Transcript CHAPTER 7: The Hydrogen Atom

CHAPTER 8
Hydrogen Atom
8.1 Spherical Coordinates
8.2 Schrödinger's Equation in Spherical
Coordinate
8.3 Separation of Variables
8.4 Three Quantum Numbers
8.5 Hydrogen Atom Wave Function
8.6 Electron Spin
8.7 Total Angular Momentum and the SpinOrbit Effect
8.8 Zeeman Effect
Werner Heisenberg
(1901-1976)
The atom of modern physics can be symbolized only through a partial differential
equation in an abstract space of many dimensions. All its qualities are inferential; no
material properties can be directly attributed to it. An understanding of the atomic world
in that primary sensuous fashion…is impossible.
- Werner Heisenberg
Schrödinger Equation For Hydrogen
Atom
The potential energy of the electron-proton system is electrostatic:
Use the three-dimensional time-independent Schrödinger Equation.
For Hydrogen-like atoms (He+ or Li++)
Replace e2 with Ze2 (Z is the atomic number).
Replace m with the reduced mass, m.
Spherical Coordinates
The potential (central force)
V(r) depends on the distance r
between the proton and
electron.
Transform to spherical polar
coordinates because of the
radial symmetry.
The Schrödinger Equation in Spherical
Coordinates
1
  2
r sin 
2

  2 
sin  r  r r



  

 sin 

  
1  2 


2
sin





Transformed into spherical coordinates, the Schrödinger
equation becomes:
See Appendix E of Text for Details
Separation of Variables
The wave function  is a function of r, , . This is a potentially
complicated function.
Assume instead that  is separable, that is, a product of three
functions, each of one variable only:
This would make life much simpler, and it turns out to work.
Solution of the Schrödinger Equation
for Hydrogen
Separate the resulting equation into three equations: R(r), f(θ), and g().
The derivatives:
Substitute:
Multiply both sides by r2 sin2 θ / R f g:
Solution of the Schrödinger Equation for H
r and θ appear only on the left side and  appears only on the right side
The left side of the equation cannot change as  changes.
The right side cannot change with either r or θ.
Each side needs to be equal to a constant for the equation to be true.
Set the constant to be −mℓ2
azimuthal equation
It is convenient to choose the solution to be
.
Solution of the Schrödinger Equation for H
satisfies the azimuthal equation for any value of mℓ.
The solution must be single valued to be a valid solution for any :
mℓ must be an integer (positive or negative) for this to be true.
Now set the left side equal to −mℓ2 and rearrange it [divide by sin2()].
Now, the left side depends only on r, and the right side depends only
on θ. We can use the same trick again!
Solution of the Schrödinger Equation for H
Set each side equal to the constant ℓ(ℓ + 1).
Radial equation
Angular equation
We’ve separated the Schrödinger equation into three ordinary secondorder differential equations, each containing only one variable.
Solution of the Radial Equation for H
The radial equation is called the associated Laguerre equation and the
solutions R are called associated Laguerre functions. There are
infinitely many of them, for values of n = 1, 2, 3, …
Assume that the ground state has n = 1 and ℓ = 0. Let’s find this solution.
The radial equation becomes:
The derivative of
This yields:
yields two terms.
Solution of the Radial
Equation for H
Try a solution
A is a normalization constant.
a0 is a constant with the dimension of length.
Take derivatives of R and insert them into the radial
equation.

To satisfy this equation for any r, both expressions in
parentheses must be zero.
Set the second expression
equal to zero and solve for a0:
Set the first expression equal
to zero and solve for E:
Both are equal to the Bohr results!
Hydrogen Atom Radial Wave Functions
First few
radial
wave
functions
Rnℓ
Subscripts
on R
specify
the
values of
n and ℓ.
Solution of the Angular and Azimuthal
Equations
The solutions to the azimuthal equation are:
Solutions to the angular and azimuthal equations are linked
because both have mℓ.
Physicists usually group these solutions together into
functions called Spherical Harmonics:
spherical harmonics
Normalized Spherical Harmonics
Solution of the Angular and Azimuthal
Equations
The radial wave function R and the spherical harmonics Y determine
the probability density for the various quantum states. The total wave
function
depends on n, ℓ, and mℓ. The wave function
becomes
The Hydrogen Atom Wave Functions
im
 e
  flm ( )  Legendre Functions
flm ( ) g ( )  Ylm ( , )  Spherical Harmonics
Rnl (r )  Laguerre polynomials
 nlm (r, , )  Cnlm Rnl (r) flm ( ) gm ( )
The Ground State
Exercise 1
1.
2.
3.
Determine the normalization constant of the ground state wave
function of the Hydrogen atom
2. Determine the ground state wave function of the hydrogen atom
Calculate <r> for the electron in the Ground State of Hydrogen.
The Excited States
Quantum Numbers
The three quantum numbers:
n: Principal quantum number
ℓ: Orbital angular momentum quantum number
mℓ: Magnetic (azimuthal) quantum number
The restrictions for the quantum numbers:
n = 1, 2, 3, 4, . . .
ℓ = 0, 1, 2, 3, . . . , n − 1
mℓ = −ℓ, −ℓ + 1, . . . , 0, 1, . . . , ℓ − 1, ℓ
Equivalently:
n>0
ℓ<n
|mℓ| ≤ ℓ
The energy levels are:
Principal
Quantum
Number n
There are many solutions to the radial wave equation, one for
each positive integer, n.
The result for the quantized energy is:
A negative energy means that the electron and proton are bound
together.
Orbital Angular Momentum Quantum
Number ℓ
Energy levels are degenerate with respect to ℓ (the energy is
independent of ℓ).
Physicists use letter names for the various ℓ values:
ℓ=
0
1
2
3
4
Letter =
s
p
d
f
g
5...
h...
Atomic states are usualy referred to by their values of n and ℓ.
A state with n = 2 and ℓ = 1 is called a 2p state.
Orbital Angular Momentum Quantum
Number ℓ
It’s associated with the R(r) and f(θ) parts of the wave function.
Classically, the orbital angular momentum
with L = mvorbitalr.
L is related to ℓ by
In an ℓ = 0 state,
This disagrees with Bohr’s
semi-classical “planetary”
model of electrons orbiting
a nucleus L = nħ.
Classical orbits—which do not
exist in quantum mechanics
Rough derivation of ‹L2› = ℓ(ℓ+1)ħ2
We expect the average of the angular momentum components
squared to be the same due to spherical symmetry:
But
Averaging over all mℓ values (assuming each is equally likely):
because:

n
m 2  (  1)(2  1) / 3
Magnetic Quantum Number mℓ
The angle  is the angle from the z axis.
The solution for g() specifies that mℓ is an integer and is related to the
z component of L:
Example: ℓ = 2:
Only certain orientations
of are possible. This is
called space
quantization.
And (except when ℓ = 0)
we just don’t know Lx and
Ly!
Exercise 2
If a system has angular momentum characterized by the quantum
number l=2, what are the possible values of L z, what is the magnitude
L, and what is the smallest possible angle between L and z axis?
Exercise 3
Find the energies E311, E222, and E321 and construct an energy –level
diagram for the three-dimensional cubic well that includes the third,
fourth, and fifth excited states. Which of your states on your diagram
are degenerate?
Exercise 4
Write down the wave function for the hydrogen atom when the
electron’s quantum numbers are n=3, l=2 and ml=-1. Check to be sure
that the wave function is normalized.
Electron Angular Momentum
Consider the atom to behave like a small magnet.
Think of an electron as an orbiting circular current loop of I = dq / dt
around the nucleus. If the period is T = 2p r / v,
then
I = -e/T = -e/(2p r / v) = -e v /(2p r).
The current loop has a magnetic moment m = IA = [-e v /(2p r)] p r2 =
[-e/2m] mrv:
where L = mvr is the magnitude of the orbital angular momentum.
m
e
L
2m
Electron Angular Momentum
The ratio –e/2m is called gyro-magnetic ratio g and the Bohr
magneton is
e
mB 
2m
 9.274  1024 A.m 2  5.788  10 5 eV / T
If the magnetic field is in the z-direction, we only care about the zcomponent of m:
e
e
mz  
Lz  
(m )   m B m
2m
2m
A magnet place in a magnetic field experiences a torque
dL
e
 
 mB 
LB
dt
2m
If angular momentum is not perpendicular to the magnetic field
we get precession with Larmor frequency
eB
L 
The potential energy due to the
magnetic field is:
2m
VB  Lml
Intrinsic Spin
In 1925, grad students, Samuel
Goudsmit and George Uhlenbeck,
in Holland proposed that the
electron must have an
intrinsic angular momentum
and therefore a magnetic moment.
Paul Ehrenfest showed that, if so, the surface of the spinning
electron should be moving faster than the speed of light!
In order to explain experimental data, Goudsmit and Uhlenbeck
proposed that the electron must have an intrinsic spin
quantum number s = ½.
The Stern-Gerlach Experiment
From the potential energy of the magnetic dipole in an external
magnetic field we find that the shift in energy is zero for l=0 ml=0
however Stern-Gerlach experiment proves this to be wrong.
Inhomogeneous
Field in z direction
 mB m (dB / dz)
The Stern-Gerlach Experiment
Since L is quantized, Q.M predicts μz to have 2l+1 values
corresponding to 2l+1 values of m see figure a
m=0; no deflection
m=1 ;3 deflections
μ is oriented
along z axis
Figure b for l=1
Figure c for l=0
Ground state of
silver and
hydrogen are
known to be in
l=0 state
 mB m (dB / dz)
The Stern-Gerlach Experiment
Stern-Gerlach expected a single line for silver but found two lines.
Since orbital angular momentum is zero, the total angular momentum
is simply due to spin of the electron. Stern and Gerlach were the first
to make the direct observation of the electron spin.
Figure a with
no magnetic
filed
Figure b for l=0
With magnetic
filed on
 mB m (dB / dz)
Intrinsic Spin
The spinning electron reacts similarly to
the orbiting electron in a magnetic field.
The magnetic spin quantum number ms
has only two values, ms = ±½.
The electron’s spin will be either “up” or
“down” and can never be spinning with its
magnetic moment μs exactly along the z
axis.
Intrinsic Spin
Recall: m L  
e
L
2m
The magnetic moment is
The coefficient of
is −2μB and is a
consequence of relativistic quantum mechanics.
.
Writing in terms of the gyromagnetic ratio, g: gℓ = 1 and gs = 2:
and
The z component of
In an ℓ = 0 state:
no splitting due to
.
.
there is space quantization due to the intrinsic spin.
Apply ms and the potential energy becomes:
Total Angular Momentum and Spin-Orbit Effect
Orbital angular momentum
Spin angular momentum
Total angular momentum
L, Lz, S, Sz, J, and Jz are quantized.
Total Angular Momentum
If j and mj are quantum numbers for the single-electron hydrogen atom:
Quantization of the magnitudes:
The total angular momentum quantum number for the single electron
can only have the values
Exercise 5
Two electrons have zero orbital angular momentum. What are the
possible quantum numbers for the total angular momentum of the twoelectron system?
Spin-Orbit Coupling
Atomic states with the same n and l values but different j values
have slightly different energies because of the interaction of the spin
of the electron with its orbital motion. This is called spin-orbit
coupling.
Figure a) L is up due
to orbit of electron.
Figure b) B is up due
to apparent motion of
proton.
When spin is parallel
to L, the magnetic
moment is anitparallel
to L an B, so the spinorbit energy has its
largest value
Spin-Orbit Coupling
The potential energy of a magnetic moment in a magnetic field
depends on its orientation relatvie to the field direction and is given
by
1
S  L   J ( J  1)  L( L  1)  S ( S  1)  2
2
V  m  B  KS  L
Fine structure
energy level
diagram,
Left: no B
Right: B due to
relativee motion
of nucleus
Because of
spin-orbit
coupling the P
level is split.
Exercise 6
The total angular momentum is J=L+S, show that;
1 2
S  L   J  L2  S 2 
2
Selection rules for Transitions
Now the selection rules for a single-electron atom become
Δn = anything
Δms = 0,
Δℓ = ±1
Δmj = 0, ±1
Hydrogen energy-level diagram for
n = 2 and n = 3 with
spin-orbit
splitting.
Δml = 0, ±1
Δj = 0, ±1
The Zeeman Effect
An atomic beam of particles in the ℓ = 1 state pass through a
magnetic field along the z direction.
 mB m (dB / dz)
The mℓ = +1 state will be deflected down, the mℓ = −1 state up, and
the mℓ = 0 state will be undeflected.
The Zeeman Effect
A magnetic field splits the mℓ levels. The potential energy is quantized
and now also depends on the magnetic quantum number mℓ.
When a magnetic field is applied, the 2p level of atomic hydrogen is
split into three different energy states with energy difference of ΔE =
mBB Δmℓ.
mℓ
Energy
1
E0 + μBB
0
E0
−1
E0 − μBB
The
Zeeman
Effect
The transition
from 2p to 1s,
split by a
magnetic field.
Exercise 7
An atomic beam of particles pass through a magnetic field along the
z direction. Find the number of splits in the ℓ = 1 and ℓ = 2 state.
How many transitions are allowed? What are the different transition
energies and their corresponding Δml values?