SME
Grade 10 Science · Problem type
A photon’s frequency and energy
Use this for the frequency and energy of a photon from its wavelength.
Diagram
λ
=
c/f
=
3 ×
10
8
/(4.57
×
10
14
)
λ = 656 nm
E = hf
E
=
6.626 ×
10
-34
×
4.57 ×
10
14
E
=
3.03 ×
10
-19
J
In
electronvolts:
3.03 ×
10
-19
/(1.602
×
10
-19
)
=
1.89
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
=
3.00
×
10
8
×
10
−9
E
J
=
6.626
×
10
−34
J·s
×
f
Hz
Formulas
f
=
(3.00
×
10
8
)/(
λ
×
10
-9
)
4.57 ×
10
14
=
(3.00
×
10
8
)/(656
×
10
-9
)
E
=
6.626 ×
10
-34
×
f
3.03 ×
10
-19
=
6.626 ×
10
-34
×
(4.57
×
10
14
)
Assumptions
•
Light travels at c = 3.00 × 10⁸ m/s, and c = λf.
•
Each photon carries E = hf, with h = 6.626 × 10⁻³⁴ J·s.
•
Shorter waves have higher frequencies and carry more energy.
Step-by-step
WE KNOW
λ = 656 nm
FIND
frequency (f), energy of one photon (E)
Step 1 · Find frequency (f)
Use f = (3.00 × 10⁸)/(λ × 10⁻⁹)
Light
travels
at
c
=
3.00 ×
10
8
m/s:
divide
c
by
the
wavelength
in
meters.
f
=
(3.00
×
10
8
)/(656
×
10
-9
)
f
=
4.57 ×
10
14
Hz
Step 2 · Find energy of one photon (E)
Use E = 6.626 × 10⁻³⁴ × f
Each photon carries Planck’s constant times its frequency.
E
=
6.626 ×
10
-34
×
(4.57
×
10
14
)
E
=
3.03 ×
10
-19
J
Check
Put the numbers back in. Both sides should match.
4.57 ×
10
14
≈
(3.00
×
10
8
)/(656
×
10
-9
)
✓
3.03 ×
10
-19
≈
6.626 ×
10
-34
×
(4.57
×
10
14
)
✓
Topic check
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