SME
Remote Sensing · Problem type

Frequency and photon energy of a band

Use this for “A sensor band is centred at 850 nm in the near infrared. What are its frequency and the energy of one photon?”

Diagram

850 nm (infrared)400 nm500600700
λ
=
c/f
=
3 ×
10
8
/(3.53
×
10
14
)
λ = 850 nm
E = hf
E
=
6.626 ×
10
-34
×
3.53 ×
10
14
E
=
2.34 ×
10
-19
J
In
electronvolts:
2.34 ×
10
-19
/(1.602
×
10
-19
)
=
1.46
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
÷
(
λ
×
10
−9
)
Solving the picture’s λ = c ÷ f for f gives f = c ÷ λ, with λ in metres.
E
Photon energy
calculated · eV

Formulas

f
=
3.00 ×
10
8
λ
×
10
-9
3.5294 ×
10
14
=
3.00 ×
10
8
850 ×
10
-9
E
=
6.626 ×
10
-34
×
f
1.602 ×
10
-19
1.4598
=
6.626 ×
10
-34
×
3.5294 ×
10
14
1.602 ×
10
-19

Assumptions

•
Light travels at c = 3.00 × 10⁸ m/s, and 1 nm is 10⁻⁹ m.
•
A photon carries E = hf, with Planck’s constant h = 6.626 × 10⁻³⁴ J·s; 1 eV is 1.602 × 10⁻¹⁹ J.
•
Together these give the shortcut E ≈ 1,240 ÷ λ, with λ in nm and E in eV.

Step-by-step

WE KNOW
λ = 850 nm
FIND
frequency (f), photon energy (E)
Step 1 · Find frequency (f)
Use f = 3.00 × 10⁸ ÷ (λ × 10⁻⁹)
The speed of light over the wavelength in metres is the number of waves passing each second.
f
=
3.00 ×
10
8
850 ×
10
-9
f
=
3.5294 ×
10
14
Hz
Step 2 · Find photon energy (E)
Use E = 6.626 × 10⁻³⁴ × f ÷ (1.602 × 10⁻¹⁹)
Planck’s constant times the frequency is the photon’s energy in joules; dividing by the joules in one electronvolt gives eV.
E
=
6.626 ×
10
-34
×
3.5294 ×
10
14
1.602 ×
10
-19
E
=
2.3386 ×
10
-19
1.602 ×
10
-19
E
=
1.4598
eV
Check
Put the numbers back in. Both sides should match.
3.5294 ×
10
14
=
3.00 ×
10
8
850 ×
10
-9
✓
1.4598
=
6.626 ×
10
-34
×
3.5294 ×
10
14
1.602 ×
10
-19
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

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