SME
Quantum Mechanics
Solving the Schrödinger equation
Use this for “An electron is trapped in a 0.8 nm wide well with infinitely high walls. Find the energies of n = 3 and n = 2, and the wavelength of the photon emitted when it falls from 3 to 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.
Emitted:
λ
=
hc
ΔE
.
This is an example. Type your own values to start a new problem.
h
2
8
×
m (kg)
(
L (nm)
)
2
=
E₁ (eV)
ΔE
=
(
n
2
−
n′
2
)
×
E₁
=
ΔE (eV)
Eₙ
Upper level’s energy
worked out (not typed) · eV
Eₙ′
Lower level’s energy
worked out (not typed) · eV
λ
Wavelength
calculated · nm
Formulas
E₁
=
(6.626 ×
10
-34
)
2
÷
(8
×
m
×
(
L
×
10
-9
)
2
)
÷
(1.602
×
10
-19
)
0.3761
=
(6.626 ×
10
-34
)
2
÷
(8
×
(9.109
×
10
-31
)
×
(1 ×
10
-9
)
2
)
÷
(1.602
×
10
-19
)
Eₙ
=
n
2
×
E₁
1.504
=
2
2
×
0.3761
Eₙ′
=
n
′²
×
E₁
0.3761
=
1
2
×
0.3761
ΔE = Eₙ − Eₙ′
1.128 = 1.504 − 0.3761
ΔE
=
(
n
2
−
n
′²)
×
(6.626 ×
10
-34
)
2
÷
(8
×
m
×
(
L
×
10
-9
)
2
)
÷
(1.602
×
10
-19
)
1.128
=
(
2
2
−
1
2
)
×
(6.626 ×
10
-34
)
2
÷
(8
×
(9.109
×
10
-31
)
×
(1 ×
10
-9
)
2
)
÷
(1.602
×
10
-19
)
λ
=
1,240
ΔE
1,099
=
1,240
1.128
Assumptions
•
The walls are infinitely high, so ψ = 0 there and only whole half-waves fit: L = nλ ÷ 2. Then Eₙ = n²h² ÷ (8mL²), with h = 6.626 × 10⁻³⁴ J·s.
•
E grows as n² and shrinks as 1 ÷ L²; joules are turned into eV by dividing by 1.602 × 10⁻¹⁹.
•
A fall from n to n′ gives one photon of ΔE = Eₙ − Eₙ′, and λ = hc ÷ ΔE with hc = 1240 eV·nm.
•
Masses: electron 9.109 × 10⁻³¹ kg, proton 1.673 × 10⁻²⁷ kg.
Step-by-step
WE KNOW
m = 9.109 × 10⁻³¹ kg, L = 1 nm, n = 2, n′ = 1
FIND
ground energy (E₁), upper level’s energy (Eₙ), lower level’s energy (Eₙ′), photon energy (ΔE), wavelength (λ)
Step 1 · Find ground energy (E₁)
Use E₁ = (6.626 × 10⁻³⁴)² ÷ (8 × m × (L × 10⁻⁹)²) ÷ (1.602 × 10⁻¹⁹)
The
ground
level:
square
h,
divide
by
8
m
and
L
2
in
meters
for
joules,
then
by
1.602 ×
10
-19
for
eV.
E₁
=
(6.626 ×
10
-34
)
2
÷
(8
×
(9.109
×
10
-31
)
×
(
10
-9
)
2
)
÷
(1.602
×
10
-19
)
E₁
=
(6.626 ×
10
-34
)
2
÷
(8
×
(9.109
×
10
-31
)
×
10
-18
)
÷
(1.602
×
10
-19
)
E₁
=
(6.626 ×
10
-34
)
2
÷
(7.2872 ×
10
-30
) ×
10
-18
1.602 ×
10
-19
E₁
=
(6.626 ×
10
-34
)
2
÷
7.2872 ×
10
-48
1.602 ×
10
-19
E₁
=
4.3904 ×
10
-67
7.2872 ×
10
-48
÷
(1.602
×
10
-19
)
E₁
=
6.0248 ×
10
-20
1.602 ×
10
-19
E₁
=
0.3761
eV
Step 2 · Find upper level’s energy (Eₙ)
Use Eₙ = n² × E₁
Levels
go
as
n
2
:
level
n
is
n
2
times
the
ground
energy.
Eₙ
=
2
2
×
0.3761
Eₙ = 4 × 0.3761
Eₙ = 1.504 eV
Step 3 · Find lower level’s energy (Eₙ′)
Use Eₙ′ = n′² × E₁
The
lower
level
the
same
way:
n
′²
times
the
ground
energy.
Eₙ′
=
1
2
×
0.3761
Eₙ′ = 0.3761 eV
Step 4 · Find photon energy (ΔE)
Use ΔE = Eₙ − Eₙ′
The photon carries off the difference between the two levels.
ΔE = 1.504 − 0.3761
ΔE = 1.128 eV
Step 5 · 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
1.128
λ
=
1,099
nm
Check
Put the numbers back in. Both sides should match.
0.3761
=
(6.626 ×
10
-34
)
2
÷
(8
×
(9.109
×
10
-31
)
×
(
10
-9
)
2
)
÷
(1.602
×
10
-19
)
✓
1.504
=
2
2
×
0.3761
✓
0.3761
=
1
2
×
0.3761
✓
1.128 = 1.504 − 0.3761 ✓
1.128
=
(
2
2
−
1
2
)
×
(6.626 ×
10
-34
)
2
÷
(8
×
(9.109
×
10
-31
)
×
(
10
-9
)
2
)
÷
(1.602
×
10
-19
)
✓
1,099
=
1,240
1.128
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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