HYDROGEN ATOM.ppt

variable separation method to solve three dimensional schrodinger eqn for hydrogen atom

1. 7.1Application of the Schrödinger Equation to the
Hydrogen Atom
7.2Solution of the Schrödinger Equation for Hydrogen
7.3Quantum Numbers
7.4Magnetic Effects on Atomic Spectra –The Normal
Zeeman Effect
7.5Intrinsic Spin
7.6Energy Levels and Electron Probabilities
CHAPTER 7
The Hydrogen Atom
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

2. 7.1: Application of the Schrödinger
Equation to the Hydrogen Atom
The approximation of the potential energy of the electron-proton
system is electrostatic:
Rewrite the three-dimensional time-independent Schrödinger
Equation.
For Hydrogen-like atoms (He
+
or Li
++
)
Replace e
2
with Ze
2
(Zis the atomic number).
Use appropriate reduced mass μ.

3. 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.
Equation 7.3

4. Application of the Schrödinger Equation
The wave function ψis a function of r, θ, .
Equation is separable.
Solution may be a product of three functions.
We can separate Equation 7.3 into three separate differential
equations, each depending on one coordinate: r, θ, or .
Equation 7.4

5. 7.2: Solution of the Schrödinger Equation
for Hydrogen
Substitute Eq (7.4) into Eq (7.3) and separate the resulting
equation into three equations: R(r), f(θ), and g( ).
Separation of Variables
The derivatives from Eq (7.4)
Substitute them into Eq (7.3)
Multiply both sides by r
2
sin
2
θ/ Rfg
Equation 7.7

6. Solution of the Schrödinger Equation
Only rand θappear on the left side and only appears on the right
side of Eq (7.7)
The left side of the equation cannot change as changes.
The right side cannot change with either ror θ.
Each side needs to be equal to a constant for the equation to be true.
Set the constant −m
ℓ
2
equal to the right side of Eq (7.7)
It is convenient to choose a solution to be .
--------azimuthal equationEquation 7.8

7. Solution of the Schrödinger Equation
 satisfies Eq (7.8) for any value of m
ℓ.
The solution must be single valued in order to have a valid
solution for any , which is
m
ℓto be zero or an integer (positive or negative) for this to
be true.
If Eq (7.8) were positive, the solution would not be normalized.

8. Solution of the Schrödinger Equation
Set the left side of Eq (7.7) equal to −m
ℓ
2
and rearrange it.
Everything depends on ron the left side and θon the right side of
the equation.

9. Solution of the Schrödinger Equation
Set each side of Eq (7.9) equal to constant ℓ(ℓ + 1).
Schrödinger equation has been separated into three ordinary
second-order differential equations [Eq (7.8), (7.10), and (7.11)],
each containing only one variable.
----Radial equation
----Angular equation
Equation 7.11
Equation 7.10

10. Solution of the Radial Equation
The radial equation is called the associated Laguerre equation
and the solutions Rthat satisfy the appropriate boundary
conditions are called associated Laguerre functions.
Assume the ground state has ℓ = 0 and this requires m
ℓ= 0.
Eq (7.10) becomes
The derivative of yields two terms and V is the Coulomb
potentials.
Write those terms:
Equation 7.13

11. Solution of the Radial Equation
Try a solution
Ais a normalized constant.
a
0is a constant with the dimension of length.
Take derivatives of Rand insert them into Eq (7.13).
To satisfy Eq (7.14) for any ris for each of the two expressions in
parentheses to be zero.
Set the second parentheses equal to zero and solve for a
0.
Set the first parentheses equal to zero and solve for E.
Both equal to the Bohr result.
Equation 7.14

12. Quantum Numbers
The appropriate boundary conditions to Eq (7.10) and (7.11)
leads to the following restrictions on the quantum numbers ℓ
and m
ℓ:
ℓ = 0, 1, 2, 3, . . .
m
ℓ= −ℓ, −ℓ + 1, . . . , −2, −1, 0, 1, 2, . ℓ . , ℓ − 1, ℓ
|m
ℓ| ≤ ℓand ℓ< 0.
The predicted energy level is
n = 1,2,3…..
n > l

13. Hydrogen Atom Radial Wave Functions
First few radial wave functions R
nℓ
Subscripts on Rspecify the values of nand ℓ.

14. Solution of the Angular and Azimuthal
Equations
The solutions for Eq (7.8) are .
Solutions to the angular and azimuthal equations are linked
because both have m
ℓ.
Group these solutions together into functions.
----spherical harmonics

15. Normalized Spherical Harmonics

16. Solution of the Angular and Azimuthal
Equations
The radial wave function Rand the spherical harmonics Y
determine the probability density for the various quantum
states. The total wave function depends on n, ℓ,
and m
ℓ. The wave function becomes

17. 7.3: Quantum Numbers
The three quantum numbers:
n Principal quantum number
ℓ Orbital angular momentum quantum number
m
ℓ Magnetic quantum number
The boundary conditions:
n= 1, 2, 3, 4, . . . Integer
ℓ = 0, 1, 2, 3, . . . , n− 1 Integer
m
ℓ= −ℓ, −ℓ + 1, . . . , 0, 1, . . . , ℓ − 1, ℓInteger
The restrictions for quantum numbers:
n> 0
ℓ < n
|m
ℓ| ≤ ℓ

18. Principal Quantum Number n
It results from the solution of R(r) in Eq (7.4) because R(r) includes
the potential energy V(r).
The result for this quantized energy is
The negative means the energy Eindicates that the electron and
proton are bound together.

19. 
3*
2
3
0
( ) ( )
nlm nlm
nl
r d r
dr r R r






r r r Orbitaln l<r> (au)
1s 1 0 1.5
2s 2 0 6.0
2p 2 1 5.0
3s 3 0 13.522
()
nl
r R r
Radial probability density
Radial probability density
n
2
a
o

20. Shapes of the spherical harmonics
z
x
y00
0, 0
1
4
lm
Y


 11
1, 1
3
sin exp( )
8
lm
Yi 


 10
1, 0
3
cos
4
lm
Y 


 11
Y 11
Re[ ]Y 10
Y 00
Y
(Images from http://odin.math.nau.edu/~jws/dpgraph/Yellm.html)

21. Shapes of spherical harmonics (2)
(Images from http://odin.math.nau.edu/~jws/dpgraph/Yellm.html)
z
x
y2
20
2, 0
5
(3cos 1)
16
lm
Y 


 21
2, 1
15
sin cos exp(i )
8
lm
Y   


 2
22
2, 2
15
sin exp(2i )
32
lm
Y 


 21
Y 21
Re[ ]Y 22
Y 22
Re[ ]Y 20
Y

22. 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 with L=
mv
orbitalr.
ℓ is related to Lby .
In an ℓ= 0 state, .
It disagrees with Bohr’s semiclassical “planetary” model of
electrons orbiting a nucleus L= nħ.

23. Orbital Angular Momentum Quantum
Number ℓ
A certain energy level is degeneratewith respect to ℓwhen the
energy is independent of ℓ.
Use letter names for the various ℓvalues.
ℓ = 0 1 2 3 4 5 . . .
Letter =s p d f g h . . .
Atomic states are referred to by their nand ℓ.
A state with n= 2 and ℓ = 1 is called a 2pstate.
The boundary conditions require n> ℓ.

24. x z y
y x z
z y x
L yp zp
L zp xp
L xp yp


 For a classicalparticle, the angular momentum is
defined byx y z
L L L

  
L r p
i j k d d d d
()
d d d d
( ) ( ) 0.
t t t t
m
    
    
L r p
r p p r×
p
p r F
Angular momentum is very important in problems
involving a central force(one that is always
directed towards or away from a central point)
because in that case it is conserved
Same origin for rand F
Classical angular momentum
L
r
p
FIn components

25. The relationship of L, L
z, ℓ, and
m
ℓfor ℓ = 2.
 is fixed
because L
zis quantized.
Only certain orientations of
are possible and this is called
space quantization.
Magnetic Quantum Number m
ℓ
The angle is a measure of the rotation about the z axis.
The solution forspecifies that m
ℓis an integer and related to
the zcomponent of L.

26. The vector model
The z-component can have the 2l+1 values
corresponding to
In the vector model this means that only
particular special angles between the angular
momentum vector and the z-axis are allowed
A given quantum number l determines the
magnitude of the vector Lvia
This is a useful semi-classical model of the quantum results.
Imagine L precesses around the z-axis. Hence the magnitude of Land
the z-component L
zare constant while the x and y components can take a
range of values and average to zero, just like the quantum eigenfunctions. 22
( 1)
( 1)
L l l
ll

L ,
z
L m l m l   
z
L
θ

27. Magnetic Quantum Number m
ℓ
Quantum mechanics allows to be quantized along only one
direction in space. Because of the relation L
2
= L
x
2
+ L
y
2
+ L
z
2
the
knowledge of a second component would imply a knowledge of the
third component because we know .
We expect the average of the angular momentum components
squared to be .

28. The Dutch physicist Pieter Zeeman showed the spectral lines
emitted by atoms in a magnetic field split into multiple energy
levels. It is called the Zeeman effect.
A spectral line is split into three lines.
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.
The current loop has a magnetic moment μ= IAand the period T=
2πr/ v.
 where L= mvris the magnitude of the orbital
angular momentum.
7.4: Magnetic Effects on Atomic Spectra—The
Normal Zeeman Effect

29. 7.4: Magnetic Effects on Atomic Spectra—The
Normal Zeeman Effect
t = mx B = dL/dt

30. The angular momentum is aligned with the magnetic moment, and
the torque between and causes a precession of .
Where μ
B= eħ/ 2mis called a Bohr magneton.
 cannot align exactly in the z direction and
has only certain allowed quantized orientations.
Since there is no magnetic field to
align them, point in random
directions. The dipole has a
potential energy
The Normal Zeeman Effect

31. The Normal Zeeman Effect
The potential energy is quantized due to the magnetic quantum
number m
ℓ.
When a magnetic field is applied, the 2plevel of atomic hydrogen
is split into three different energy states with energy difference of
ΔE= μ
BB Δm
ℓ.
m
ℓEnergy
1 E
0+ μ
BB
0 E
0
−1E
0−μ
BB

32. The Normal Zeeman Effect
A transition from 2pto 1s.

33. An atomic beam of particles in the ℓ = 1 state pass through a magnetic
field along the zdirection.


The m
ℓ= +1 state will be deflected down, the m
ℓ= −1state up, and the
m
ℓ= 0state will be undeflected.
If the space quantization were due to the magnetic quantum number
m
ℓ, m
ℓstates is always odd (2ℓ + 1) and should have produced an odd
number of lines.
The Stern-Gerlach Experiment

34. 7.5: Intrinsic Spin
Samuel Goudsmit and George Uhlenbeck in Holland proposed that
the electron must have an intrinsic angular momentumand
therefore a magnetic moment.
Paul Ehrenfest showed that 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
numbers= ½.

35. Intrinsic Spin
The spinning electron reacts similarly to the orbiting electron in a
magnetic field.
We should try to find L, L
z, ℓ, and m
ℓ.
The magnetic spin quantum numberm
shas only two values,
m
s= ±½.
The electron’s spin will be either “up” or
“down” and can never be spinning with its
magnetic moment μ
sexactly along the z axis.
The intrinsic spin angular momentum
vector .

36. Intrinsic Spin
The magnetic moment is .
The coefficient of is −2μ
Bas with is a consequence of theory
of relativity.
The gyromagnetic ratio(ℓor s).
g
ℓ= 1and g
s= 2, then
The z component of .
In ℓ= 0 state
Apply m
ℓand the potential energy becomes
no splitting due to.
there is space quantization due to the
intrinsic spin.
and

37. 7.6: Energy Levels and Electron Probabilities
For hydrogen, the energy level depends on the principle quantum
number n.
In ground state an atom cannot emit
radiation. It can absorb
electromagnetic radiation, or gain
energy through inelastic
bombardment by particles.

38. Selection Rules
We can use the wave functions to calculate transition
probabilities for the electron to change from one state to another.
Allowed transitions:
Electrons absorbing or emitting photons to change states when
Δℓ= ±1.
Forbidden transitions:
Other transitions possible but occur with much smaller
probabilities when Δℓ≠ ±1.

39. 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 dτis .

40. 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) = r
2
|R(r)|
2
and it depends
only on nand l.

41. R(r) and P(r) for the
lowest-lying states of
the hydrogen atom.
Probability Distribution Functions

42. Probability Distribution Functions
The probability density for the hydrogen atom for three different
electron states.