Atomic Physics
Step Potential
Consider a particle of energy E moving in region in which the
potential energy is the step function
U(x) = 0,
x<0
U(x) = V0,
x>0
What happened when
a particle moving from
left to right encounters
the step?
The classical answer is
simple: to the left of the
step, the particle moves
with a speed v = √2E/m
Step Potential
At x =0, an impulsive force
act on the particle. If the
initial energy E is less than
V0, the particle will be
turned around and will then
move to the left at its
original speed; that is, the
particle will be reflected by
the step. If E is greater than
V0, the particle will continue
to move to the right but with
reduced speed given by
v = √2(E – U0)/m
Step Potential
We can picture this classical problem as a ball
rolling along a level surface and coming to a steep
hill of height h given by mgh=V0.
If the initial kinetic energy of the ball is less than
mgh, the ball will roll part way up the hill and then
back down and to the left along the lower surface at
it original speed. If E is greater than mgh, the ball
will roll up the hill and proceed to the right at a
lesser speed.
The quantum mechanical result is similar when E is
less than V0. If E<V0 the wave function does not go to
zero at x=0 but rather decays exponentially. The wave
penetrates slightly into the classically forbidden region
x>0, but it is eventually completely reflected.
Step Potential
This problem is somewhat similar to that of total
internal reflection in optics.
For E>V0, the quantum mechanical result differs
from the classical result. At x=0, the wavelength
changes from
λ1=h/p1 = h/√2mE
to
λ2=h/p2 = h/√2m(E-V0).
When the wavelength changes suddenly, part of
the wave is reflected and part of the wave is
transmitted.
Reflection Coefficient
Since a motion of an electron (or other
particle) is governed by a wave equation, the
electron sometimes will be transmitted and
sometimes will be reflected.
The probabilities of reflection and
transmission can be calculated by solving the
Schrödinger equation in each region of space
and comparing the amplitudes of transmitted
waves and reflected waves with that of the
incident wave.
Reflection Coefficient
This calculation and its result are similar to finding the
fraction of light reflected from the air-glass interface. If R
is the probability of reflection, called the reflection
coefficient, this calculation gives:
(k1  k 2 ) 2
R
2
(k1  k 2 )
where k1 is the wave number for the incident wave and k2
is the wave number for the transmitted wave.
Transmission Coefficient
The result is the same as the result in optics for the
reflection of light at normal incidence from the boundary
between two media having different indexes of
refraction n.
The probability of transmission T, called the
transmission coefficient, can be calculated from the
reflection coefficient, since the probability of
transmission plus the probability of reflection must
equal 1:
T+R=1
In the quantum mechanics, a localized particle is
represented by the wave packet, which has a maximum
at the most probable position of the particle.
Time development of a one dimensional wave packet
representing a particle incident on a step potential for E>V0.
The position of a classical particle is indicated by the dot. Note
that part of the packet is transmitted and part is reflected.
Reflection coefficient R and transmission coefficient T for a
potential step V0 high versus energy E (in units V0).
A particle of energy E0 traveling in a region in
which the potential energy is zero is incident on
a potential barrier of height V0=0.2E0. Find the
probability that the particle will be reflected.
Lets consider a rectangular potential barrier of height V0 and
with a given by:
U(x) = 0,
x<0
U(x) = V0,
0<x<a
U(x) = 0,
x>a
Barrier Potential
We consider a particle of
energy E , which is slightly
less than V0, that is incident
on the barrier from the left.
Classically, the particle
would always be reflected.
However, a wave incident
from the left does not
decrease immediately to
zero at the barrier, but it
will instead decay exponentially in the classically forbidden
region 0<x<a. Upon reaching the far wall of the barrier
(x=a), the wave function must join smoothly to a sinusoidal
wave function to the right of barrier.
Barrier Potential
If we have a beam of particle incident from left, all with the
same energy E<V0, the general solution of the wave equation
are, following the example for a potential step,
1 ( x)  Ae
ik1 x
2 ( x)  Ce
x
 Be
ik1 x
 De
x
3 ( x)  Feik1x  Geik1x
x0
0 xa
xa
where k1 =√2mE/ħ and α = √2m(V0-E)/ħ
This implies that there is some probability of the particle (which
is represented by the wave function) being found on the far side
of the barrier even though, classically, it should never pass
through the barrier.
Barrier Potential
For the case in which the quantity
αa = √2ma2(V0 – E)/ħ2
is much greater than 1, the transmission coefficient
is proportional to e-2αa, with
α = √2m(V0 – E)/ħ2
The probability of penetration of the barrier thus
decreases exponentially with the barrier thickness
a and with the square root of the relative barrier
height (V0-E). This phenomenon is called barrier
penetration or tunneling. The relative probability of
its occurrence in any given situation is given by the
transmission coefficient.
A wave packet representing a particle incident on two
barriers of height just slightly greater than the energy of the
particle. At each encounter, part of the packet is transmitted
and part reflected, resulting in part of the packet being
trapped between the barriers from same time.
A 30-eV electron is incident on a square barrier of
height 40 eV. What is the probability that the electron will
tunnel through the barrier if its width is (a) 1.0 nm?
(b) 0.1nm?
The penetration of the barrier is not unique to quantum
mechanics. When light is totally reflected from the glass-air
interface, the light wave can penetrate the air barrier if a
second peace of glass is brought within a few wavelengths of
the first, even when the angle of incidence in the first prism is
greater than the critical angle. This effect can be demonstrated
with a laser beam and two 45° prisms.
α- Decay
The theory of barrier penetration was used by George Gamov in 1928 to
explain the enormous variation of the half-lives for α decay of radioactive
nuclei. Potential well shown on the diagram for an α particle in a
radioactive nucleus approximately describes a strong attractive force
when r is less than the nuclear radius R. Outside the nucleus the strong
nuclear force is negligible, and the potential is given by the Coulomb’s
law, U(r) = +k(2e)(Ze)/r, where Z is the nuclear charge and 2e is the
charge of α particle.
α- Decay
An α-particle inside the nucleus oscillates back and forth, being reflected
at the barrier at R. Because of its wave properties, when the α-particle
hits the barrier there is a small chance that it will penetrate and appear
outside the well at r = r0. The wave function is similar to that for a square
barrier potential.
The probability that an α-particle will tunnel through the
barrier is given by
T e

2 2 m (V0  E )a

which is a very small number, i.e., the α particle is usually
reflected. The number of times per second N that the α
particle approaches the barrier is given by
v
N
2R
where v equals the particle’s speed inside the nucleus.
The decay rate, or the probability per second that the nucleus
will emit an α particle, which is also the reciprocal of the mean
life time
, is given by

decay
1
v
rate  
e
 2R

2 2 m (V0  E )a

The decay rate for emission of α particles from radioactive
nuclei of Po212. The solid curve is the prediction of equation
decay
1
v
rate  
e
 2R

2 2 m (V0  E )a
The points are the experimental results.

Applications of Tunneling
• Nanotechnology refers to the design and application of
devices having dimensions ranging from 1 to 100 nm
• Nanotechnology uses the idea of trapping particles in
potential wells
• One area of nanotechnology of interest to researchers is the
quantum dot
– A quantum dot is a small region that is grown in a silicon
crystal that acts as a potential well
• Nuclear fusion
– Protons can tunnel through the barrier caused by their
mutual electrostatic repulsion
Resonant Tunneling Device
• Electrons travel in the gallium arsenide
semiconductor
• They strike the barrier of the quantum dot from the
left
• The electrons can tunnel through the barrier and
produce a current in the device
Scanning Tunneling Microscope
• An electrically conducting
probe with a very sharp
edge is brought near the
surface to be studied
• The empty space
between the tip and the
surface represents the
“barrier”
• The tip and the surface
are two walls of the
“potential well”
Scanning Tunneling Microscope
• The STM allows
highly detailed
images of surfaces
with resolutions
comparable to the
size of a single atom
• At right is the surface
of graphite “viewed”
with the STM
Scanning Tunneling Microscope
• The STM is very sensitive to the distance from
the tip to the surface
– This is the thickness of the barrier
• STM has one very serious limitation
– Its operation is dependent on the electrical
conductivity of the sample and the tip
– Most materials are not electrically conductive at their
surfaces
– The atomic force microscope (AFM) overcomes this
limitation by tracking the sample surface maintaining
a constant interatomic force between the atoms on
the scanner tip and the sample’s surface atoms.
SUMMARY
1. Time-independent Schrödinger equation:
 2 d 2  ( x)

 U ( x) ( x)  E ( x)
2
2m dx
2.In the simple harmonic oscillator:
1

E n   n    0
2

the ground wave function is given:
0 ( x)  Ae
 ax2
where A0 is the normalization constant and a=mω0/2ħ.
3. In a finite square well of height V0, there are only a finite
number of allowed energies.
SUMMARY
4.Reflection and barrier penetration:
When the potentials changes abruptly over a
small distance, a particle may be reflected even
though E>U(x). A particle may penetrate a
region in which E<U(x). Reflection and
penetration of electron waves are similar for
those for other kinds of waves.
The Schrödinger Equation in Three Dimensions
The one-dimensional time-independent Schrödinger
equation
 d  ( x)

 U ( x) ( x)  E ( x)
2
2m dx
2
2
(1)
is easily extended to three dimensions. In rectangular
coordinates, it is
 2  d 2 d 2 d 2 
 2  2  2   U  E

2m  dx
dy
dz 
(2)
where the wave function ψ and the potential energy U are
generally functions of all three coordinates , x, y, and z.
The Schrödinger Equation in Three
Dimensions
To illustrate some of the features of problems
in three dimensions, we consider a particle in
three-dimensional infinity square well given by
U(x,y,z) =0 for 0<x<L, 0<y<L, and 0<z<L.
Outside this cubical region, U(x,y,z)=∞. For
this problem, the wave function must be zero at
the edges of the well.
The Schrödinger Equation in Three Dimensions
The standard method for solving this partial differential
equation is guess the form of the solution using the probability.
For a one-dimensional box along the x axis, we have found the
probability that the particle is in the region dx at x to be
A12sin2(k1x)dx
(3)
where A1 is the normalization constant, and k1=nπ/L is the
wave number. Similarly, for a box along y axis, the probability of
a particle being in a region dy at y is
A22sin2(k2y)dy
(4)
The probability of two independent events occurring is the
product of probabilities of each event occurring.
The Schrödinger Equation in Three Dimensions
So, the probability of a particle being in region dx at x and
in region dy at y is
A12 sin 2 (k1x)dx  A22 sin 2 (k2 y)dy  A12 sin 2 (k1x)  A22 sin 2 (k2 y)dxdy
The probability of a particle being in the region dx, dy,
and dz is ψ2(x,y,x)dxdydz, where ψ(x,y,z) is the solution of
equation
2
2
2

 d  d  d 
 2  2  2   U ( x, y, z )  E (2)

2m  dx
dy
dz 
2
The Schrödinger Equation in Three Dimensions
 2  d 2 d 2 d 2 
 2  2  2   U ( x, y, z )  E

2m  dx
dy
dz 
(2)
The solution is of the form
( x, y, z)  Asin(k1x) sin(k2 y) sin(k3z)dxdydz
where the constant A is determined by normalization.
Inserting this solution in the equation (2), we obtain for the
energy:
2 2
2
2
E
(k1  k 2  k3 )
2m
E
p x2  p y2  p z2
2m
which is equivalent to:
with px=ħk1 and so on.
The Schrödinger Equation in Three
Dimensions
The wave function will be zero at x=L if k1=n1π/L,
where n1 is the integer. Similarly, the wave function will be
zero at y=L if k2=n2π/L, and the wave function will be zero
at z=L if k3=n3π/L. It is also zero at x=0, y=0, and z=0. The
energy is thus quantized to the values
En1n2n3
 2 2 2
2
2
2
2
2

(n1  n2  n3 )  E1 (n1  n2  n3 )
2
2mL
where n1, n2, and n3 are integers and E1 is the ground-state
energy of the one dimensional well.
The lowest energy state (the ground state) for the
cubical well occurs when n12=n22=n32=1 and has the value
3 2 2
E1,1,1 
 3E1
2
2mL
The Schrödinger Equation in Three Dimensions
The first excited energy level can be obtained in three
different ways:
1) n1=2, n2=n3=1; 2) n2=2, n1=n3=1; 3) n3=2, n1=n2=1.
Each way has a different wave function . For example, the
wave function for n1=2, n2=n3=1 is:
2x
y
z
2,1,1  A sin
sin
sin
L
L
L
There are thus three different quantum states as described
by three different wave functions corresponding to the same
energy level. The energy level with more than one wave
function are associate is said to be degenerate. In this case,
there is threefold degeneracy.
The Schrödinger Equation in Three Dimensions
Degeneracy is related to the spatial symmetry of
the system. If, for example, we consider a noncubic
well, where U=0 for 0<x<L1, 0<y<L2, and 0<z<L3,
the boundary conditions at the edges would lead to
the quantum conditions k1L1=n1π, k2L2=n2π, and
k3L3=n3π and the total energy would be:


2m
2
En1n2n3
2
n n
n 
   
L L L 
2
1
2
1
2
2
2
2
2
3
2
3
This energy level are not degenerate if L1, L2, and L3 are all
different.
Figure shows the energy levels for the ground state and
first two excited levels for an infinity cubic well in which the
excited states are degenerated and for a noncubic infinity well in
which L1, L2, and L3 are all slightly different so that the excited
levels are slightly split apart and the degeneracy is removed.
The Degenerate States
The ground state is the state where the quantum
numbers n1, n2, and n3 are all equal to 1. Non of
the three quantum numbers can be zero. If any one
of n1, n2, and n3 were zero, the corresponding
wave number k would also equal to zero and
corresponding wave function would equal to zero
for all values of x,y, and z.
Example 1.
A particle is in three-dimensional box with
L3=L2=2L1. Give the quantum numbers n1,
n2, and n3 that correspond to the thirteen
quantum states of this box that have the
lowest energies.
Example 2.
Write the degenerate wave function for the
fourth and fifth excited states (level 5 and
6) from the Example1.
The Schrödinger Equation for Two Identical Particles
Thus far our quantum mechanical consideration
was limited to situation in which a single particle
moves in some force field characterized by a
potential energy function U.
The most important physical problem of this
type is the hydrogen atom, in which a single
electron moves in the Coulomb potential of the
proton nucleus.
The Schrödinger Equation for Two Identical Particles
This problem is actually a two-body problem,
since the proton also moves in the field of electron.
However, the motion of the much more massive
proton requires only a very small correction to the
energy of the atom that is easily made in both
classical and quantum mechanics.
When we consider more complicated problems,
such as the helium atom, we must apply the
quantum mechanics to two or more electrons
moving in an external field.
The Schrödinger Equation for Two Identical Particles
The interaction of two electrons with each
other is electromagnetic and is essentially the
same, that the classical interaction of two charged
particles.
The Schrödinger equation for an atom with
two or more electrons cannot be solved exactly, so
approximation method must be used. This is not
very different from classical problem with three or
more particles, however, the complications arising
from the identity of electrons.
The Schrödinger Equation for Two Identical Particles
There are due to the fact that it is impossible to
keep track of which electron is which.
Classically, identical particles can be identified
by their position, which can be determined with
unlimited accuracy.
This is impossible quantum mechanically
because of the uncertainty principle.
The Schrödinger Equation for Two Identical Particles
The undistinguishability of identical particles
has important consequence. For instance,
consider the very simple case of two identical,
noninteracting particles in one-dimensional infinity
square well.
The time independent Schrödinger equation
for two particles, each mass m, is
 d  ( x1 , x2 )  d  ( x1 , x2 )


 U ( x1 x2 )  E ( x1 x2 )
2
2
2m
dx1
2m
dx2
2
2
2
2
where x1 and x2 are the coordinates of the two
particles.
The Schrödinger Equation for Two Identical Particles
If the particles interact, the potential energy U
contains terms with both x1 and x2 that can not be
separated. For example, the electrostatic repulsion of
two electrons in one dimension is represented by
potential energy ke2/(x2-x1).
However if the particles do not interact, as we
assuming here, we can write U = U(x1) + U(x2).
For the infinity square well, we need only solve the
Shrödinger equation inside the well where U=0, and
require that the wave function be zero at the walls of
the well.
The Schrödinger Equation for Two Identical Particles
With U=0, equation
 2 d 2  ( x1 , x2 )  2 d 2  ( x1 , x2 )


 U ( x1 x2 )  E ( x1 x2 )
2
2
2m
dx1
2m
dx2
looks just like the expression for a two-dimensional
well
 2  d 2  d 2  d 2  


 2  U  E
2
2
2m  dx
dy
dz 
with no z and with y replaced by x2.
The Schrödinger Equation for Two Identical Particles
 2  d 2  d 2  d 2  


 2  U  E
2
2
2m  dx
dy
dz 
Solution of this equation can be written in the
form
Ψn,m= ψn(x1)ψm(x2)
where ψn and ψm are the single particle wave
function for a particle in the infinity well and n and
m are the quantum numbers of particles 1 and 2.
For example, for n=1 and m=2 the wave function
is
x1
2x2
1, 2  A sin
sin
L
L
The Schrödinger Equation for Two Identical Particles
The probability of finding particle 1 in dx1 and particle 2 in
dx2 is ψ2n,m(x1,x2)dx1dx2, which is just a product of separate
probabilities ψ2n(x1)dx1 and ψ2m(x2)dx2. However, even though
we label the particles 1 and 2, we can not distinguish which is in
dx1 and which is in dx2 if they are identical. The mathematical
description of identical particles must be the same if we
interchange the labels. Therefore, the probability density
ψ2(x2,x1) = ψ2(x1,x2)
This equation is satisfied if ψ is either symmetric or
antisymmetric:
ψ2(x2,x1) = ψ2(x1,x2),
symmetric
or
ψ2(x2,x1) = -ψ2(x1,x2),
antisymmetric
The Schrödinger Equation for Two Identical Particles
For example, the symmetric and antisymmetric
wave function for the first exited state of two
identical particles in a infinity square well:
2x 2
x2
2x1 
 x1
S  A sin
sin
 sin
sin

L
L
L
L 

and
2x 2
x2
2x1 
 x1
A  A sin
sin
 sin
sin

L
L
L
L 

The Schrödinger Equation for Two Identical Particles
There is an important difference between
antisymmetric and symmetric wave functions. If
n=m, the antisymmetric wave function is identically
zero for all values of x1 and x2 , whereas the
symmetric function is not. Thus, the quantum
numbers n and m can not be the same for
antisymmetric function.
Pauli exclusion principle:
No two electrons in an atom can have same
quantum numbers.
The Schrödinger Equation in Spherical Coordinates
In quantum theory, the electron is described by
its wave function ψ. The probability of finding the
electron in some volume dV of space is equals the
product of absolute square of the electron wave
function |ψ|2 and dV.
Boundary conditions on the wave function lead
to quantization of the wavelengths and frequencies
and thereby to the quantization of the electron
energy.
The Schrödinger Equation in Spherical Coordinates
Consider a single electron of mass m moving in
three dimensions in a region in which the potential
energy is V. The time independent Schrödinger
Equation for such a particle:
 2  d 2  d 2  d 2  


 2  V  E
2
2
2m  dx
dy
dz 
For a single isolated atom, the potential energy V
depends only on the radial distance r = √x2 + y2 + z2 .
The problem is then most conveniently treated using
the spherical coordinates.
The Schrödinger Equation in Spherical Coordinates
We will use coordinates r, θ, and φ, which related to the
rectangular coordinates x, y, and z by
z = r cosθ, x = r sinθcosφ, and y = r sinθsinφ
The Schrödinger Equation in Spherical Coordinates
 2  d 2  d 2  d 2  


 2  U  E
(1)
2
2


2m  dx
dy
dz 
The transformation of the wave term in the equation:
d 2  d 2  d 2  1 d  2 d  1 1 d 
d  1 1 d 2 
 2  2  2
r
 2
 sin 
 2
2
dx
dy
dz
r dr  dr  r sin  d 
d  r sin 2  d 2
Substitution in equation (1) gives (2):
 2 d  2 d 
2

r

2
2mr dr  dr  2mr 2
 1 d 
d 
1 d 2 
2
sin



U
(
r
)


E




2
2 
sin

d

d

sin

d





The Schrödinger Equation in Spherical Coordinates
The first step in solving this partial differential
equation is to separate the variables by writing the
wave function ψ(r,θ,φ) as a product of functions of
each single variable:
ψ(r,θ,φ) = R(r) f(θ) g(φ),
where R represent only the radial coordinate r; f
depends only of θ, and g depends only of φ. When
this form of ψ(r,θ,φ) is substituted into equation (2) the
partial differential equation can be transformed into
three ordinary differential equations, one for R(r), one
for f(θ) and one for g(φ).
The potential energy U(r) appears only in equation for
R(r), which is called the radial equation
The Schrödinger Equation in Spherical Coordinates
In three dimensions, the requirement that the wave function
be continuous and normalizable introduces three quantum
numbers, one associated with each spatial dimension. In
spherical coordinates the quantum number associated with r is
labeled n, that associated with θ is labeled ℓ, and that
associated with φ is labeled mℓ.
For the rectangular coordinates x,y, and z the
corresponded quantum numbers n1, n2, and n3 for a particle in
a three-dimensional square well were independent of one other,
but the quantum numbers associated with wave function for
spherical coordinates are interdependent.
Summary of the Quantum Numbers
The possible values of this quantum numbers are:
n = 1,2,3,…..
ℓ = 0,1,2,3,…,(n-1)
mℓ = -ℓ, (-ℓ +1),……,-2,-1,0,1,2,….,(ℓ +1),ℓ=0, ±1, ±2,…,±ℓ
The number n is called the principal quantum number. It is
associated with the dependence of the wave function on the
distance r and therefore with the probability of finding the
electron at various distances from the nucleus.
The quantum numbers ℓ and mℓ are associated with the
angular momentum of the electron and with the angular
dependence of the electron wave function.
The quantum number ℓ is called the orbital quantum number.
The magnitude L of the orbital angular momentum is related to
ℓ by
│L│ = √ ℓ (ℓ +1) ħ
Summary of the Quantum Numbers
The quantum number mℓ is called the magnetic quantum
number, it is relate to the z-component of angular momentum.
Since there is not preferred direction for the z-axis for any
central force, all spatial directions are equivalent for an isolated
atom.
However, if we will place the atom in an external magnetic
field the direction of the field will be separated out from the
other directions. If z-direction is chosen for the magnetic field
direction, than z-component of the angular momentum of the
electron is given by the quantum condition:
LZ= mℓħ
This quantum condition arises from the boundary
conditions on the azimuth coordinate φ that the probability of
finding the electron at some angle φ1 must be the same as that
of finding the electron at angle φ1+2π because these are the
same points in the space.
Summary of the Quantum Numbers
If we measure the
angular momentum of the
electron in units of ħ, we
see that the angularmomentum magnitude is
quantized to the value
√ ℓ (ℓ+1) units and that its
component along any
direction can have only
the 2ℓ + 1 values ranging
from -ℓ to +ℓ units. On
the figure we can see the
possible orientation of
angular
momentum
vector for ℓ=2.
Vector-model diagram illustrating
the possible values of z-component
of the angular momentum vector for
the case ℓ=2. The magnitude of
L=ħ√6.
The direction of the angular momentum:
If the angular momentum is characterized by the
quantum number ℓ = 2, what are the possible values of
LZ, and what is the smallest possible angle between L
and the z axis?
Summary of the Quantum Numbers
That is, n can be any positive integer; ℓ can be zero or
any positive integer up to (n-1); and mℓ can have (2ℓ+1)
positive values, ranging from -ℓ to +ℓ in integral steps.
In order to explain the fine structure and to clear up
some difficulties with explanation the table of elements Pauli
suggested that in addition to the quantum numbers n, ℓ, and
mℓ the electron should have a fourth quantum number, which
could take on just two values.
This fourth quantum number is the z-component, mz, of
an intrinsic angular momentum of the electron, called spin.
The spin vector S relate to this fourth quantum number s by
│S│ = √s(s+1) ħ and can take only two values ±½.
Quantum Theory of the Hydrogen Atom
We can treat the simplest hydrogen atom as a stationary
nucleus, a proton, that has a single moving particle, an
electron, with kinetic energy p2/2m. The potential energy U(r)
due to the electrostatic attraction between the electron and
the proton is
2
kZe
U (r )  
r
In the lowest energy state, which is the ground state, the
principal quantum number n=1, ℓ=0, and mℓ=0.
The allowed energies
En  
E0 
Z 2mk 2e4
2 2
2 n
 mk 2e4
2
2
 Z
2
 13.6eV
E0
n
2
n  1,2,3,....
Potential energy of an electron in a hydrogen atom. If the
total energy is greater than zero, as E’, the electron is not
bound and the energy is not quantized. If the total energy is
less than zero, as E, the electron is bound, than, as in onedimensional problems, only certain discrete values of the
total energy lead to well-behaved wave function.
Energy-level diagram for
hydrogen. The diagonal
lines show transitions
that involve emission or
absorption of radiation
that obey the selection
rules Δℓ = ±1, mℓ=0 or
±1. States with the same
value of n but with
different values of ℓ have
the same energy –E0/n2,
where E0=13.6 eV.
The wavelength of the
light emitted by the atom
relate to the energy
levels by
hf = hc =Ei-Ef
Energy-level diagram for
the hydrogen atom,
showing transitions
obeying the selection rule
Δl = ±1. States with the
same n value but
different l value have the
same energy, -E1/n2,
where E1=13.6 eV, as in
the Bohr theory. The
wavelength of the Lyman
α(n = 2 → n =1) and
Balmer α(n = 3 → n = 2)
lines are shown in nm.
Note that the latter has
two possible transactions
due to the l degeneracy.
Wave Function and the probability density
The ground state: In the lowest energy state, the ground
state of the hydrogen, n=1, ℓ=0, and mℓ=0, E0=13.6eV, and
the angular momentum is zero. The wave function for the
ground state is
1,0,0  C1,0,0e
Zr

a0
where
2
a0 
 0.0529nm
2
mke
is the Bohr radius and C1,0,0 is a constant that is determined
by the normalization. In three dimensions, the normalization
condition is
∫│ψ│2dV = 1
where dV is a volume element and the integration is
performed over all space.
Volume Element in Spherical Coordinates
Volume element in spherical coordinates
In spherical coordinates the volume element is
dV = (r sinθdφ)(rdθ)dr = r2 sinθdθdφdr
We integrate over φ, from φ=0 to φ=2π; over θ, from θ=0 to
θ=π; and over r from r=0 to r=∞. The normalization conditions
is thus
 2


 
2
2 2
  dV  0 0  0  r sin d d dr 


 2 Zr
    2


      C12, 0, 0 e a0 r 2 sin d d dr  1



0 0  0
 

Since there is no θ or φ dependence in Ψ1,0,0 the triple
integral can be factored in a product of three integrals
This gives

2 Zr

 2  
 



2
a0
2
2 
 dV   d  sin d  C1,0,0e
r dr  




 0
 0
 0




 2 Zr



2 
2 a0
 2  2  C1,0,0  r e
dr   1
0




The remaining integral is of the form

n  ax
x
 e dx
0
with n a positive integer and with a>0.
This integral can be looked up in a table of integrals

x e
n  ax
0
n!
dx  n 1
a
so

2 Zr
a0
2
r e
0
3
0
a
dr 
4Z 3
Than
4C
2
1, 0, 0
 a 
 3   1
 4Z 
3
0
so
C1, 0, 0
1 Z
 

  a0 
3
2
Probability Densities
The normalized ground-state wave function is thus
3
1,0,0
1 Z
  e

  a0 
2

Zr
a0
The probability of finding the electron in a volume dV is
|ψ|2dV
Computer generated probability density |ψ|2 for the ground state
of the hydrogen. The quantity -e |ψ|2 can be though of as the
electron charge density in the atom. The density is spherically
symmetric (it depends only on r and independent of θ or φ), is
greatest at the origin , and decrease exponentially with r.
Probability Densities
We are more often interested in the probability of finding
the electron at some radial distance r between r and r+dr. This
radial probability P(r)dr is |ψ|2dV, where dV is the volume of
the spherical shell of thickness dr, which is dV=4πr2dr.
The probability of finding the electron in the range from r to
r+dr is thus
P (r )dr   4r 2 dr
2
and the radial probability density is
P(r )  4r 
2
2
Probability Densities
For the hydrogen atom in the ground state, the
radial probability density is
2 Zr

a0
2
P(r )  4r   4C1, 0, 0 r e
2
2
3
Z 2
 4  r e
 a0 

2 Zr
a0
Radial Probability Density
Radial probability density
P(r) versus r / a0 for the
ground state of the hydrogen
atom. P(r) is proportional to
r2Ψ2. The value of r for
which P(r) is maximum is the
most probable distance r=a0,
which is the first Bohr radius.
Radial probability density P(r) vs. r/a0 for the n=2 states
in hydrogen. P(r) for l=1 has a maximum at the Bohr
value 22a0. For l = 0 there is a maximum near this value
and a smaller submaximum near the origin. The markers
on the r/a0 axis denote the values of (r/a0).
P(r) vs. r/a0 for the n = 3 state in hydrogen.
Probability density Ψ*Ψ for the n=2 states in hydrogen. The
probability is spherically symmetric for l=0. It is proportional to
cos2θ for l=1, m=0, and to sin2 for l=1, m=±1. The probability
densities have rotational symmetry about the z axis. Thus, the
three-dimensional charge density for l=1, m=0 state is shaped
roughly like a dumbbell, while that for the l=1, m=±1 states
resembles a doughnut, or toroid. The shapes of these
distributions are typical for all atoms in S states (l=0) and P
states (l=1) and play an important role in molecular bounding.
A particle moving in a
circle has angular
momentum L. If the
particle have a positive
charge, the magnetic
moment due to the
current is parallel to L.
Bar-magnet model of
magnetic moment.
(a) In an external
magnetic field, the
moment experiences a
torque which tends to
align it with the field. If
the magnet is spinning
(b), the torque caused
the system to precess
around the external
field.
Example: Probability that electron is in a thin
spherical shell
Find the probability of finding the electron in a thin
spherical shell of radius r and thickness Δr=0.06a0
at (a) r=a0 and (b) r=2a0 for the ground state of the
hydrogen atom.
The spin-orbit effect and fine structure
The total angular momentum of an electron in
an atom is a combination of the orbital angular
momentum and spin angular momentum. It is
characterized by the quantum number j, which can
be either │ℓ - ½│ or │ℓ + ½│. Because of
interaction of the orbital and spin magnetic
moments, the state j = │ℓ - ½│ has lower energy
than the state j = │ℓ + ½│, for ℓ ≥1. This small
splitting of the energy states gives rise to a small
splitting of the spectral lines called fine structure.
The Table of Elements
We can treat the simplest hydrogen atom as a stationary
nucleus, a proton, that has a single moving particle, an
electron, with kinetic energy p2/2m. The potential energy U(r)
due to the electrostatic attraction between the electron and
the proton is
2
kZe
U (r )  
r
In the lowest energy state, which is the ground state, the
principal quantum number n=1, ℓ=0, and mℓ=0.
The allowed energies
mk 2 e 4
2 E0
En  2 2   Z 2
2 n
n
 mk 2 e 4
E0 
 13.6eV
2
2
n  1,2,3,....
The Table of Elements
For atoms with more than one electron, the
Schrödinger equation cannot be solved exactly.
However, the approximation methods allow to
determine the energy levels of the atoms and
wave functions of the electrons with high
accuracy.
As a first approximation, the Z electrons in
an atom are assumed to be noninteracting. The
Schrödinger equation can then be solved, and
the resulting wave function used to calculate the
interaction of the electrons.
The Table of Elements
The state of each electron in an atom is described by four
quantum numbers n,l,m, and ms.
Beginning with hydrogen, each larger neutral atom adds
one electron. The electrons go into those states that will give
the lowest energy consistent with the Pauli exclusion principle:
No two electrons in an atom can have the same set of
values for the quantum numbers n, ℓ, m, and ms
The energy of the electron is determined mainly by the
principal quantum number n, which is relate to the radial
dependence of the wave function, and by the orbital angularmomentum quantum number ℓ.
The dependence of the energy on ℓ is due to the
interaction of the electrons in the atoms with each other.
The Table of Elements
The specification of n and ℓ for each electron in an atom is
called the electron configuration.
The ℓ values are specified by a code:
s p d f g h
ℓ values
0 1 2 3 4 5
The n values are referred as shells, which are identified by
another letter code:
shell
K L M N …..
n values
1 2 3 4 …..
Using the exclusion principle and the restriction of the
quantum numbers (n is a positive integer, ℓ ranged from
0 to n-1, m changed from -ℓ to ℓ in integral steps, and ms can
be either +½ or -½), we can understand much of the structure
of the periodic table.
The Periodic Table
The energy required to remove the most loosely
electron from an atom in the ground state is called the
ionization energy. This energy is the binding energy of the
last electron placed in the atom. The ionization energy
can be found from:
En  
2
Z eff
n
2
E0  
2
Z eff
n
2
(13.6eV )
Hydrogen (Z = 1): n=1, ℓ = 0, m = 0, ms = ±½ - 1s
Helium (Z = 2): two electrons, in the ground state both
electrons are in the K shell,
n=1, ℓ=0, m=0,
ms1=+½, ms2=-½ - 1s2
Lithium (Z=3): K shell (n=1) is completely full, one electron
on the L-shell – 2p1
States of Hydrogen Atom
n
1
ℓ
0
0
1
0
1
2
m
0
0
0,±1
0
0 ,±1
0, ±1, ±2
ms
±
½
2
±½
±½
±½
±½
±½
2
6
2
6
10
Sub
shell
Total
States
2
2
8
3
18
The Structure of Atom
The electrons in atom that have same
principal quantum number n form an
electron shell:
total
K
n=1
2
L
n=2
8
2n2
M
n=3
18
N
n=4
32
O
n=5
50
}
Depending from orbital quantum number ℓ the
electrons forms subshells:
n Shell
Number of electrons in subshell
s
p
d
f
g
(ℓ=0) (ℓ =1) (ℓ =2) (ℓ =3) (ℓ =4)
2
-
Total
elect.
number
1
K
2
2
L
2
6
-
-
-
8
3
M
2
6
10
-
-
18
4
N
2
6
10
14
-
32
5
O
2
6
10
14
18
50
Distribution Electrons in atoms
Z
Element
K
L
M
1s
2s 2p
3s3p3d
nℓZ
1
H
1
-
-
1s
2
He
2
-
-
1s2
3
Li
2
1
-
1s2, 2s
4
Be
2
2
-
1s2, 2s2
5
B
2
2
1
1s2, 2s2, 2p
6
C
2
2
2
1s2, 2s2, 2p2
7
N
2
2
3
1s2, 2s2, 2p3
8
O
2
2
4
1s2, 2s2, 2p4
9
F
2
2
5
1s2, 2s2, 2p5
10
Ne
2
2
6
1s2, 2s2, 2p6
For example the structure for oxygen,
O, 1s2, 2s2, 2p4 – it mean that 2 electrons
are in the state with n=1 and ℓ=0;
2 electrons in the state with n=2 and ℓ=0;
and 4 electrons in the state with n=2 and
ℓ=1.
1. Effective Nuclear Charge for an Outer
Electron
Suppose the electron cloud of the outer electron in
the lithium atom in the ground state were completely
outside the electron clouds of the two inner electrons,
the nuclear charge would be shielded by the two inner
electrons and the effective nuclear charge would be
Z'e=1e. Then the energy of the outer electron would
be –(13.6eV)/22=-3.4eV. However, the ionization
energy of lithium is 5.39eV, not 3.4eV. Use this fact to
calculate the effective nuclear charge Zeff seen by the
outer electron in lithium.
2. The Effective Charge of the Rb Ion
The 5s electron in rubidium sees an effective
charge of 2.771e. Calculate the ionization
energy of this electron.
3. Determining Zeff experimentally
The measured energy of a 3s state of sodium is
-5.138eV. Calculate the value of Zeff.
Example
The double charged ion N+2 is formed by removing
two electrons from a nitrogen atom. (a) What is the
ground state electron configuration for the N+2 ion?
(b) Estimate the energy of the least strongly bond level
in the L shell of N+2.
The double charged ion P+2 is formed by removing
two electrons from a phosphorus atom. (c) What is the
ground-state electron configuration for the P+2 ion?
(d) estimate the energy of the least strongly bound
level in the M shell of P+2.
Electron Interaction Energy in Helium
The ionization energy for helium is 24.6 eV.
(a) Use this value to calculate the energy of
interaction of the two electrons in the ground state
of the helium atom.
(b) Use your result to estimate the average
separation of the two electrons.
Angular Momentum of the Exited Level Of Hydrogen
Consider the n=4 state of hydrogen. (a) What is
the maximum magnitude L of the orbital angular
momentum? (b) What is the maximum value of
LZ? (c) What is the minimum angle between L and
Z-axis? Give your answers to (a) and (b) in terms
of  .
A Hydrogen Wave Function.
The ground–state wave function for the hydrogen (1s
state) is:
1s (r ) 
1
a 3
e

r
a
(a) Verify that this function is normalized. (b) What is
the probability that the electron will be found at a
distance less than a from the nucleus?
Atomic Spectra
Atomic spectra include optical spectra and X-ray
spectra. Optical spectra result from transmissions
between energy levels of a single outer electron
moving in the field of the nucleus and core
electrons of the atom.
Characteristic X-ray spectra result from the
excitation of a inner core electron and the
subsequent filling of the vacancy by other electrons
in the atom.
Selection Rules
Transition between energy states with the
emission of a photon are governed by the
following selection rules
Δmℓ = 0 or ±1
Δℓ = ±1