2.50 × 10⁻⁷ = 7.95 × 10⁻¹⁹ J

KE_max = E − φ = 7.95 × 10⁻¹⁹ − 3.65 × 10⁻¹⁹ = 4.30 × 10⁻¹⁹ J (2.68 eV)

Worked example 3: maximum speed

Using the answer to example 2, find the maximum speed of the photoelectrons.

KE = ½mv² → v = √(2KE ÷ m)

v = √(2 × 4.30 × 10⁻¹⁹ ÷ 9.109 × 10⁻³¹) = √(9.44 × 10¹¹) = 9.72 × 10⁵ m/s

That’s nearly a million metres per second, about 0.3% of the speed of light.

Worked example 4: working in electronvolts

Electronvolts make many calculations quicker.

Light of wavelength 400 nm falls on a metal with work function 2.10 eV. Find KE_max in eV.

Photon energy: E ≈ 1240 ÷ 400 = 3.10 eV

KE_max = 3.10 − 2.10 = 1.00 eV

Worked example 5: stopping potential

In an experiment, the photocurrent stops when the collector is held at −1.00 V relative to the metal. What was KE_max?

eV_s = KE_max → KE_max = 1.602 × 10⁻¹⁹ × 1.00 = 1.60 × 10⁻¹⁹ J (1.00 eV)

The stopping potential in volts equals KE_max in electronvolts, a very convenient result.

Worked example 6: finding the work function

Light of frequency 1.20 × 10¹⁵ Hz ejects electrons with KE_max = 3.20 × 10⁻¹⁹ J. Find the work function and identify a likely metal from the table below.

Photon energy: hf = 6.626 × 10⁻³⁴ × 1.20 × 10¹⁵ = 7.95 × 10⁻¹⁹ J

φ = hf − KE_max = 7.95 × 10⁻¹⁹ − 3.20 × 10⁻¹⁹ = 4.75 × 10⁻¹⁹ J

In eV: 4.75 × 10⁻¹⁹ ÷ 1.602 × 10⁻¹⁹ = 2.97 eV

Metal	φ (eV, approx.)
Caesium	2.1
Sodium	2.3
Calcium	2.9
Magnesium	3.7
Zinc	4.3

The closest match is calcium.

Worked example 7: no emission

Red light of wavelength 700 nm falls on zinc (φ = 4.3 eV). Are electrons emitted?

Photon energy: 1240 ÷ 700 = 1.77 eV

1.77 eV < 4.3 eV, so no electrons are emitted. Increasing the intensity only increases the number of photons, each of which still has only 1.77 eV. This is the key observation that classical wave theory couldn’t explain.

Worked example 8: the KE_max vs frequency graph

Rearranging the photoelectric equation:

KE_max = hf − φ

This has the form y = mx + c, so a graph of KE_max (y-axis) against frequency f (x-axis) is a straight line with:

• gradient = h (Planck’s constant)
• x-intercept = f₀ (threshold frequency)
• y-intercept = −φ (extrapolating back to f = 0)

A student obtains these results for a metal:

f (× 10¹⁴ Hz)	7.0	8.0	9.0	10.0
KE_max (× 10⁻¹⁹ J)	1.12	1.78	2.44	3.10

Gradient = (3.10 − 1.12) × 10⁻¹⁹ ÷ (10.0 − 7.0) × 10¹⁴ = 1.98 × 10⁻¹⁹ ÷ 3.0 × 10¹⁴ = 6.6 × 10⁻³⁴ J s (Planck’s constant ✓)

x-intercept: set KE_max = 0. Using the point (7.0 × 10¹⁴, 1.12 × 10⁻¹⁹): f₀ = 7.0 × 10¹⁴ − (1.12 × 10⁻¹⁹ ÷ 6.6 × 10⁻³⁴) = 7.0 × 10¹⁴ − 1.70 × 10¹⁴ = 5.3 × 10¹⁴ Hz

φ = h × f₀ = 6.6 × 10⁻³⁴ × 5.3 × 10¹⁴ = 3.5 × 10⁻¹⁹ J (about 2.2 eV, close to sodium or potassium)

Every metal gives a line with the same gradient (h), but a different intercept (its own φ). That universal gradient was Millikan’s strongest evidence for Einstein’s theory.

Worked example 9: intensity and current

A metal is illuminated with light above its threshold frequency. The intensity is doubled while the frequency stays the same. What happens to (a) KE_max, (b) the number of electrons emitted per second, (c) the photocurrent?

• (a) KE_max is unchanged: each photon has the same energy.
• (b) The number of electrons per second doubles: twice as many photons arrive.
• (c) The photocurrent (charge per second) doubles.

A step-by-step routine

1. List what you know and convert everything to consistent units (all J or all eV; wavelengths in metres if using joules).
2. Find the photon energy from the frequency (E = hf) or wavelength (E = hc/λ).
3. Compare the photon energy with the work function. If it’s smaller, stop: no electrons are emitted.
4. Subtract to find KE_max = E − φ.
5. Convert as needed: to speed with v = √(2KE/m), to stopping potential with V_s = KE_max/e, or to electronvolts.
6. Sense-check: photoelectron speeds are typically 10⁵ to 10⁶ m/s, and work functions are usually between 2 and 6 eV.

Following the same routine every time makes these calculations quick and reliable, even in unfamiliar contexts.

Common mistakes

1. Mixing J and eV in the same subtraction. Convert everything to the same unit first.
2. Forgetting nm → m when using E = hc/λ in joules.
3. Thinking brighter light gives faster electrons. Only frequency affects KE_max.
4. Taking the square root too early or forgetting the factor of 2 in v = √(2KE/m).
5. Getting a negative KE_max and reporting it. A negative value simply means the photon energy is below the work function: no emission.

Practice questions

1. Potassium has φ = 2.30 eV. Find its threshold wavelength.
2. Light of 300 nm falls on a metal with φ = 3.00 eV. Find KE_max in eV and J.
3. The stopping potential for a metal illuminated with 5.00 eV photons is 1.50 V. Find φ.

Answers:

1. λ₀ ≈ 1240 ÷ 2.30 = 539 nm
2. E ≈ 1240 ÷ 300 = 4.13 eV; KE_max = 1.13 eV = 1.81 × 10⁻¹⁹ J
3. KE_max = 1.50 eV, so φ = 5.00 − 1.50 = 3.50 eV

For more photon conversions, see wavelength, frequency and photon energy calculations.

Key takeaways

• hf = φ + KE_max: photon energy equals work function plus the maximum kinetic energy of the electron.
• Threshold frequency f₀ = φ/h and threshold wavelength λ₀ = hc/φ.
• Stopping potential in volts equals KE_max in electronvolts.
• A KE_max vs frequency graph has gradient h and x-intercept f₀ for every metal.
• Intensity changes the number of electrons, not their maximum energy.