Download Lecture 19: The Hydrogen Atom

Lecture 19: The Hydrogen Atom
• Reading: Zumdahl 12.7-12.9
• Outline
– The wavefunction for the H atom
• Know what wave functions look like for a particle
trapped in a “box”; now we need to know what they
look like for an electron attracted to a nucleus; and
the energy of each wave function.
– Quantum numbers and nomenclature
– Orbital (i.e. wavefunction) shapes and energies
• Problems (Chapter 12, 5th Ed.)
– 48, 49, 50, 52, 54, 55, 56, 57, 60
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H-atom wavefunctions
• Recap: The Hamiltonian is a sum of kinetic (KE,
or T) and potential (PE, or V) energy.
Hˆ = Tˆ + Vˆ
The ‘bar’ means average
over the position of the
electron.
E = T + V = 12 V
er
P+
V (Potential E.)
• The hydrogen atom potential energy is given by:
r
0
−e
ˆ
V = V (r) =
r
2
2
The Coulombic PE (V) can be generalized
− Ze 2
V (r ) =
( 4πε o ) r
F
e=−
NA
e′2 =
e-
e2
( 4πε o )
2
p
T=
where p = mv
2m
r
Z
+
P
• Z = atomic number (= 1 for hydrogen)
• r is the distance between the electron and the nucleus
• Only one electron allowed (for now).
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H-atom Coordinates Frame
• The radial dependence of the potential suggests
that we should from Cartesian coordinates to spherical
polar coordinates.
r = interparticle distance
(0 ≤ r ≤ ∞)
e-
p+
Major (azimuthal) angle
θ = angle from z to“xy plane”
(0 ≤ θ ≤ π)
Minor angle
φ = rotation in “xy plane”
(0 ≤ φ ≤ 2π)
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H-atom Allowed Energies
When we solve the Schrodinger equation using the Coulomb
potential, we find that the bound-state energy levels are
quantized or discrete:
4
2
⎛
⎞
⎛
Z
me
Z ⎞
En = − 2 ⎜ 2 2 ⎟ = − ⎜ 2 ⎟ ⋅ 2.178 x10−18 J
n ⎝ 8ε 0 h ⎠
⎝n ⎠
2
• n (an integer counter) is the principal quantum number,
and ranges from 1 to infinity. n=1 is the lowest energy
(level) or ground state for an electron bound to a
hydrogen-like nucleus.
•This is the same formula Bohr gave us.
•Compare and contrast these energy levels with those of
the particle in a box.
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Solve the Wave Equation for the Electron bound to the
Nucleus
• Set up the Schrödinger equation (SE) for the wave function
in terms of x,y and z coordinates, then rewrite in polar
coordinates (because V depends only on r).
• Solve the SE the same way Schrödinger did: Look the
answer up in a math book (Courant and Hilbert, in his
case).
• The solution gives a set of wave functions, and the energy
of each wave function.
• The wave functions (and energies) are distinct and
countable (although in principle there are an infinite
number of wavefunctions).
• The wavefunctions are now called orbitals as they describe
the probability of the electron in the vicinity of the nucleus.
They are not orbits but regions of space wherein the
electron orbits, hence orbitals.
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Form of WaveFunctions (for Orbitals)
• Like the particle in a box the wave function
depends on the coordinate and a quantum number
(like x and n).
• There are three coordinates so the wave function is
a product of a part that
– Depends only on r and has n (and l) with it
– Depends only on theta and has l (and m) with it
– Depends only on phi (and has m with it)
• The total wave function has the form:
Ψ n ,l ,m ( r , θ , φ ) = Rn ,l ( r ) Θl ,m (θ ) Φ m (φ )
x = r sin θ cos φ
Relation between Cartesian and
polar coordinates.
y = r sin θ sin φ
z = r cos θ
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Orbitals
• Orbitals are a description of where the
electron resides (like a house)
• Quantum numbers are like the address of
the house.
• The orbital does exist even without the
electron (so an empty orbital is called a
virtual orbital).
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Energy levels
The energy expression for the QM result is the same as
Bohrs, because the Virial Relation (which is also true for
planets going around the sun) is also true for Quantum
Mechanics and embodies the balance between potential
and kinetic energy.
p2
Ze 2
KE = T =
>0 PE = V = −
<0
2m
4πε o r
V = −2T ⇐ This is the Virial Relation
1
1 V 2 −1 V 2
E = T +V = V =
=
2
2V
4 T
rp = n=
⎛ Ze ⎞
2
2
⎜
⎟
2
2
2
−1 V
−1 ⎝ 4πε o r ⎠
− m ⎛ Ze ⎞
− m ⎛ Ze ⎞
=
=
=
⎜
⎟ =
⎜
⎟
2
p
4 T
4
2 ⎝ 4πε o rp ⎠
2 ⎝ 2ε o nh ⎠
2m
2
EBohr
Bohr’s suggestion:
2
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H-atom quantum numbers
• n is called the principal quantum number.
In solving the Schrodinger Equation, two other
quantum numbers become evident:
A is the orbital angular momentum quantum number.
Ranges in value from 0 to (n-1). This tell us
how much energy is in the rotating part of the electron.
mA is the “z component” of orbital angular momentum.
Ranges in value from
−A to + A
Because of the range of ml values there are a
possible 2A + 1
different values of m for a given value of
A
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H-atom Quantum Numbers (cont.)
• We can then characterize the wavefunctions based on
the three quantum numbers: n, A, mA
We went from Cartesian to polar coordinates now we go from
polar coordinates to integer counters, we call quantum numbers
⎧ x ⎫ ⎧ r ⎫ ⎧ n ⎫ ⎧ principal ⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪
⎨ y ⎬ ⇒ ⎨θ ⎬ ⇒ ⎨ A ⎬ = ⎨ angular ⎬
⎪ z ⎪ ⎪φ ⎪ ⎪m ⎪ ⎪ magnetic ⎪
⎩ ⎭ ⎩ ⎭ ⎩ A⎭ ⎩
⎭
When we “look at” orbitals, the angular
quantum number is the best one to see
the shape.
A value 0 1
symbol s
2
3
p d
f
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Orbital Numbers
• Naming orbitals using QNs is done as follows
– n is simply referred to as the quantum number
– l (0 to (n-1)) is given a letter value (named by
the spectroscopists) as follows:
name
A value symbol
0
s
sharp
1
p
principal
2
d
diffuse
3
f
fine
- ml (-l…0…l) is usually “dropped”
The Payoff: the three QNs help us understand the structure of
the periodic table and the chemistry of the elements (Aufbau
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Principle). These QNs are the numbers of chemistry.
1s Orbital Shapes
• For each set of 3 quantum numbers there is a specific wave
function or orbital with a unique shape.
• Let’s take a look at the lowest energy orbital (or ground
state), the “1s” orbital (n = 1, l = 0, m = 0)
3
3
1 ⎛Z⎞
1 ⎛ Z ⎞ 2 −σ
ψ 1s =
=
⎜ ⎟ e
⎜ ⎟ e
π ⎝ ao ⎠
π ⎝ ao ⎠
Z
σ = r A reduced or dimensionless distance
a0
•
2
Z
− r
a0
a0 is referred to as the Bohr radius, and = 0.529 Å
1
1
2⎞
⎛
Z
−18
E n = −2.178x10 J⎜ 2 ⎟ = −2.178x10−18 J
⎝n ⎠
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1s Orbital Shapes
• Note that the “1s” wavefunction has no angular
dependence (i.e., Θ and Φ do not appear).
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1 ⎛Z⎞
ψ1s =
⎜ ⎟ e
π ⎝ ao ⎠
2 −Z r
a0
Probability Density =
3
1 ⎛ Z ⎞ 2 −σ
=
⎜ ⎟ e
π ⎝ ao ⎠
ψψ
*
Probability is spherical,
depends only on r
EVERY orbital has the factor:
(Z12.55), why 90% not 100%
e
−σ
n
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Counting Orbitals
Three Rules: n = 1, 2,3"
0≤A<n
− A ≤ mA ≤ A
• Table 12.3: Quantum Numbers and Orbitals
n l
Orbital
1 0
2 0
1
3 0
1
2
1s
2s
2p
3s
3p
3d
ml
0
0
-1, 0, 1
0
-1, 0, 1
-2, -1, 0, 1, 2
2
# of Orb. n
1
1
3
1
3
5
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Naming Orbital
• Example: Write down the orbitals associated with n = 4.
Ans: n = 4
l = 0 to (n-1)
= 0, 1, 2, and 3
= 4s, 4p, 4d, and 4f
Number of orbitals, or
degeneracy:
4s (1 ml sublevel)
4p (3 ml sublevels)
4d (5 ml sublevels
4f (7 ml sublevels)
Total number of orbitals for any n: n2 (=16 for n=4)
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Identify Orbital Names
• Z12.49: Which designations (Orbital Addresses) are
incorrect (or correct) and why?
– 1s, 1p, 7d, 9s, 3f, 4f, 2d
• Z12.52 How many orbitals can have the
Designation 5 p,3d z 2 , 4d , n = 5, n = 4
• Z12.50: Which of the follwing QNs are not allowed for H
atom? What is wrong?
n A m Answer
a
2 1 −1
2p
b
1 1
c
d
e
8 7 −6
X
8j
1 0
3 2
2
2
X
3d
f
4 3
4
X
g
0 0
0
X
0
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S Family of Orbital Shapes
S or (l = 0) orbitals; increasing n
• r dependence only
Nodes: Zeros of
Polynomial
As n increases, orbitals
demonstrate n-1 nodes.
No node at the origin (?)
What is an orbital?
What are we seeing?
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All 3 are identical in shape and size they just
point along x,y and z (respectively)
P Orbital Shapes
2p (l = 1) orbitals
All have a planar node through the middle,
normal to the direction
x = r sin θ cos φ
• not spherical, but lobed.
ψ2p
z
y = r sin θ sin φ
1
=
4 2π
z = r cosθ
3
⎛ Z ⎞ 2 −σ 2
⎜ ⎟ e ⋅ σ cos θ
⎝ ao ⎠
• Labeled with respect to orientation along x, y, and z.
•Orbitals have spatial direction (nodes help describe that)
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P Orbital Shapes increase n (Z12.54, 56)
3p orbitals; contrast with 2p
Why no 1p orbitals?
ψ3p
z
2 ⎛Z ⎞
=
⎜ ⎟
81 π ⎝ a o ⎠
3
2
(6σ − σ )e
2
−σ
3
z = r cos θ
σ = Zr a
o
Where does this node
occur? σ = ? (Z12.60)
• more nodes (one planar and now one radial) as
compared to 2p (expected.). Why is one nodal surface
Planar and the other spherical? (Z12.57)
• still can be represented by a “dumbbell” contour:
(angular part stays the same)
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cos θ
D Orbital Shapes
3d (l = 2) orbitals
xy = r 2 sin 2 θ cos φ sin φ
• labeled as dxz, dyz, dxy, dx2-y2 and dz2.
e.g. think about dxy, it literally is x time y.
2 planar nodes21
F Orbital Shapes
4f (l = 3) orbitals
• exceedingly complex probability distributions.
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3 planar nodes
How to look at Orbital Shapes
Begin with a ball (sphere);
Don’t cut it at all: s orbital (l=0), only one types
Cut it (in half) with a single plane: This generates a
p orbital (l=1), three different ones; generated by
cutting on three different planes. To see it better
sometimes we color two halves differently (y/b).
Cut it (into quarters) with two perpendicular
(orthogonal) planes. This generates a d orbital
(l=2), 5 different ones;
Cut it (into eights) with three perpendicular planes.
(eg. One in xy, one in xz and one in yz plane) This
generates f orbitals (l=2), 7 different ones
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Orbital Energies (H Atom)
Continuum
0 ------------------------------------
• energy increases as -1/n2
• orbitals of same n, but different
l are considered to be of equal
energy (“degenerate”).
• the “ground” or lowest energy
orbital is the 1s.
All discrete or bound states are
below the E=zero line.
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