SME
Quantum Mechanics
The hydrogen atom
Use this for “For hydrogen’s n = 3, ℓ = 1 state find the energy, the number of states in the shell, ⟨r⟩ and |L|.”
Diagram
H 3p: P(r) = r²R²(r), the chance per a₀ of finding the electron at r; the shaded area is 1.
Radial nodes n − l − 1 = 1 (at 6 a₀).
⟨r⟩
=
3n² − l(l + 1)
2Z
⟨r⟩
=
3 ×
3
2
− 1 × 2
2
=
12.5
a₀
⟨r⟩ = 0.661 nm
rₘₚ
=
12
a₀
=
0.635
nm,
the
tallest
hump
(
n
2
a₀
Z
holds
only
when
l
=
n
−
1).
Energy of the shell Eₙ = −1.511 eV
States in the shell states = 18
Size of L |L| = 1.414 ħ
This is an example. Type your own values to start a new problem.
Eₙ
=
−13.6
eV
÷
n
2
=
Eₙ (eV)
ℓ
Orbital quantum number
example
states
States in the shell
calculated
⟨r⟩
Mean radius
worked out (not typed) · a₀
⟨r⟩
Mean radius
worked out (not typed) · nm
|L|
Size of L
worked out (not typed) · ħ
Formulas
Eₙ
=
−
13.6
n
2
−1.511
=
−
13.6
3
2
states
=
2
×
n
2
18
=
2
×
3
2
⟨r⟩
=
3 ×
n
2
−
ℓ
× (
ℓ
+ 1)
2
12.5
=
3 ×
3
2
− 1 × (1 + 1)
2
⟨r⟩ = ⟨r⟩ × 0.0529
0.6613 = 12.5 × 0.0529
|L|
=
√(
ℓ
×
(
ℓ
+
1))
1.414 = √(1 × (1 + 1))
Assumptions
•
Hydrogen: one electron round a proton held still, so Eₙ = −13.6 eV ÷ n², with a₀ = 0.0529 nm.
•
The energy depends on n only (fine structure is ignored), so all 2n² states of a shell share it, spin included.
•
ℓ runs over the whole numbers from 0 to n − 1; ⟨r⟩ is the mean distance from the nucleus, and |L| is in units of ħ.
•
For ℓ = n − 1 the most probable radius is n²a₀, a little less than ⟨r⟩. A photon between shells: he.physics.university-3#3~bohr.
Step-by-step
WE KNOW
n = 3, ℓ = 1
FIND
energy of the shell (Eₙ), states in the shell (states), mean radius (⟨r⟩), mean radius (⟨r⟩), size of L (|L|)
Step 1 · Find energy of the shell (Eₙ)
Use Eₙ = −13.6 ÷ n²
The
ground
state
is
13.6
eV
deep;
each
shell
is
bound
n
2
times
more
weakly.
Eₙ
=
−
13.6
3
2
Eₙ
=
−
13.6
9
Eₙ = −1.511 eV
Step 2 · Find states in the shell (states)
Use states = 2 × n²
Each
ℓ
from
0
to
n
−
1
has
2
ℓ
+
1
orbitals,
n
2
in
all,
and
each
holds
two
spins.
states
=
2
×
3
2
states = 2 × 9
states = 18
Step 3 · Find mean radius (⟨r⟩)
Use ⟨r⟩ = (3 × n² − ℓ × (ℓ + 1)) ÷ 2
The
mean
radius
in
a₀
grows
as
n
2
;
for
the
same
n
,
a
higher
ℓ
keeps
the
electron
a
little
closer.
⟨r⟩
=
3 ×
3
2
− 1 × (1 + 1)
2
⟨r⟩
=
3 ×
3
2
− 2
2
⟨r⟩
=
3 × 9 − 2
2
⟨r⟩
=
27 − 2
2
⟨r⟩
=
25
2
⟨r⟩ = 12.5 a₀
Step 4 · Find mean radius (⟨r⟩)
Use ⟨r⟩ = ⟨r⟩ × 0.0529
One Bohr radius a₀ is 0.0529 nm.
⟨r⟩ = 12.5 × 0.0529
⟨r⟩ = 0.6613 nm
Step 5 · Find size of L (|L|)
Use |L| = √(ℓ × (ℓ + 1))
The
size
of
L
in
ħ
is
the
root
of
ℓ
(
ℓ
+
1);
an
s
state
(
ℓ
=
0)
has
none.
|L| = √(1 × (1 + 1))
|L|
=
√
2
|L| = 1.414 ħ
Check
Put the numbers back in. Both sides should match.
−1.511
=
−
13.6
3
2
✓
18
=
2
×
3
2
✓
12.5
=
3 ×
3
2
− 1 × (1 + 1)
2
✓
0.6613 = 12.5 × 0.0529 ✓
1.414 = √(1 × (1 + 1)) ✓
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