Grade 11 Science · Problem type

Hydrogen’s spectral lines

Use this for “An electron in hydrogen drops from n = 3 to n = 2. What are the photon’s energy and wavelength?”

Diagram

0 eV (free)n = 2−3.4 eVn = 3−1.51 eVn = 4−0.85 eVBelow: n = 1 (−13.6 eV)1.89 eV380 nm750 nm656 nm
The electron drops from n = 3 to n = 2 (the Balmer series).
Drawn to scale from n = 2 up; the lower levels are listed below the break.
Photon energy = 13.6 × (1/2² − 1/3²) = 1.889 eV.
Wavelength
=
1240
1.889
=
656
nm:
visible.
This is an example. Type your own values to start a new problem.
E (eV)
=
13.6
eV
×
1
nₗ
2
−
1
nᵤ
2
λ
Wavelength
calculated · nm

Formulas

E
=
13.6
×
(1/
nₗ
2
−
1/
nᵤ
2
)
1.89
=
13.6
×
(
1
2
²
−
1
3
²)
λ
=
1,240
E
656 = 1240/1.89

Assumptions

•
Hydrogen’s levels have energies Eₙ = −13.6/n² eV; the electron can only be on a level.
•
A drop from the upper level to the lower one gives off one photon with the difference in energy.
•
Drops to n = 1 are ultraviolet, to n = 2 visible, to n = 3 infrared. λ = 1240/E nm.

Step-by-step

WE KNOW
nᵤ = 3, nₗ = 2
FIND
photon energy (E), wavelength (λ)
Step 1 · Find photon energy (E)
Use E = 13.6 × (1/nₗ² − 1/nᵤ²)
The photon carries the energy between the two levels.
E
=
13.6
×
(
1
2
²
−
1
3
²)
E
=
13.6
×
(0.25
−
1
9
)
E
=
13.6
×
5
36
E
=
1.89
eV
Step 2 · Find wavelength (λ)
Use λ = 1,240/E
hc = 1240 eV·nm, so the wavelength in nm is 1240 divided by the energy in eV.
λ
=
1240/1.89
λ
=
656
nm
Check
Put the numbers back in. Both sides should match.
1.89
≈
13.6
×
(
1
2
²
−
1
3
²)
✓
656 ≈ 1240/1.89 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

Refresh

Related lessons

© 2026 SME. All rights reserved.