SME
Grade 10 Science · Problem type
A line in hydrogen’s spectrum
Use this for the energy and wavelength of the light a hydrogen atom gives off when its electron drops a level.
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.
E
eV
=
13.6
eV
×
1
n₁
2
−
1
n₂
2
λ
nm
=
1240
eV·nm
eV
Formulas
E
=
13.6
×
(1/
n₁
2
−
1/
n₂
2
)
1.889
=
13.6
×
(
1
2
²
−
1
3
²)
λ
=
1,240
E
656 = 1240/1.889
Assumptions
•
Hydrogen’s levels have energies Eₙ = −13.6/n² eV.
•
The electron drops from the upper level to a lower one and gives off one photon.
•
Drops to n = 2 give the visible lines; drops to n = 1 give ultraviolet.
Step-by-step
WE KNOW
n₂ = 3, n₁ = 2
FIND
photon energy (E), wavelength (λ)
Step 1 · Find photon energy (E)
Use E = 13.6 × (1/n₁² − 1/n₂²)
The
photon
carries
the
energy
between
the
two
levels,
Eₙ
=
−13.6/
n
2
eV.
E
=
13.6
×
(
1
2
²
−
1
3
²)
1
4
−
1
9
=
5
36
13.6
×
5
36
=
1.8889
E
=
1.889
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
=
1240/1.889
λ
=
656
nm
Check
Put the numbers back in. Both sides should match.
1.889
≈
13.6
×
(
1
2
²
−
1
3
²)
✓
656 ≈ 1240/1.889 ✓
Topic check
See what you know: 5 questions on this topic
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