CH 28 – Atomic Physics
Our present understanding of the atom is that it consists of a nucleus consisting of
protons and neutrons with electrons in orbits about the nucleus. The neutron has no
charge and the number of protons and electrons are the same so that the atom has no net
charge. The diameter of the nucleus is much smaller than the diameter of the atom, but
the nucleus contains most of the mass of the atom.
This model of the atom was not accepted until the early part of the 1900s. An earlier
model proposed by J.J. Thomson did not have a nucleus. Instead, it had electrons
embedded in a cloud of positive charge. The model is sometimes called the ‘plum
pudding model’, where the pudding was the positive charge and the plums the electrons.
The nucleus was first discovered by Ernest Rutherford. He did an experiment in which
alpha particles were projected against a thin metal foil and the deflection of the alpha
particles by the foil was measured. The experiment showed that most of the alpha
particles traveled through the foil with very little deflection. However, some of the alpha
particles were deflected by large angles, and some were even deflected backwards. If the
atoms were like a plum pudding, then the scattering would be more uniform and none
would be expected to be completely deflected backwards. Rutherford’s interpretation
was that the atom must have a massive, positively charged nucleus.
Bohr Theory of the Hydrogen Atom
If a tube is filled with a gas such as hydrogen, helium, or neon at low pressure and a high
voltage is applied between two electrodes in the gas, then a current will be produced and
the gas will emit light. If this light is examined using a prism or diffraction grating, it can
be seen to consist of discrete wavelengths. These spectral lines are evidence of quantized
energy levels in the atoms of the gas.
The first successful quantum theory of the atom was developed by Niels Bohr to explain
the hydrogen spectrum. The basic assumptions in the Bohr model of the hydrogen atom
are as follows:
1. The electron orbits the proton under the influence of the Coulomb force between
the two charges. This is much like the moon orbiting the earth under the
influence of the gravitational force.
2. Certain orbits are stable and no radiation is emitted while the electron is in these
orbits. (Classically, a charge moving in a circle would radiate electromagnetic
energy.)
3. If an electron jumps from one orbit to another, the energy difference would appear
as an emitted photon. That is,
Ei  E f  hf
(1)
1
4. The angular momentum of the orbiting electron can only have certain discrete
values given by
mvr  n , n  1, 2,3, ...
(2)
where   h / 2 .
The above assumptions lead to quantized values for the energy and radius of orbit.
The force on the electron is given by
F k
e2
m
r2
v2
r
(3)
By multiplying each side of this equation by r/2, we get the kinetic energy 1 mv 2
2

ke2
2r
(4)
Combining this equation with the equation for the quantization of the angular momentum,
we can algebraically solve for r and get
rn 
n 2 2
mke2
, n  1, 2, 3, ...
(5)
Eq. (5) gives the allowed orbits of the electron. The lowest orbit obtained for n = 1 is
called the Bohr radius ao and is
ao 
2
2
 0.0529 nm
(6)
mke
We can write the allowed radii in terms of the Bohr radius as
rn  n 2 a0 , n  1, 2, 3, ...
(7)
The four lowest radii are
r1  a0  0.053 nm
r2  4a0  0.21 nm
r3  9a0  0.48 nm
r4  16a0  0.85 nm
2
The energy of the electron is
E  KE  PE 
1 mv 2
2
e2
k
r
(8)
Using the expression for KE from Eq. (4), we can write the energy as
E
ke2 ke2
ke2


2r
r
2r
(9)
Using the value of radius from Eq. (5), we get
En  
mk 2 e 4  1 
  , n  1, 2, 3, ...
2 2  n 2 
(10)
Eq. (10) gives the allowed values of the energy of the hydrogen atom. The negative sign
means that the electron is bound the proton. If E  0, then the electron is unbound.
Numerically, we can write the energy as
En  
13.6ev
n2
, n  1, 2, 3, ...
(11)
The four lowest energy values are
13.6ev
 13.6 ev
1
13.6ev

 3.40 ev
4
13.6ev

 1.51 ev
9
13.6ev

 0.85 ev
16
E1  
E2
E3
E4
The ground state energy (n = 1) is -13.6 ev. This means that it would take 13.6 ev to
remove the electron from the atom. This is the ionization energy.
3
The allowed energies are shown in the energy level diagram below.
When the atom is in an excited state (n > 1), it can make transitions to a lower level and
lose its energy by emitting a photon with frequency
f 
Ei  E f
h

mk 2 e 4  1
1 

4  3  n f 2 ni 2 
The inverse wavelength is
1


where RH 
 1
f mk 2 e 4  1
1 
1 


 RH 

,
n 2 n 2 
c 4 c 3  n f 2 ni 2 
i 


 f
mk 2 e 4
4 c
3
 1.097 x10 7 m 1
is the Rydberg constant.
Conversely, a hydrogen atom in a lower energy level can absorb a photon and make a
transition to a higher level if the energy of the photon is equal the energy difference
between the two levels.
4
It is conventional to classify the emission lines according to the final state, nf, in the
transition as shown in the energy level diagram. The series are given as follows.
nf = 1:
Lyman series
nf = 2:
Balmer series
nf = 3:
Paschen series
etc.
The emission spectra can be observed using a prism or grating spectrometer. Only the
Balmer series would have wavelengths in the visible part of the spectrum. These visible
lines would correspond to transitions from states 3  2, 4  2, 5  2, and 6  2. All
other transitions would be in the ultraviolet or infrared region. The spectral lines for the
Balmer series are shown below. The wavelengths of the measured spectral lines for
hydrogen are in good agreement with those calculated by the Bohr theory.
Example:
Calculate the wavelength corresponding to the transition ni = 3 to nf = 2.
Solution:
 1
1 
 ( 1.097 x10 7 / m )  

 2 2 32 
1
  6.56 x10  7 m  656 nm ( red )
5
Example:
What is the shortest wavelength in the emission spectrum of hydrogen?
Solution:
This would correspond to ni =  to nf = 1.
1
1 
  1.097 x10 7 / m
 ( 1.097 x10 7 / m ) 
2
2

 
1
  91 nm ( uv )
1
The Bohr Theory assumes one orbital electron. Of course, the only one-electron atom is
hydrogen. However, the model would also apply to singly ionized helium, He+, doubly
ionized lithium, Li++, etc., which also have only one orbital electron but for which the
atomic number Z is greater than one. We would need to modify the model to account for
the fact that the charge of the nucleus is Ze and replace e2 with (Ze)(e) = Ze2 in the
Coulomb force and energy terms. This would then give the following for the energy,
emission wavelength, and radius of orbit.
En  
Z 2 ( 13.6 ev )
mk 2 Z 2 e 4  1 
   
, n  1, 2, 3, ...
2 2  n 2 
n2
mk 2 Z 2 e 4  1
1 


,

4 c 3  n f 2 ni 2 
1
rn 
n 2 2
mkZe2

n 2 a0
, n  1, 2, 3, ...
Z
Example:
What is the energy required to remove the ‘last’ electron from lithium?
Solution:
For lithium, Z = 3. So
E | E1 | Z 2 ( 13.6ev )  ( 3 )2 ( 13.6ev )  122 ev
6
Example:
What is the radius of orbit of doubly-ionized lithium?
Solution:
a
0.053nm
r1  0 
 0.018 nm
Z
3
De Broglie Waves and the Bohr Model
One of the assumptions in the Bohr model is that angular momentum is quantized:
mvr  n , n  1, 2,3, ...
This condition is equivalent to assuming that the
orbiting electron is a wave with momentum p = h/
and that there are an integral number of wavelengths
in one complete orbit.
circumference  2 r  n 
mvr 
nh nh

p mv
nh
2
Hydrogen Atom Wave Function and Spatial Quantum Numbers
The Bohr Theory is very successful in predicting the energy levels of the hydrogen atom.
It also suggests that the electron has a wave-like nature as described above. However, it
does not completely describe other quantum effects and it does not sufficiently describe
the wave nature of the electron. In addition, the Bohr Theory does not accurately
describe the nature of multi-electron atoms.
In 1926 Erwin Schrödinger published a paper in which he described a differential
equation which could be used to solve for the wave function  of the hydrogen atom.
The physical meaning of  is that its square gives the probability of finding the particle.
Schrödinger’s equation is not limited to the hydrogen atom, but can in principle be used
to determine the wave function of any physical system. In the case of the hydrogen atom,
meaningful solutions for  only occur for certain quantum numbers n, l, and ml. The
quantum number n is called the principal quantum number, l is the orbital quantum
number, and ml is the orbital magnetic quantum number. The allowed values of these
‘spatial’ quantum numbers are
7
n = 1, 2, 3, …
l = 0, 1, 2, …, n-1
ml = - l, - l + 1, …, l - 1, l
For each value of n, there are n allowed values of l, and for each value of l there are
2 l+1 allowed values of ml.
Schrödinger’s equation predicts that the energy of the hydrogen atom is the same as given
by the Bohr Theory, that is
mk 2 e 4  1 
 
En  
2 2  n 2 
It predicts that the magnitude of the angular momentum is
L  (   1 )
and the component of the angular momentum along an axis is
Lz  m 
The angular momentum given by the Schrödinger equation is different from the Bohr
Theory, which is L  n . Also, in the Bohr Theory Lz is not quantized.
Example:
A hydrogen atom is in a state with principal quantum number n = 2. What are the
allowed values of the magnitude and z-component of the angular momentum?
Solution:
For n = 2, the allowed values of l are 0 and 1. So,
  0, m  0 : L  0(0  1)  0, L z  (0)  0
  1, m  0,  1 : L  1(1  1)  2, Lz  0,  
In the Bohr model, L  n  2 and Lz could have any value between -2 and 2.
Schrödinger’s equation does not predict discrete orbits, as in the Bohr Theory. Rather the
wave function suggests that the electron can be envisioned as a ‘cloud’ surrounding the
nucleus, with the density of the cloud giving the probability of finding the electron. The
8
electron cloud and the most probable location of the electron depend on the quantum
numbers n, l, and ml. The electron cloud and the probability of finding the electron as a
function of radius are given below for the lowest state (n, l, ml = 1, 0, 0).
While the electron has a chance of being found anywhere, the most probable radius for
the (1, 0, 0) state is the same as the Bohr radius. All states for which l = 0 have
spherically symmetrical wave functions, somewhat like that shown above for n = 1, and
their most probable radius is the same as given in the Bohr Theory; that is, r  n 2 a0 .
Those states for which l > 0 do not have spherical wave functions.
Electron Spin
Schrödinger’s equation predicts that the electron has an orbital angular momentum. In
addition, the electron has an intrinsic ‘spin’ angular momentum. Although this is not a
classical concept, the spin of the electron is analogous to the angular momentum of the
earth as it rotates about its axis. The electron spin angular momentum has a fixed
magnitude and can be measured either ‘up’ or ‘down’. The up or down component has
the value  ½ . In general, we can write
S z  m s , m s   1
2
Allowed Quantum States of the Hydrogen Atom
For each set of spatial quantum numbers n, l, ml we can have ms =  ½. The first 28
allowed states are summarized in the following table
9
n
1
2
3
l
0
0
1
0
1
2
ml
0
0
-1,0,1
0
-1,0,1
-2,-1,0,1,2
ms
½
½
½
½
½
½
no. l
2
2
6
2
6
10
no. n
2
total no.
2
8
10
18
28
Shell and Subshell Notation
Those states which have the same principal quantum number n are said to lie in the same
‘shell’. The shells corresponding to n = 1, 2, 3, 4, … are referred to as the K, L, M, N,
…shells. So, there are 2 states in the K shell, 8 states in the L, shell, 18 states in the M
shell, etc.
Those states which have the same orbital quantum number l are said to lie in the same
‘subshell’. The subshells corresponding to l = 0, 1, 2, 3, … are referred to as s, p, d, f, ….
There are 2 states in the s subshell, 6 states in the p subshell, 10 states in the d subshell,
etc.
Shorthand notation for describing a specific subshell corresponding to a given values of n
and l is as follows:
n, l = 1, 0:
= 2, 0:
= 2, 1:
= 3, 0:
= 3, 1:
= 3, 2:
etc.
1s
2s
2p
3s
3p
3d
Multi-electron Atoms
In principle the Schrödinger equation can also be used to solve for the wave function,
energies, and angular momentum of atoms that contain more than one electron. The
problem is much more complex than for hydrogen because of the interactions of the
electrons with each other. These interactions include both Coulomb interactions and
magnetic interactions. The magnetic interactions occur since there are magnetic
moments associated with the orbital and spin angular momentum of the electrons. These
interactions give different energy values than would be obtained by treating each electron
independently. However, it is found that the states of the electrons can still be described
using the four quantum numbers (n, l, ml, ms).
10
The Pauli Exclusion Principle states that no two electrons in an atom can have the same
set of quantum numbers. So, we could imagine building up a multi-electron atom by
filling up the available states given in the table above in order of increasing energy.
So, in the ‘ground’ state, the two electrons in helium would have the quantum numbers
(n, l, ml, ms) = (1, 0, 0, ½) and (n, l, ml, ms) = (1, 0, 0, -½). We describe the ‘electronic
configuration’ of helium as 2s2, where the superscript refers to the number of electrons
which occupy the 2s subshell. Likewise, for lithium, we would have (n, l, ml, ms) = (1, 0,
0, ½), (1, 0, 0, -½), and (2, 0, 0, ½), and the electronic configuration would be 1s22s1.
The table below gives the electronic configurations of the elements up through Kr. The
states are filled in order of increasing quantum number through Ar. The outer electron
for K, however, is 4s1 instead of 3d1. This is because the interactions between electrons
cause the 4s1 state to have lower energy than the 3d1 state. There are other departures in
the filling sequence for higher Z numbers for a similar reason.
Z
1
2
3
4
5
6
7
8
9
10
11
12
Sym
bol
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Elect.
Conf.
1s1
1s2
[He]2s1
2s2
2s22p1
2s22p2
2s22p3
2s22p4
2s22p5
2s22p6
[Ne]3s1
3s2
Z
13
14
15
16
17
18
19
20
21
22
23
24
Sym
bol
Al
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Elect.
Conf.
3s23p1
3s23p2
3s23p3
3s23p4
3s23p5
3s23p6
[Ar]4s1
4s2
3d4s2
3d24s2
3d34s2
3d54s1
Z
25
26
27
28
29
30
31
32
33
34
35
36
Sym
bol
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
Elect.
Conf.
3d54s2
3d64s2
3d74s2
3d84s2
3d104s1
3d104s2
3d104s24p1
3d104s24p2
3d104s24p3
3d104s24p4
3d104s24p5
3d104s24p6
Energy Levels of Multi-electron Atoms
The calculation of the energy levels of multi-electron atoms is complex because of the
interactions between the electrons. However, in some cases the Bohr model can be used
to estimate the energy levels. One example would be the excited states of the Alkali
metals such as Na or K. The electronic configuration of Na, for example, is 1s22s22p63s1.
The first 10 electrons form closed shells like that of Ne, and the 3s1 valence electron is
loosely bound compared to the inner electrons. When the 3s1 electron is in an excited
state, then it ‘sees’ a spherically symmetric charge distribution consisting of a nucleus
with charge +11e shielded by an electronic charge distribution with charge -10e. Thus,
the effective nuclear charge is +e, the same as for hydrogen. Indeed, the energy levels of
Na in the excited states are similar to that of hydrogen in its excited states.
11
Similarly, a core electron, say, in a K shell (n = 1), ‘sees’ a nuclear charge +Ze that is
shielded by approximately one electronic charge –e due to the other electron in the shell.
So, the effective nuclear charge is Zeff = (Z-1)e. Thus, the energy of the K electrons can
be estimated from the Bohr model to give
En  
E1  
Z 2 ( 13.6 ev )
n2
Z eff 2 ( 13.6 ev )
(1 )
2
 ( Z  1 ) 2 ( 13.6 ev )
Example:
Estimate the energy of a K shell electron in Cu.
Solution:
For Cu, Z = 29. So
E1  (29  1) 2 (13.6 ev)  10,662 ev
So, the energy required to extract a core electron from Cu is huge compared to the energy
to ionize hydrogen.
X-rays are produced in x-ray tubes when high-energy electrons strike a metal target in the
tube. One of the sources of the x-rays results from electronic transitions between core
levels. The high energy electron can penetrate the core of the atom and dislodge one of
the inner core electrons. Then an electron in an upper level makes a transition to this
lower level and loses its energy in the form of a high energy photon - an x-ray.
Example:
What is the approximate energy of an electron in the M (n = 3) shell of Cu when there is
a vacancy in the K shell?
Solution:
The M shell electron would be shielded by one electron in the K shell and 8 electrons in
the L shell. So, the effective nuclear charge would be Zeff e = (Z-9)e. Thus, for Cu
E3  
( 29  9 ) 2 ( 13.6 ev )
( 3 )2

( 20 ) 2 ( 13.6 ev )
 604 ev
9
12
Example:
What would be the wavelength of the photon if an electron in Cu makes a transition
between the M shell and a vacancy in the K shell?
Solution:
From the above,
E  E3  E1  604  ( 10662 )  10058 ev 

hc

hc ( 6.6 x10  34 J  s )( 3x108 m / w )

 1.23 m  0.123 nm
E
( 10058ev )( 1.6 x10 19 ev / J )
Energy Levels in Solids
If atoms are brought together to form a solid, then the interactions between the atoms
causes the discrete electronic energy levels of the atoms to split into closely spaces levels
grouped together in bands. The number of levels in a band is of the order of the number
of atoms in the solid, so the energy levels in a band are so closely spaced as to be nearly
continuous. In general, the bands are separated by energy gaps. The figure below
illustrates this splitting of the 1s and 2s levels as two, five, and then a very large number
of atoms are brought together.
13
The nature of the upper most bands determines whether a solid is a conductor, an
insulator, or a semiconductor. In order for a solid to conduct, there must be empty states
into which the electrons can move.
If the upper band is partly filled, then there are empty states which can be easily accessed
and the solid is a conductor. A partially filled band will exist if a filled band overlaps
with an empty band.
If all bands are either completely filled or completely empty, then the solid may be either
an insulator or a semiconductor, depending on the energy gap Eg between the filled
valence band and the empty conduction band. If Eg is sufficiently small, then the solid
will be a semiconductor. In this case, electrons in the valence band can be thermally
activated to occupy levels in the bottom of the conduction band, leaving empty states
(holes) in the top of the valence band. Conduction can then take place in both the
conduction band and in the valence band. The electron conduction in the valence band is
equivalent to the holes moving in a direction opposite to the conduction electrons. The
holes behave as positive electronic charges. Common semiconductors are silicon and
germanium. For silicon, Eg = 1.1 ev and for germanium Eg = 0.66 ev.
If the energy gap is very large, then no thermal activation of electrons from the valence
band into the conduction band can take place and the solid is an insulator. Diamond is a
very good insulator and has an energy gap of about 5.5 ev.
14