LECTURE 21 - UMD Department of Physics

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Transcript LECTURE 21 - UMD Department of Physics

PHYSICS 420
SPRING 2006
Dennis Papadopoulos
LECTURE 21
THE HYDROGEN AND
HYDROGENIC ATOMS
The Hydrogen Atom
• 1
Application of the Schrödinger Equation to the
Hydrogen Atom
•
•
•
•
Solution of the Schrödinger Equation for Hydrogen
Quantum Numbers
Energy Levels and Electron Probabilities
Magnetic Effects on Atomic Spectra – The Normal
Zeeman Effect
Intrinsic Spin
2
3
4.
5
• 6
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
1: Application of the Schrödinger
Equation to the Hydrogen Atom
• The approximation of the potential energy of the electronproton system is electrostatic:
• Rewrite the three-dimensional time-independent
Schrödinger Equation.
For Hydrogen-like atoms (He+ or Li++)
• Replace e2 with Ze2 (Z is the atomic number).
Application of the Schrödinger Equation
• 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.
Insert the Coulomb potential
into the transformed
Schrödinger equation.
reduced mass   1/   (1/ me )  (1/ mi )
Fig. 8-5, p.266
Application of the Schrödinger
Equation
• The wave function ψ is a function of r, θ, .
• Equation is separable.
• Solution may be a product of three functions.
• We can separate the SE into three separate differential
equations, each depending on one coordinate: r, θ, or .
Solution of the Schrödinger
Equation
• Eqs (8.11) to eqs (8.15) yield
----Radial equation
----Angular equation
d 2 g ( )
2


m
l g ( ) --------Azimuthal equation
2
d
g:
eiml
• SE has been separated into three ordinary second-order
differential equations each containing only one variable.
Solution of the Radial Equation
• The radial equation is called the associated Laguerre
equation and the solutions R that satisfy the appropriate
boundary conditions are called associated Laguerre
functions.
• Assume the ground state has ℓ = 0 and this requires mℓ =
0. The radial equation becomes
or
Solution of the Radial Equation
• Try a solution
Take derivatives of R and insert them into the SE equation.
• To satisfy it for any r each of the two expressions in parentheses
to be zero.
Set the second parentheses equal to zero and solve for a0.
Set the first parentheses equal to zero and solve for E.
Both equal to the Bohr result.
Quantum Numbers
• The appropriate boundary conditions to the radial and
angular equations leads to the following restrictions on the
quantum numbers ℓ and mℓ:
– ℓ = 0, 1, 2, 3, . . .
– mℓ = −ℓ, −ℓ + 1, . . . , −2, −1, 0, 1, 2, . ℓ . , ℓ − 1, ℓ
– |mℓ| ≤ ℓ and ℓ < 0.
• The predicted energy level is
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 for azimuthal equation are
.
• Solutions to the angular and azimuthal equations are
linked because both have mℓ.
• Group these solutions together into functions.
---- 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
Normalized Spherical Harmonics
Table 8-2, p.269
Table 8-3, p.269
Probability Distribution Functions
• We must use wave functions to calculate the probability
distributions of the electrons.
• The “position” of the electron is spread over space and is not well
defined.
• We may use the radial wave function R(r) to calculate radial
probability distributions of the electron.
• The probability of finding the electron in a differential volume
element dτ is
.
Probability Distribution Functions
• The differential volume element in spherical polar
coordinates is
Therefore,
• We are only interested in the radial dependence.
• The radial probability density is P(r) = r2|R(r)|2 and it
depends only on n and l.
P (r )dr   4 r 2 dr
2
Isotropic
States only
l=0
Fig. 8-9, p.282
P(r )   4 r 2
2

1   P(r )dr
0

f   f (r ) P(r )dr
0
Average vs. most
probable distance

r   rP(r )dr  (3 / 2)ao
0
Fig. 8-10a, p.283
Fig. 8-10, p.283
Probability Distribution Functions

R(r) and P(r) for the
lowest-lying states of
the hydrogen atom.
Table 8-5, p.280
Fig. 8-11b, p.285
l=2
Fig. 8-11c, p.285
Probability Densities
Symmetric about z-axis
Fig. 8-12, p.286
3: Quantum Numbers
The three quantum numbers:
– n
– ℓ
– mℓ
Principal quantum number
Orbital angular momentum quantum number
Magnetic quantum number
The boundary conditions:
– n = 1, 2, 3, 4, . . .
– ℓ = 0, 1, 2, 3, . . . , n − 1
– mℓ = −ℓ, −ℓ + 1, . . . , 0, 1, . . . , ℓ − 1, ℓ
The restrictions for quantum numbers:
– n>0
– ℓ<n
– |mℓ| ≤ ℓ
Integer
Integer
Integer
Principal Quantum Number n
• It results from the solution of R(r) in because R(r) includes
the potential energy V(r).
The result for this quantized energy is
• The negative means the energy E indicates that the
electron and proton are bound together.
Orbital Angular Momentum
Quantum Number ℓ
• It is associated with the R(r) and f(θ) parts of the wave function.
• Classically, the orbital angular momentum
mvorbitalr.
• ℓ is related to L by
• In an ℓ = 0 state,
with L =
.
.
It disagrees with Bohr’s semiclassical “planetary” model of electrons
orbiting a nucleus L = nħ.
n  1, 2, 3,
The allowed energy
levels are quantized
much like or particle in
a box. Since the
energy level decreases
a the square of n, these
levels get closer
together as n gets
larger.
l  0,1, 2,K n 1
Chemical properties of an atom are determined by the least tightly
bound electrons.
Factors:
•Occupancy of subshell
•Energy separation between the subshell and the next higher
subshell.
Pauli principle and Minimum Energy Principle
•
•
•
•
•
He Z=2, n=1, l=0, m=0.
Two electron with opposite spin
Zero angular momentum
High ionization energy 54.4 eV
Inert
Z2
E  13.6 2  4 x13.6  54.4eV
n
• Li Z=3, n=1 full, go to n=2, L-shell
• Bigger atom, 4 times ao (~n2)
• Nuclear charge partially screened by n=1
electrons
• Low ionization potential
• Energy of outer electrons
1 kZ e 2
E
2 r
Z   1.1  1.2
Hund’s
Rule
Unpaired
spins
Fig. 9-15, p.321
Table 9-2a, p.322
Table 9-2b, p.323
Fig. 9-16, p.324