SME
Grade 11 Science · Problem type
The photoelectric effect
Use this for “Light of 250 nm falls on a metal with a work function of 2.3 eV. What is the most kinetic energy an electron can leave with? What is the threshold wavelength?”
Diagram
Photon energy: E = 1240/λ = 1240/250 = 4.96 eV
Kinetic energy: Kₘₐₓ = E − φ = 4.96 − 2.3 = 2.66 eV
Each photon frees at most one electron; a shorter wavelength gives each electron more energy.
Threshold: λ₀ = 1240/φ = 1240/2.3 = 539 nm
This is an example. Type your own values to start a new problem.
Kₘₐₓ (eV)
=
1240
eV·nm
λ (nm)
−
φ (eV)
f
Frequency of the light
calculated · Hz
E
Photon energy
calculated · eV
λ₀
Threshold wavelength
calculated · nm
f₀
Threshold frequency
calculated · Hz
V₀
Stopping voltage
calculated
Formulas
E
=
1,240
λ
4.96
=
1240
250
E
=
4.136 ×
10
-15
×
f
4.96
=
4.136 ×
10
-15
×
(1.20
×
10
15
)
Kₘₐₓ
=
E
−
φ
2.66 = 4.96 − 2.3
V₀
=
Kₘₐₓ
2.66 = 2.66
λ₀
=
1,240
φ
539 = 1240/2.3
f₀
=
φ
/(4.136
×
10
-15
)
5.56 ×
10
14
=
2.3/(4.136
×
10
-15
)
Assumptions
•
Light comes in photons of E = hc/λ = 1240/λ eV, with λ in nm.
•
One photon frees at most one electron; the work function φ is the least energy that takes.
•
Past the threshold wavelength λ₀ = 1240/φ (longer waves, less energy per photon) no electron leaves, however bright the light.
•
By frequency: E = hf with h = 4.136 × 10⁻¹⁵ eV·s, and the threshold frequency is f₀ = φ/h.
Step-by-step
WE KNOW
λ = 250 nm, φ = 2.3 eV
FIND
photon energy (E), frequency of the light (f), greatest kinetic energy of an electron (Kₘₐₓ), stopping voltage (V₀), threshold wavelength (λ₀), threshold frequency (f₀)
Step 1 · Find photon energy (E)
Use E = 1,240/λ
hc in eV·nm over the wavelength in nm.
E
=
1240
250
E
=
4.96
eV
Step 2 · Find frequency of the light (f)
Use E = 4.136 × 10⁻¹⁵ × f
The photon energy over h.
f
=
E
/(4.136
×
10
-15
)
f
=
4.96/(4.136
×
10
-15
)
f
=
1.20 ×
10
15
Hz
Step 3 · Find greatest kinetic energy of an electron (Kₘₐₓ)
Use Kₘₐₓ = E − φ
What is left of the photon’s energy after freeing the electron.
Kₘₐₓ = 4.96 − 2.3
Kₘₐₓ = 2.66 eV
Step 4 · Find stopping voltage (V₀)
Use V₀ = Kₘₐₓ
Each volt stops 1 eV: the voltage that stops the fastest electron is Kₘₐₓ in eV.
V₀
=
2.66
eV
e
=
2.66
V
V₀
=
2.66
V
Step 5 · Find threshold wavelength (λ₀)
Use λ₀ = 1,240/φ
The
wavelength
whose
photons
carry
just
φ
.
λ₀
=
1240/2.3
λ₀
=
539
nm
Step 6 · Find threshold frequency (f₀)
Use f₀ = φ/(4.136 × 10⁻¹⁵)
The
frequency
whose
photons
carry
just
φ
:
below
it
no
electron
leaves.
f₀
=
2.3/(4.136
×
10
-15
)
f₀
=
5.56 ×
10
14
Hz
Check
Put the numbers back in. Both sides should match.
4.96
=
1240
250
✓
4.96
≈
4.136 ×
10
-15
×
(1.20
×
10
15
)
✓
2.66 = 4.96 − 2.3 ✓
2.66 = 2.66 ✓
539 ≈ 1240/2.3 ✓
5.56 ×
10
14
≈
2.3/(4.136
×
10
-15
)
✓
Topic check
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