SME
Grade 12 Science · Earth & Space Science
Light, spectra, telescopes and distances to stars
Use this for “A star’s surface is 5,772 K. At what wavelength does its light peak?” with Wien’s law.
Diagram
λ = 502 nm: visible light (green).
c = λ × f
300,000,000 m/s = 502 nm × 5.98 × 10¹⁴ Hz
This is an example. Type your own values to start a new problem.
λ (nm)
=
2.898
×
10⁶
÷
T (K)
f
Frequency of the peak wavelength
calculated · Hz
Formulas
λ
=
2.898 ×
10
6
T
502
=
2.898 ×
10
6
5,772
f
=
3.00 ×
10
8
λ
×
10
-9
5.98 ×
10
14
=
3.00 ×
10
8
502 ×
10
-9
Assumptions
•
Hotter stars peak at shorter wavelengths, so they look bluer.
•
A star glows at every wavelength; λ is only the brightest one.
•
c = 3.00 × 10⁸ m/s.
•
f is the frequency of the peak wavelength; graphed by frequency, the peak falls elsewhere.
Step-by-step
WE KNOW
T = 5,772 K
FIND
peak wavelength (λ), frequency of the peak wavelength (f)
Step 1 · Find peak wavelength (λ)
Use λ = 2.898 × 10⁶ ÷ T
Wien’s
law:
the
peak
wavelength
in
nm
is
2.898 ×
10
6
over
the
temperature
in
K.
λ
=
2.898 ×
10
6
5,772
λ
=
502
nm
Step 2 · Find frequency of the peak wavelength (f)
Use f = 3.00 × 10⁸ ÷ (λ × 10⁻⁹)
Light’s
speed
over
the
wavelength
in
meters:
a
nanometer
is
10
-9
m.
f
=
3.00 ×
10
8
502 ×
10
-9
f
=
5.98 ×
10
14
Hz
Check
Put the numbers back in. Both sides should match.
502
≈
2.898 ×
10
6
5,772
✓
5.98 ×
10
14
≈
3.00 ×
10
8
502 ×
10
-9
✓
Topic check
See what you know: 5 questions on this topic
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