SME
University Physics III: Waves, Optics & Modern · Problem type
Hydrogen’s levels and spectral lines (Bohr)
Use this for “An electron in a hydrogen atom falls from n = 4 to n = 1. Find the energy of each level, the photon’s wavelength and the series it belongs to.”
Diagram
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.
nᵤ
Upper level
example
nₗ
Lower level
example
Eᵤ
Energy of the upper level
calculated · eV
Eₗ
Energy of the lower level
calculated · eV
ΔE
Photon energy
calculated · eV
λ
Wavelength
calculated · nm
Series
Series of the line
worked out (not typed)
Formulas
Eᵤ
=
−
13.6
nᵤ
2
−1.5111
=
−
13.6
3
2
Eₗ
=
−
13.6
nₗ
2
−3.4
=
−
13.6
2
2
ΔE = Eᵤ − Eₗ
1.8889 = −1.5111 − (−3.4)
λ
=
1,240
ΔE
656.4706
=
1,240
1.8889
Series = the series that ends on nₗ
2 = 2, so Balmer
Assumptions
•
Bohr’s hydrogen: the electron sits only on levels Eₙ = −13.6 eV ÷ n², n = 1, 2, 3, …; 0 is the electron set free.
•
A drop from nᵤ to nₗ gives off one photon carrying the gap, ΔE = Eᵤ − Eₗ, with λ = hc ÷ ΔE and hc = 1240 eV·nm.
•
Lines that end on one level form a series: Lyman (nₗ = 1, ultraviolet), Balmer (2, visible), Paschen (3), Brackett (4), Pfund (5), Humphreys (6).
Step-by-step
WE KNOW
nᵤ = 3, nₗ = 2
FIND
energy of the upper level (Eᵤ), energy of the lower level (Eₗ), photon energy (ΔE), wavelength (λ), series of the line (Series)
Step 1 · Find energy of the upper level (Eᵤ)
Use Eᵤ = −13.6 ÷ nᵤ²
Bohr’s
level
nᵤ:
−13.6
eV
over
nᵤ
2
.
Eᵤ
=
−
13.6
3
2
Eᵤ
=
−
13.6
9
Eᵤ = −1.51111 eV
Step 2 · Find energy of the lower level (Eₗ)
Use Eₗ = −13.6 ÷ nₗ²
The same for the lower level nₗ.
Eₗ
=
−
13.6
2
2
Eₗ
=
−
13.6
4
Eₗ = −3.4 eV
Step 3 · Find photon energy (ΔE)
Use ΔE = Eᵤ − Eₗ
The photon carries the gap between the two levels.
ΔE = −1.51111 − (−3.4)
ΔE = 1.8888889 eV
Step 4 · Find wavelength (λ)
Use λ = 1,240 ÷ ΔE
hc
=
1240
eV·nm
over
the
photon’s
energy
in
eV
gives
λ
in
nm.
λ
=
1,240
1.8888889
λ
=
656.4706
nm
Step 5 · Find series of the line (Series)
Use Series = the series that ends on nₗ
Each series is named for the level its lines end on.
2 = 2: Balmer
Series = the series that ends on 2
Series = Balmer
Check
Put the numbers back in. Both sides should match.
−1.51111
=
−
13.6
3
2
✓
−3.4
=
−
13.6
2
2
✓
1.8888889 = −1.51111 − (−3.4) ✓
656.4706
=
1,240
1.8888889
✓
2 = 2, so Balmer ✓
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