SME
Grade 11 Science · Physics
Modern physics: photons, atomic spectra and relativity
Use this for “Green light has a wavelength of 532 nm. What are its frequency and photon energy in joules and electronvolts?”
Diagram
λ
=
c/f
=
3 ×
10
8
/(5.64
×
10
14
)
λ = 532 nm
E = hf
E
=
6.626 ×
10
-34
×
5.64 ×
10
14
E
=
3.74 ×
10
-19
J
In
electronvolts:
3.74 ×
10
-19
/(1.602
×
10
-19
)
=
2.33
eV
Higher frequency, shorter wavelength, more energy per photon: blue photons carry more than red.
This is an example. Type your own values to start a new problem.
f (Hz)
=
3
×
10⁸
×
10
−9
EJ
Photon energy in joules
worked out (not typed) · J
E
Photon energy
calculated · eV
Formulas
f
=
3 ×
10
8
/(
λ
×
10
-9
)
5.64 ×
10
14
=
3 ×
10
8
/(532
×
10
-9
)
EJ
=
6.626 ×
10
-34
×
f
3.74 ×
10
-19
=
6.626 ×
10
-34
×
(5.64
×
10
14
)
E
=
EJ/(1.602
×
10
-19
)
2.33
=
3.74 ×
10
-19
/(1.602
×
10
-19
)
Assumptions
•
Light comes in photons, each with energy E = hf, h = 6.626 × 10⁻³⁴ J·s.
•
A shorter wavelength means a higher frequency and more energy per photon; c = fλ = 3.00 × 10⁸ m/s.
•
Brighter light has more photons, not more energetic ones. 1 eV = 1.602 × 10⁻¹⁹ J.
Step-by-step
WE KNOW
λ = 532 nm
FIND
frequency (f), photon energy in joules (EJ), photon energy (E)
Step 1 · Find frequency (f)
Use f = 3 × 10⁸/(λ × 10⁻⁹)
c
=
fλ:
the
speed
of
light
over
the
wavelength
in
meters
(1
nm
=
10
-9
m).
f
=
3 ×
10
8
/(532
×
10
-9
)
λ
=
532
nm
=
5.32 ×
10
-7
m
f
=
5.64 ×
10
14
Hz
Step 2 · Find photon energy in joules (EJ)
Use EJ = 6.626 × 10⁻³⁴ × f
Planck’s constant times the frequency, in joules.
EJ
=
6.626 ×
10
-34
×
(5.64
×
10
14
)
EJ
=
3.74 ×
10
-19
J
Step 3 · Find photon energy (E)
Use E = EJ/(1.602 × 10⁻¹⁹)
Each
electronvolt
is
1.602 ×
10
-19
J.
E
=
3.74 ×
10
-19
/(1.602
×
10
-19
)
E
=
2.33
eV
Check
Put the numbers back in. Both sides should match.
5.64 ×
10
14
≈
3 ×
10
8
/(532
×
10
-9
)
✓
3.74 ×
10
-19
≈
6.626 ×
10
-34
×
(5.64
×
10
14
)
✓
2.33
≈
3.74 ×
10
-19
/(1.602
×
10
-19
)
✓
Topic check
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