Modern Physics 342
References :
1. Modern Physics by Kenneth S. Krane. , 2nd Ed.
John Wiley & Sons, Inc.
2. Concepts of Modern Physics by A. Beiser, 6th Ed.
(2002), McGraw Hill Com.
3. Modern Physics for Scientists and Engineers by J.
Taylor, C. Zafiratos and M. Dubson, 2nd Ed, 2003.
Chapters: 5 (Revision), 6 (6.4), 7, 8, 10, 11 and 12
Ch. 5 (Revision)
Schrödinger Equation
Schrödinger Equation Requirements
1. Conservation of energy is necessary:
K U E
Kinetic energy
Potential energy
Total energy
The kinetic energy K is conveniently given by
2
1
P
K  mv 2 
2
2m
Where P is the momentum = m v
(2)
Schrödinger Equation Requirements
(continued)
2. Consistency with de Broglie hypothesis
h
P 

and
P  k
Where, λ and k are, respectively, the wavelength and the wave number.
(k)2
K
2m
(3)
Schrödinger Equation Requirements
(continued)
3. Validity of the equation
The solution of this equation must be valid everywhere, single valued and
linear.
By linear we meant that the equation must allow de Broglie waves to
superimpose properly.
The following is a mathematical form of the wave associating the particle.
(x)  A sin( k x)
(4)
To make sure that the solution is continuous, its
derivative must have a value everywhere.
d
 A k cos( k x )
dx
d2 
2
2


A
k
sin(
k
x
)


k

2
dx
k2 
(5)
2mK
2
d2 
2m


K
2
2
dx

(6)
Time-Independent Schrödinger Equation
d2 
2m


(E  U) 
2
2
dx


 2 d2 

 U   E
2
2 m dx
(7)
Probability, Normalization and Average
• The probability density is given by
2
P( x ) dx  ( x ) dx
Which is the probability of finding a particle in
the space dx.
The probability of finding this particle in a region
between x1 and x2 is
P( x ) 
x2
 (x)
x1
2
dx
By normalization we mean the total probability
allover the space is 1, so that,

 (x )

2
dx  1
The mean value ( the expectation value) of x,

x 
 p(x) x dx
 p(x) dx

 (x)


2
 (x)
2
x dx
dx

If the wave function is normalized, therefore,

x 
 (x)

2
x dx
Applications
• The Free Particle ( a particle is moving with no forces acting on it)
• U(x)= constant anywhere=0 (arbitrarily)
 d

 E
2
2m dx
2
2
d2 
2


k

2
dx
d2 
2mE
 2 
2
dx

2mE
k  2

2
The solution of such deferential equation is
given in the form
Ψ(x)=A sin(kx)+B cos(kx)
(8)
Particle in a one dimensional box
0
(x)  
A sin(k x)  B cos(k x)
at x  0 and at x  L
at 0  x  L
Finding A
Ψ(0)=A sin(kx) + B cos(kx)=0 and for this, B=0
Therefore,
ψ(x)=A sin(kx)
(9)
Ψ(L)=0, therefore
A sin(kL)=0
since A≠0, sin(kL)=0
np
kL=p, 2p, 3p, .. .., np
(10)
k
L
np
0 A sin ( L x )  1 Normalization condition
L
2np
np 

  sin( L x ) L x 
np


  2
4
2 
A L


0
L
2
2
A
2
L
The wave function is now given by,
2
np
(x) 
sin( x)
L
L
(11)
The energy is given by using (8) and (10)
together,
np 2 2mE
( )  2
L

n2 p 2  2
En 
2mL2
(12)
The ground state (lowest ) energy EO is given by
p2 2
EO 
2mL2
We used n=1
The allowed energies for this particle are
En  EO , 4EO , 9EO ,.....
The wave function is shown here for n=1 and n=2
Ψ(x)
n=1
x
n=2
Example 5.2 P149
An electron is trapped in a one-dimensional region of length
1X10-10 m. How much energy must be supplied to excite the
electron from the ground state to the first excited state? In the
ground state, what is the probability of finding the electron in
the region from 0.09 X 10-10 m to 0.11 X 10-10 m? In the first
excited state, what is the probability of finding the electron
between x=0 and x=0.25 X 10-10 m?
Example 5.3 P151
Show that the average value of x is L/2, for a particle in a box of
length L, independent of the quantum state (not quantized).
Since the wave function is
(x) 
2
np
sin( x)
L
L

And the average value is defined by
x 
 (x)
2
x dx

2
np 2
 x  [
sin( x)] x dx
L
L
0
L
L
2np
np


x sin(
x ) cos(2
x)
2

2
x
L
L
L
 x  

  
np
np 2
L
4
2
4
8( )
L
L

0
A particle in a two dimensional box
The Schrödinger equation in two dimensions is
 2   2(x, y)  2(x, y) 


 U(x, y) (x, y)  E (x, y)

2
2m  x 2
y

U(x,y)=0 inside the box (0≤x≤L) & (0≤y≤L)
U(x,y)=∞ outside the box
The wave function ψ(x,y) is written as a product of two functions in
x and y,
(x, y)  f (x) g(y)
f (x)  A sink x x  B cosk x x
g(y)  A sink y y  B cosk y y
Since ψ(x,y) must be zero at the boundaries,
ψ(0,y) =0
ψ(L,y) =0
ψ(x,0) =0
ψ(x,L) =0
Therefore, A sin kx (0)+ B cos kx (0)=0 which requires B=0

f (x)  A sink x x
In the same way
g(y)  C sink y y
For x=L, f(x)=0 and y=L, g(y)=0
A sink xL  0

This requires kx L=nxp with n=1,2,3
ny p
nx p
f ( x )  A sin
x and g(y)  C sin
y
L
L
ny p
nx p
(x, y)  f (x ) g(y)  A' sin
x  sin
y
L
L
To find the constant A’, the wave function should
be normalized
L L
ny p 2
nx p
0 0 ( A' sin L x  sin L y) dx dy  1
This integration gives
A' 
2
L
ny p
nx p
2
(x, y)  sin
x  sin
y
L
L
L
The energy states of a particle in a two dimensional box
Substituting about the wave function ψ(x,y) in
Schrödinger equation, we find
 2 2p 2

2m L3
ny p  2
ny p 

2
pnx
pnx
2
sin( L x)  sin( L y)(nx  ny )   L sin( L x)  sin( L y)  E




Which after simplification becomes
 2p
2
2
E
(nx  ny )
3
2m L
2
2
Chapter 7
The Hydrogen Atom Wave Functions
• The Schrödinger Equation in Spherical Coordinates
The Schrödinger equation in three dimensions is
 2
2m
  2  2  2 
 2  2  2   U( x, y, z)  E
y
z 
 x
The potential energy for the force between the nucleus and the electron is
1 e2
1
U

4pO r 4pO
e2
x 2  y2  z2
This form does not allow to separate wave function Ψ into functions in terms of x, y and z, so we
have to express the whole equation of Schrödinger in terms of spherical coordinates, r, θ, and
φ.
Cartesian and spherical coordinates
r cos θ
z
electron
θ
φ
x
r
y
x= r sin θ cos φ
y= r sin θ sin φ
z= r cos θ
And Schrödinger equation becomes
 2   2 2 
1
 
 
1
 2 

 2
 U(r, , )  E
 sin 
 2 2
 2 
2 
2m  r
r r r sin   
  r sin   
  (r, , )

This wave function can be written in terms of 3 functions in
their corresponding variables, r, θ and φ
(r, , )  R (r)()()
Hydrogen wave functions in spherical coordinates
R(r) is called radial function
Θ(θ ) is called polar function and
Φ(φ) is called azimuthal function
when solving the three differential equations in
R(r), Θ(θ ) and Φ(φ), l and ml quantum numbers
were obtained in addition to the previous
principal quantum number n obtained before.
n the principal quantum number
l angular momentum quantum number
ml magnetic quantum number
n
1
2
2
2
1, 2, 3, …
0, 1, 2, ……±(n-1)
0, ±1, ±2, ……± l
The energy levels of the hydrogen atom
m e4
1
En  
32 p 2 O2  2 n2
The allowed values of the radius r around the nucleus are given by
4pO  2 2
rn 
n
2
me
Bohr radius (r at n=1) is denoted by aO and is given by
4pO  2
aO 
me 2
The Radial Probability Density P(r)
The radial probability density of finding the electron at a given location is
determined by
P(r )  r Rn, (r )
2
2
The total probability of finding the electron anywhere around the nucleus is


 P(r ) dr   r
0
2
2
R n, (r ) dr
0
The limits of the integration depend on the conditions of the problem
Example 7.1
Prove that the most likely distance from the origin of an electron in the
n=2, l=1 state is 4aO .
At n=2 and l =1, R2,1 (r) is given by
1
r
e
3/2
3 (2a O ) a O
R 2,1(r ) 

r
2 aO
The most likely distance means the most probable position. The maximum
value of the probability is obtained if r=4aO .
To prove that, the first derivative of P(r) with respect to r is zero at this value.

1
r
P(r )  r 
e
3/ 2
 3 (2aO ) aO
2
r

2aO



2
2
r
 



d 2
1
r

2 aO
e
 0
r 
3/2
dr   3 (2a O ) a O
 
 


r
aO

r
aO
r3 e
r4 e

0
5
6
6 aO
24 aO
Simplifying this result we get
r  4 aO
Example 7.2
An electron in the n=1, l=0 state. What is the probability of finding the
electron closer to the nucleus than the Bohr radius aO ?
The probability is given by


 P(r ) dr   r Rn, (r ) dr
2
2
0
0
R1,0 (r ) 
2
a3O/ 2
e r / aO
2
aO

r 
2r
r  aO
2

 2 aO 
2r 2r   aO 
2
r 0 r  3 e  dr  1  aO  aO2  e 
 a2


0
 o

 0.323
32.3 % of the time the electron is closer than 1 Bohr radius to the nucleus.
Angular Momentum
We discussed the radial part R(r) of Schrodinger equation. In this section we will
discuss the angular parts of the Schrodinger equation.
The classical angular momentum vector is given by
L  r p
During the variables separation of wave functions in Schrodinger equation,
angular momentum quantum number l was produced. The length of the
angular momentum vector L is given by
L  (  1)
The z-components of L are given by
L  m 
where ml is the magnetic quantum number 0, ±l
The angular momentum vector components
For l=2, ml =0, ±1, ±2
The angle  is given by
cos  
m 
(  1)

m
(  1)
Intrinsic Spin
 iA
i

q
T
A  pr 2
Angular momentum vector
2 pr
T
v
linear momentum p  mv
q

rp
2m
Using q=-e the charge of the electron, and
rp= L , we get

Magnetic moment due to electric current i
e
L  
L
2m
The negative singe indicates that µL and L work in opposite directions.
When the angular momentum vector L is inclined to the direction of the z-axis, the
magnetic moment µL has a z-component given by
 L,z  
e
e
Lz  
m 
2m
2m
e
2m
Bohr Mgneton
B 
L,z  m B
 B  9.274 X 10 24
J
T
Remember, ml =0, ±l
An electric dipole in a uniform and non-uniform electric field
A magnetic dipole in a non-uniform magnetic field
The electric dipole has its moment p rotates to align with the
direction of the electric field
Two opposite dipoles in the same non-uniform electric field are affected by
opposite net forces that lead to displacing each dipole up and down according to
their respective alignments.
Similarly, the magnetic dipoles are affected in the same way.
When an electron with an angular momentum inclined to the magnetic
filed, it may move up or down according to the direction of rotation around
the nucleus.
Stern-Gerlach Experiment
A beam of hydrogen atoms is in the n=2, l= 1
state. The beam contains equal numbers of
atoms in the ml = -1, 0, and +1 states. When the
beam passes a region of non-uniform magnetic
field, the atoms with ml =+1 experience a net
upward force and are deflected upward, the
atoms with ml =-1 are deflected downward,
while the atoms with ml =0 are undeflected.
After passing through the field, the beam strikes a screen where it makes a visible
image.
1. When the filed is off, we expect to see one image of the slit in the center of
the screen
2. When the field is on, three images of the slit on the screen were expected –
one in the center, one above the center (ml =+1 ) and one below (ml =-1).
The number of images is the number of ml values = 2l+1= 3 in our example.
• In the Stern - Gerlach experiment, a beam, of silver atoms is
used instead of hydrogen.
• While the field is off, and instead of observing a single image of
the slit, they observed two separate images.
• A new quantum number is introduced, which is specifies the
electron, the spin quantum number s. It may have two values
±½. The magnitude of its vector is given by
S  s(s  1)
Its z component to the magnetic field direction is
Sz  m 
The experiment
Example 7.6
In a Stern – Gerlach type of experiment, the
magnetic field varies with distance in the z
dB
T

1
.
4
direction according to dz
mm The silver
atoms travel a distance x=3.5 cm through the
magnet. The most probable speed of the atoms
emerging from the oven is v=750 m/s. Find the
separation of the two beams as they leave the
magnet. The mass of a silver atom is 1.8 X 10-25
kg, and its magnetic moment is about 1 Bohr
magneton.
z
3.5 cm
vO =750 m/s. The force applied to the beam must be obtained
The force is the change of potential energy U with distance z. The potential
energy U is given by
U  μ  B   zBz
Fz  (9.27X10 24

Fz  
dBz
dU
 z
dz
dz
 z  B
J
T
J
) (1.4 X103 )  1.29X10 20
T
m
m
This is the vertical force due to the effect of the magnetic field on the electron
magnetic dipoles.
Using the law of motion at constant acceleration, Dz=vO t + ½ a t2. The initial
vertical speed was zero before the effect of the magnetic field. We can find
the vertical deflection Dz above the horizontal level of the beam.
a
1.29X10 20
Fz

m 1.8 X 10-25
J
m  7 X10 4 m
kg
s2
x 3.5 X10 2 m
t 
 4.6 X10 5 s
m
v
750
s
Dz= ½ a t2 =7.5X10-6 m, the beam separation is 2(7.5X10-5 m)=1.6X10-4 m
Energy Levels and Spectroscopic Notation
The notation for the quantum state of an electron is now
described by the four quantum numbers n, l, ml and ms .
For example, the ground state of the hydrogen is labeled as
(n,l,ml,ms )=(1,0,0,±½) which means the are two quantum states
that can be occupied by the electron; (1,0,0,+½) or (1,0,0,-½).
Degeneracy of the atomic levels
The ground state energy level is now degenerated into two
quantum states.
The 1st excited state of the hydrogen atom has (n,l,ml,ms)=
(2,1,1,+½), (2,1,1,-½), (2,1,0,+½), (2,1,0,-½), (2,1,-1,+½), (2,1,-1,-½),
(2,0,0,+½), (2,0,0,-½). The degeneracy is 8, or 2n2 , where n is the principal
quantum number.
Some degenerate states after spin quantum number
n=1,2,..
l=0,1,2, ..,n-1
ml =0,±l
1
0
0
0
0
1
2
1
0
-1
ms =±½
+½
-½
Degeneracy 2n2
2
+½
-½
+½
-½
+½
-½
+½
-½
8
Spectroscopic Notation
Magnetic quantum numbers, ml and ms have no
necessity to be mentioned unless a magnetic field is
applied to the atom. In normal cases, a certain
notation is used to specify the energy levels of the
electron in the atom. Also, it is important to specify
the number of electrons occupying such state. This
notation depends on the l values as follows:
Value of l
0
1
2
3
4
5
notation
s
p
d
f
g
h
For the electron in the n=1 level, it is in the state s and the level occupied by this
electron is denoted as (1s1 )or generally (ns1 )
Electronic configuration of some elements
element
Number of
n
electrons
l
1
0
2
0
3
1
0
H
1
1s1
He
2
1s2
Li
3
1s2
2s1
Be
4
1s2
2s2
B
5
1s2
2s2
2p1
Ne
6
1s2
2s2
2p2
Na
11
1s2
2s2
2p6
3s1
K
18
1s2
2s2
2p6
3s2
1
3p6
2
Selection rules
Transitions between energy states of the atom are governed
by the condition of Dl=±1.
The transition from 4s state is possible to the 3p state because
Dl=+1 , but not possible between 4s and 3s, 2s or 1s.
In addition to this condition, there is a rule against ml
differences (D ml ).
D  1
Dm  0,  1
l
n
4
3
2
1
0
1
2
3
s
p
d
f
Zeeman Effect
A hydrogen atom is prepared in a 2p (l =1) level, and is placed in an external uniform
magnetic field B. The orbital angular momentum magnetic moment L interacts with
the field and the potential energy of this interaction U is given by
U  μL  B
U  μL,z  B  mBB
In addition to this, there is the energy of this level, EO , for example. So on turning
the magnetic field on, the energy of this electron is given by
E  EO  mBB
Since ml in this example has the values +1, 0, and -1, there will be 3 energy
values, 3 different wave lengths emitted.
E  EO  mBB
E  EO
E  EO  mBB
Wavelength notation

E
hc

dE  
DE  mBB

hc
d 
2

 2
D 
DE
hc
DE
  D

  D
hc
D
2

2
D  
mBB
hc
D

DE  

+1
0
2
BB
hc
0
 BB
0
-1

2
BB
hc
 BB
Problem 22 p 233
A hydrogen atom is in an excited 5g state, from which it makes a
series of transitions, ending in the 1s state. Show on an energy
levels diagram the sequence of transitions that can occur. Repeat
the last steps if the atom begins in the 5d state.
5s
5p
5d
5f
4s
4p
4d
4f
3s
3p
3d
2s
2p
1s
5g
5s
5p
5d
5f
4s
4p
4d
4f
3s
3p
3d
2s
2p
1s
5g
Problem 23 p 233
Consider the normal Zeeman effect applied to 3d to 2p
transition. (a) sketch an energy-level diagram that shows the
splitting of the 3d and 2p levels in an external magnetic field.
Indicate all possible transitions from each ml state of the 3d
level to each ml state of the 2p level. (b) which transitions satisfy
the Dml =±1 or 0 selection rule? (c) Show that there are only
three different transitions energies emitted.
+2
+1
0
-1
-2
2p
+1
0
-1
ml
3d
Since there are three different values of Dml
namely +1, 0 and -1, there will be three different
energies emitted. Let the energy of the 2p state at
ml=0 be denoted by Eop and that of the 3d state at
ml=0 be denoted by Eod .
The following equation can be used to calculate any
energy released from allowed transitions.
DE  (EOd  EOp )  DmBB
The difference Eod –Eop is constant, while Dml has 3 different values, therefore, DE
has only three different values.
Many Electron Atoms
Pauli Exclusion Principle
It was believed that different atoms in the ground states have all their
electrons dropped down in the 1s state. This means they all must have the
same physical properties. This is not the case, in fact.
A conclusion was drawn by Pauli that states that:
No two electrons in a single atom can have the same set of quantum numbers
(n, l, ml , ms).
Inert
Gases
Alkali Metals
Alkaline Earth Metals
Transition Elements (Heavy Metals)
Lanthanide Series
Actinide Series
Nonmetals
Halogen
s