SME
University Physics III: Waves, Optics & Modern · Problem type
The photoelectric effect and the stopping potential
Use this for “Ultraviolet light of 200 nm falls on a copper plate whose work function is 4.7 eV. What stopping potential halts the fastest electrons, and what is copper’s threshold wavelength?”
Diagram
Photon energy: E = 1240/λ = 1240/250 = 4.96 eV
Kinetic energy: Kₘₐₓ = E − φ = 4.96 − 2.28 = 2.68 eV
Each photon frees at most one electron; a shorter wavelength gives each electron more energy.
Threshold: λ₀ = 1240/φ = 1240/2.28 = 544 nm
This is an example. Type your own values to start a new problem.
Kₘₐₓ (eV)
=
1,240
eV·nm
λ (nm)
−
φ (eV)
E
Photon energy
calculated · eV
V₀
Stopping potential
calculated · V
λ₀
Threshold wavelength
calculated · nm
Formulas
E
=
1,240
λ
4.96
=
1,240
250
Kₘₐₓ
=
E
−
φ
2.68 = 4.96 − 2.28
V₀
=
Kₘₐₓ
2.68 = 2.68
λ₀
=
1,240
φ
543.8596
=
1,240
2.28
Assumptions
•
Light comes in photons of E = hc ÷ λ, with hc = 1240 eV·nm; one photon frees at most one electron.
•
The work function φ is the least energy that frees an electron, so the fastest ones leave with Kₘₐₓ = E − φ.
•
A reverse voltage V₀ stops even those: eV₀ = Kₘₐₓ, so V₀ in volts is Kₘₐₓ in eV. Kₘₐₓ against frequency is a line of slope h.
•
Waves longer than λ₀ = hc ÷ φ free no electron, however bright the light.
Step-by-step
WE KNOW
λ = 250 nm, φ = 2.28 eV
FIND
photon energy (E), greatest kinetic energy (Kₘₐₓ), stopping potential (V₀), threshold wavelength (λ₀)
Step 1 · Find photon energy (E)
Use E = 1,240 ÷ λ
hc = 1240 eV·nm over the wavelength in nm gives the photon’s energy in eV.
E
=
1,240
250
E
=
4.96
eV
Step 2 · Find greatest kinetic energy (Kₘₐₓ)
Use Kₘₐₓ = E − φ
Freeing
the
electron
costs
φ
;
the
fastest
ones
keep
the
rest
of
the
photon’s
energy.
Kₘₐₓ = 4.96 − 2.28
Kₘₐₓ = 2.68 eV
Step 3 · Find stopping potential (V₀)
Use V₀ = Kₘₐₓ
Each volt takes 1 eV from an electron, so the voltage that stops the fastest is Kₘₐₓ in eV.
V₀
=
2.68
V
Step 4 · Find threshold wavelength (λ₀)
Use λ₀ = 1,240 ÷ φ
The wavelength whose photons carry just the work function: longer waves free nothing.
λ₀
=
1,240
2.28
λ₀
=
543.8596
nm
Check
Put the numbers back in. Both sides should match.
4.96
=
1,240
250
✓
2.68 = 4.96 − 2.28 ✓
2.68 = 2.68 ✓
543.8596
=
1,240
2.28
✓
← Previous lesson
Photons and quantum basics
Next lesson →
The de Broglie wavelength of an accelerated particle
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Photons and quantum basics
Main lesson
The de Broglie wavelength of an accelerated particle
Use this for “Electrons in a microscope are accelerated from rest through 2500 V. What is their de Broglie wavelength?”
Hydrogen’s levels and spectral lines (Bohr)
Use this for “An electron in a hydrogen atom falls from n = 4 to n = 1. Find the energy of each level, the photon’s wavelength and the series it belongs to.”
The uncertainty principle: the least spread in momentum
Use this for “An electron is confined to a region 0.1 nm wide. What is the least uncertainty in its momentum and in its speed?”