SME
Physical Chemistry II: Quantum & Spectroscopy
Particle in a box and harmonic oscillator
Use this for “An electron is held in a box 1.00 nm long. Find E₁ and the wavelength it absorbs going from n = 1 to n = 2.”
Diagram
Eₙ
=
n
2
h
2
8mL²
:
the
levels
grow
as
n
2
,
and
ψ
=
0
at
both
walls.
ψ₂
has
1
node,
ringed.
Absorbed:
λ
=
hc
ΔE
.
This is an example. Type your own values to start a new problem.
m
Mass
example
L
Box length
example
n₁
Lower level
example
n₂
Upper level
example
E₁
Ground energy
calculated · J
E₁,eV
Ground energy in electronvolts
worked out (not typed) · eV
ΔE
Gap
calculated · J
λ
Wavelength absorbed
calculated
Formulas
E₁
=
(6.626 ×
10
-34
)
2
÷
(8
×
m
×
(
L
×
10
-9
)
2
)
6.025 ×
10
-20
=
(6.626 ×
10
-34
)
2
÷
(8
×
(9.109
×
10
-31
)
×
(1 ×
10
-9
)
2
)
E₁
,eV
=
E₁
1.602 ×
10
-19
0.3761
=
6.025 ×
10
-20
1.602 ×
10
-19
ΔE
=
(
n₂
2
−
n₁
2
)
×
E₁
1.807 ×
10
-19
=
(
2
2
−
1
2
)
×
(6.025
×
10
-20
)
λ
=
6.626 ×
10
-34
×
2.998 ×
10
8
ΔE
×
10
9
1,099
=
6.626 ×
10
-34
×
2.998 ×
10
8
1.807 ×
10
-19
×
10
9
Assumptions
•
V = 0 inside and the walls are infinitely high, so only whole half-waves fit: Eₙ = n²E₁.
•
n starts at 1, so E₁ > 0: the particle can never be still (zero-point energy).
•
For a dye with N π electrons along a chain of length L, the jump is from n₁ = N ÷ 2 to n₁ + 1.
•
h = 6.626 × 10⁻³⁴ J·s, c = 2.998 × 10⁸ m/s; an electron’s mass is 9.109 × 10⁻³¹ kg.
Step-by-step
WE KNOW
m = 9.109 × 10⁻³¹ kg, L = 1 nm, n₁ = 1, n₂ = 2
FIND
ground energy (E₁), ground energy in electronvolts (E₁,eV), gap (ΔE), wavelength absorbed (λ)
Step 1 · Find ground energy (E₁)
Use E₁ = (6.626 × 10⁻³⁴)² ÷ (8 × m × (L × 10⁻⁹)²)
Square
h,
then
divide
by
8
m
and
by
L
2
in
square
meters.
E₁
=
(6.626 ×
10
-34
)
2
÷
(8
×
(9.109
×
10
-31
)
×
(
10
-9
)
2
)
E₁
=
(6.626 ×
10
-34
)
2
÷
(8
×
(9.109
×
10
-31
)
×
10
-18
)
E₁
=
(6.626 ×
10
-34
)
2
÷
((7.2872
×
10
-30
)
×
10
-18
)
E₁
=
(6.626 ×
10
-34
)
2
÷
(7.2872
×
10
-48
)
E₁
=
4.3904 ×
10
-67
7.2872 ×
10
-48
E₁
=
6.025 ×
10
-20
J
Step 2 · Find ground energy in electronvolts (E₁,eV)
Use E₁,eV = E₁ ÷ (1.602 × 10⁻¹⁹)
One
electronvolt
is
1.602 ×
10
-19
J.
E₁
,eV
=
6.025 ×
10
-20
1.602 ×
10
-19
E₁
,eV
=
0.3761
eV
Step 3 · Find gap (ΔE)
Use ΔE = (n₂² − n₁²) × E₁
Each
level
is
n
2
E₁
:
subtract
the
squares,
then
multiply
by
E₁
.
ΔE
=
(
2
2
−
1
2
)
×
(6.025
×
10
-20
)
ΔE
=
(4
−
1)
×
(6.025
×
10
-20
)
ΔE
=
3
×
(6.025
×
10
-20
)
ΔE
=
1.807 ×
10
-19
J
Step 4 · Find wavelength absorbed (λ)
Use λ = 6.626 × 10⁻³⁴ × 2.998 × 10⁸ ÷ ΔE × 10⁹
The
photon
carries
the
gap:
divide
hc
by
ΔE
for
meters,
then
×
10
9
for
nanometers.
λ
=
6.626 ×
10
-34
×
2.998 ×
10
8
1.807 ×
10
-19
×
10
9
λ
=
(6.626
×
10
-34
)
×
2.998 ×
10
8
1.807 ×
10
-19
×
10
9
λ
=
1.9865 ×
10
-25
1.807 ×
10
-19
×
10
9
λ
=
(1.0993
×
10
-6
)
×
10
9
λ
=
1,099
nm
Check
Put the numbers back in. Both sides should match.
6.025 ×
10
-20
=
(6.626 ×
10
-34
)
2
÷
(8
×
(9.109
×
10
-31
)
×
(
10
-9
)
2
)
✓
0.3761
=
6.025 ×
10
-20
1.602 ×
10
-19
✓
1.807 ×
10
-19
=
(
2
2
−
1
2
)
×
(6.025
×
10
-20
)
✓
1,099
=
6.626 ×
10
-34
×
2.998 ×
10
8
1.807 ×
10
-19
×
10
9
✓
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