SME
General Chemistry I
Atomic structure and periodicity
Use this for “Find the wavelength and energy of the photon when hydrogen’s electron drops from n = 4 to n = 2.”
Diagram
The electron drops from n = 4 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 − 1/4²) = 2.55 eV.
Wavelength
=
1240
2.55
=
486
nm:
visible.
This is an example. Type your own values to start a new problem.
1
λ (nm)
=
R
H
1
n₁
2
−
1
n₂
2
ν
Frequency
calculated · Hz
E
Photon energy
calculated · J
Eₘ
Energy per mole of photons
calculated · kJ/mol
EeV
Photon energy in electronvolts
worked out (not typed) · eV
Formulas
λ
=
10
9
1.097 ×
10
7
× (1 ÷
n₁
2
− 1 ÷
n₂
2
)
486.2
=
10
9
1.097 ×
10
7
× (1 ÷
2
2
− 1 ÷
4
2
)
ν
=
2.998 ×
10
8
λ
×
10
-9
6.1665 ×
10
14
=
2.998 ×
10
8
486.2 ×
10
-9
E
=
6.626 ×
10
-34
×
ν
4.0859 ×
10
-19
=
6.626 ×
10
-34
×
(6.1665
×
10
14
)
Eₘ
=
E
×
6.022 ×
10
23
1,000
246.1
=
4.0859 ×
10
-19
×
6.022 ×
10
23
1,000
EeV
=
E
1.602 ×
10
-19
2.551
=
4.0859 ×
10
-19
1.602 ×
10
-19
Assumptions
•
One electron only (hydrogen). The electron drops from n₂ to a lower n₁, and one photon carries the difference.
•
n₁ names the series: 1 Lyman (ultraviolet), 2 Balmer (visible), 3 Paschen (infrared).
•
RH = 1.097 × 10⁷ m⁻¹, c = 2.998 × 10⁸ m/s, h = 6.626 × 10⁻³⁴ J·s, NA = 6.022 × 10²³ mol⁻¹.
Step-by-step
WE KNOW
n₂ = 4, n₁ = 2
FIND
wavelength (λ), frequency (ν), photon energy (E), energy per mole of photons (Eₘ), photon energy in electronvolts (EeV)
Step 1 · Find wavelength (λ)
Use λ = 10⁹ ÷ (1.097 × 10⁷ × (1 ÷ n₁² − 1 ÷ n₂²))
Rydberg:
1
λ
=
RH(1/
n₁
2
−
1/
n₂
2
)
gives
λ
in
meters;
×
10
9
turns
it
into
nanometers.
λ
=
10
9
1.097 ×
10
7
× (1 ÷
2
2
− 1 ÷
4
2
)
λ
=
10
9
(1.097 ×
10
7
) × (1 ÷ 4 − 1 ÷ 16)
λ
=
10
9
(1.097 ×
10
7
) × (0.25 − 0.0625)
λ
=
10
9
(1.097 ×
10
7
) × 0.1875
λ
=
10
9
2,056,875
λ
=
486.1744
nm
Step 2 · Find frequency (ν)
Use ν = 2.998 × 10⁸ ÷ (λ × 10⁻⁹)
Light travels at c = λν: divide c by the wavelength in meters.
ν
=
2.998 ×
10
8
486.1744 ×
10
-9
ν
=
6.16651 ×
10
14
Hz
Step 3 · Find photon energy (E)
Use E = 6.626 × 10⁻³⁴ × ν
One
photon
carries
Planck’s
constant
times
ν
.
E
=
6.626 ×
10
-34
×
(6.16651
×
10
14
)
E
=
4.08593 ×
10
-19
J
Step 4 · Find energy per mole of photons (Eₘ)
Use Eₘ = E × 6.022 × 10²³ ÷ 1,000
A mole of photons is NA of them; ÷ 1000 turns J into kJ.
Eₘ
=
4.08593 ×
10
-19
×
6.022 ×
10
23
1,000
Eₘ
=
246,054.7046
1,000
Eₘ = 246.1 kJ/mol
Step 5 · Find photon energy in electronvolts (EeV)
Use EeV = E ÷ (1.602 × 10⁻¹⁹)
One
electronvolt
is
1.602 ×
10
-19
J.
It
is
the
gap
between
the
levels:
13.6
eV
×
(1/
n₁
2
−
1/
n₂
2
)
=
2.55
eV.
EeV
=
4.08593 ×
10
-19
1.602 ×
10
-19
EeV = 2.551 eV
Check
Put the numbers back in. Both sides should match.
486.1744
=
10
9
1.097 ×
10
7
× (1 ÷
2
2
− 1 ÷
4
2
)
✓
6.16651 ×
10
14
=
2.998 ×
10
8
486.1744 ×
10
-9
✓
4.08593 ×
10
-19
=
6.626 ×
10
-34
×
(6.16651
×
10
14
)
✓
246.1
=
4.08593 ×
10
-19
×
6.022 ×
10
23
1,000
✓
2.551
=
4.08593 ×
10
-19
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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