Download Interaction of Photons with Matter

Atomic Physics
The atom consists of a central small, massive nucleus made
of Z protons and N neutrons surrounded by a cloud of Z
very light electrons.
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Units
1 Angstrom = 10-10 m
1 femtometer = 1fm = 10-15m
Mp ~ 1.672648 10-27 kg
Mn ~ 1.674928 10-27 kg
Me ~ 9.1094 10-31 kg
As an alternative to kg a more appropriate unit of mass is the ATOMIC
MASS UNIT (u): 1u = 1/12 mass of a carbon atom = 1.66054 10-27 kg
{12 gms (the atomic weight in gms) of carbon contain Avogadro’s
number, NA ~ 6.022 1023 atoms of carbon. Therefore 1u = 1gm/NA.}
Mp~ 1.0073 u
Mn~ 1.0087 u
Me ~ 0.00055 u
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The usual unit of Energy is the Joule = 1 kg m2sec-2. = 1 ampere Volt
Second. This again is a very large unit.
There is another set of units that is also used to describe Energy. As some
particles are charged ( the proton, electron) they can be accelerated
through a potential difference giving them kinetic energy.
The charge on the electron ( or proton) is 1.6 10-19 Coulomb. Therefore
when accelerated through 1 Volt an energy of
1 electron-Volt (eV) = 1.6 10-19 Coulomb Volt = 1.6 10-19 Ampere Sec
Volt =
1.6 10-19 Joules is achieved.
This and the KeV (1000 eV) and the MeV (106 eV) is used as an
alternative to the Joule.
With these energy units - and Einstein’s insight that E = mc2 - different
mass units follow - MeV/c2.
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For instance a mass of 1u has energy value:
1.66054 10-27 kg  (~3 108)2 m2sec-2 = 1.494486 10-10 joules
~ 0.934 103 MeV.
Therefore 1u ~ 934 MeV/c2
Using an accurate value for c:
Mp = 938.279 MeV/c2
Mn = 939.5731 MeV/c2
Me = 0.511 MeV/c2.
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How do we know the size of the
nucleus is a few fm?
I = Number of s incident per second in area Ab cm2
N() = Number of elastically scattered s detected per second
t = thickness of Au foil in cm, d = distance of detector from foil (cm)
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A = area of detector in cm2.
How many gold nuclei can the beam interact with?
In 197 gms of gold there are Avogadro’s number (NA~ 6 1023) of gold
nuclei. Therefore in 1 gm there are NA/197 nuclei.
If the density of gold is  gms/cm3 there are NA  /197 nuclei/cm3.
The gold foil is thin ( ie the beam goes through it). Therefore the number
of nuclei the beam can interact with is:
(NA  t/197)Ab nuclei.
Let the effective cross-sectional area of a gold nucleus the beam scatters
from be  cm2.
Then the TOTAL number of scatters per second - IN ALL DIRECTIONS
- is:
I  (NA  t/197)Ab  (/Ab) = I NA  t/197
where (/Ab) is the probability of an alpha to scatter in any direction per
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second.
Now the probability to scatter anywhere is the sum of probabilities of
scattering at every angle between 0 and 180 degrees.
We define a quantity - unit solid angle, 1 steradian (sr) - for the s to
scatter into.
If the probability for scattering into unit solid angle is the same
at all angles, say k/Ab, then:
 = 4 k
as a sphere has a solid angle 4 = 4 d2/ d2.
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The number of scatters per second into a detector at angle  would be:
N( ) = I  (NA  t/197)Ab  (k/Ab)  A/d2
where k is in cm2/sr or barns/sr or millibarns/sr (mb/sr).
1 barn = 10-24 cm2
In fact the probability is not a constant. On the basis of scattering by a
Coulomb potential Lord Rutherford calculated the probability of
scattering at angle  into unit solid angle to be:
1.296 (2Z/T)2 cosec4(/2)/ Ab (mb/sr)/cm2
where T is the beam energy (in MeV), Z is the atomic number of the
target ( e.g. Z=92 for gold).
Hence:
N( ) = I  (NA  t/197)Ab  1.296 (2Z/T)2 cosec4(/2)/Ab 10-27 A/d2
=
I  (NA  t/197)  1.296 (2Z/T)2 cosec4(/2)  10-27  A/d2
Units
(NA  t/197)
1.296 (2Z/T)2 cosec4(/2)  10-27
A/d2
cm-2
cm2/sr
sr
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The quantity 1.296 (2Z/T)2 cosec4(/2) mb/sr is called the
differential cross-section, d/d() - the small fraction of the
cross-section,  associated with the scattering into unit solid
angle at angle .
This quantity can be determined by measuring N() and I, and
knowing NA,, Z and T.
We find:
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d/d
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At the energy at which structure due to nuclear scattering, rather than
Coulomb scattering, occurs the alpha particles’ distance of closest
approach to the nucleus d = R + Rnucleus.
How do we work out the distance of closest approach?

Nucleus
The kinetic energy of the  beam is: T = 1/2 Mv2
The potential energy is :V = 2Ze2/r
The Total Energy E = 1/2 Mv2 +2Ze2/r
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At the point of closest approach:
T = 0
V=2Ze2/d
Therefore E = 2Ze2/d.
At large distances r  . Therefore: V = 0 and E=1/2Mv2
Therefore as E is a constant,
2Ze2/d = 1/2Mv2
At the beam energy where the Rutherford Scattering calculation starts
to fail:
d = R + Rnucleus.
Looking at many nuclei of different atomic species we find :
Rnucleus = r0 A1/3
where r0 = 1.2 fm and A = N + Z - the number of neutrons and protons
in a particular nucleus.
So the radius of even a heavy nucleus - say the stable lead nucleus
208 Pb containing 208 protons and neutrons, 82 protons and 120
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neutrons only has a radius of ~ 7 fm.
So a typical atom - lets say lead Pb - has a tiny massive nucleus a few
fm radius with an electric charge of + 82e where e = 1.6 10-19 coulomb,
the magnitude of the charge on a proton and on an electron.
To make the atom electrically neutral surrounding this nucleus at
distance up to several angstroms ( 10-10 m) is a cloud of - in the case of
lead - 82 electrons.
How are these electrons distributed in space and in energy?
Atomic spectra
When an electric discharge is passed through a gas, light is emitted.
When this is dispersed by a prism a spectrum of different colours is
observed. As light is an electromagnetic wave each colour is associated
with both a wavelength and a frequency.
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Light wave, velocity c m/sec, wavelength  m, frequency  sec-1
c m/sec
m
(a)
Frequency is the number of wavelengths that pass a point - say point (a) per sec.
= c / 
Associated with frequency is energy: E = h 
where h is Planck’s constant :
h = 4.136 10-21 MeV sec
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The first spectrum was observed in the visible region for light
emitted by hydrogen atoms:
It is observed that: k = 1/ = RH(1/22 -1/n2)
n = 3,4,5 ….
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Later other series were observed:
From recent
~ 1.1 107 m-1
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Other elements also exhibit series, but they overlap and are much more
difficult to sort out. However they are all of the form:
1/ = RH(1/(m-a) 2 -1/(n -b)2)
where a and b are constants for a particular series. (For hydrogen a=b=0).
How do we explain these series?
The Bohr Model
L = mvr = nh/2 (1)
 = (Ei - Ef)/h (2)
n is called a ‘quantum number’.
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In the hydrogen atom, the Coulomb attractive force between the proton
and the electron is basically:
e2/r2 = ma
where a is acceleration of the electron towards the proton. For circular
motion, a = v2/r i.e. e2/r2 = mv2/r
(3)
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Combining (1) with (3) we get: r = n2h2/42me2 (4)
for the radius of the nth orbit. Is this reasonable?
For the lowest orbit, the normal state of the hydrogen atom n=1. What
values do we substitute for h, m( the electron mass) and e ? We must use
the same system of units for each quantity.
m = 0.511 MeV/c2
h = 4.136 10-21 MeV Sec
For e2 we use the value e2 ~ 1.445 MeV fm (NOT Coulomb2) - we shall
see where this unit comes from later when we discuss fine structure of
spectral lines.
Then for n=1 we find r = 5.3 10-11m ~ 0.5 Angstroms - the right order of
magnitude.
What is the total energy of the nth orbit?
Total energy = Potential energy + Kinetic energy.
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The Kinetic energy is 1/2mv2, what is the Potential energy - the energy
it has by virtue of its position in the atom?
We assume a hydrogen atom is made by an electron being attracted to a
proton by the Coulomb force (e2/r2) and ending up in a circular orbit
around the proton? Effectively it came from an infinite distance to an
orbit radius, r under a force e2/r2.
If we raise a mass m a height h on earth there is a gravitational force =
mg that has to be overcome. When this is done the mass has a potential
energy relative to it’s starting point of mgh - force  distance, the work
done in raising it. In this case over a small distance h, the force is
roughly constant = mg.
For the electron coming from  to r the force is NOT constant - it is
e2/r2 so we have to sum the work done in little steps, ri:
i (e2/r2)i ri
As
e2/r2
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is continuous at all r we can replace this with an INTEGRAL
Potential energy = r e2/r2 dr.
The solution to this integral is Potential energy = -e2/r
The Total energy is thus = mv2/2 - e2/r
From (3) v2=e2/mr
Thus Total Energy = e2/2r - e2/r = -e2/2r
Substituting for r from(4) r = n2h2/42me2 we get
Total Energy of electron in the nth orbit En is = - 22me4/n2h2
-3.39 eV
n=2
-13.6 eV
n=1
Energy Levels
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From (2) the frequency of light emitted as an electron drops
down from a level ni, energy Ei to a level nf , energy Ef is ;
 = (Ei - Ef)/h
Thus  = 22me4/h3{1/nf2 - 1/ni2}
and k = 1/ = /c = 22me4/h3c{1/nf2 - 1/ni2}
To compare this with Balmer’s formula we simply use:
nf =2, ni = 3,4,5 …
Then we see the formula gives an approximation to Rydberg’s
constant R = 22me4/h3c ~1.1 107 m-1
It is not quite right because we have assumed the proton is so
massive, only the electron moves in orbit around it. In fact , as
the proton is ~ 1800 times the mass of an electron this is not a
bad approximation but we can improve our value of RH by 22
considering the motion of both.
Correction for finite mass of nucleus
For circular orbits of the electron about a massive nucleus : mvr = nh/2.
We sometimes write this as mr2 = nh/2  where  is the angular
velocity.
When motion of both electron and nucleus is considered we get orbits of
both about the CENTRE-OF-MASS of the system.
r

Centre-of-mass
Proton
Electron
x
r-x

To find the position of the centre of mass we balance
moments:
Mx = m(r-x)
giving:
x= mr/(M+m)
r-x = Mr/(M+m)
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The total orbital angular momentum
L = m(r-x)2  + Mx2 
= mM2r2 / (m + M)2 + Mm2r2 /(m +M)2
= mM(M+m)r2 /(m+M)2
= mMr2 /(m+M)
= r2 
where  = mM/(m+M) - the REDUCED MASS
NOW instead of mr2 = nh/2  we have r2  = nh/2.
Working through the derivation of R with this definition we then get:
R  = 22  e4/h3c
as an even better expression for Rydberg’s constant.
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Bohr’s model can be extended further by including:
• elliptical orbits rather than just circular ones for the hydrogen atom
•correcting the motion for the relativistic change in mass of the
electron with velocity.
Introducing elliptical orbits:
r

Whereas in a circle only r can vary and we only need one quantum
number to define orbits, n for an ellipse both r and  vary from
point to point and we must have TWO quantum numbers nr (for
allowed radial values) as before AND k ( for allowed values of
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ellipse shapes - including a circle).
The quantum number which determines the energy of an orbit n = nr
+ k.
Possible combinations are:
n=1
nr = 0, k =1
We cannot have k=0, nr =1 - we’d get a straight line i.e. the electron
would have to go through the proton. Bohr could not conceive this
could happen.
n= 2
nr = 0, k=2
nr = 1, k =1
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n =3
k = 3 nr =0
k = 2 nr =1
k = 1 nr= 2
En= -22  e4/n2h2 where n = nr + k
Thus generalising to elliptic orbits leaves the energy levels
unchanged. However in high resolution measurements of spectra
fine structure is observed. To try to explain this Sommerfeld
included the fact that at different velocities electron mass changes:
m(v) = m/(1-v2/c2)1/2
Incorporating this relativistic correction he found that the Energy of
a particular state depends on both n AND k.
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En,k= -22  e4/n2h2{ 1 + Z22(n/k -3/4)/n2}
where  = 2e2/hc ~ 1/137 - the fine structure constant.
N.B. This is the origin of the definition of e2 = 1.445 MeVfm check by substituting for h and c
We now have an energy level scheme:
n=3
k=3
k=2
k=1
n=2
k=2
k=1
n=1
k=1
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When a hydrogen atom is excited by heating to an n=3 state (say)
and then de-excites, emitting light, to an n=2 state it would seem that
6 frequencies of light {E(3,k) -E(2,k)}/h could be emitted. However
only THREE are observed corresponding to the green lines
below.There is a SELECTION RULE operating:
k = 1
{This is because light can be thought of as a beam of massless photons
and their intrinsic spin s = 1.}
k=3
k=2
n=3
k=1
n=2
k=2
k=1
n=1
k=1
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This model of the atom, whilst very successful for hydrogen and partially
successful for atoms ( eg Lithium) that can be treated as:
Nucleus + electron cloud + outer single electron
fails conceptually for such atoms as the outer electron is moving quite
slowly - i.e. is non-relativistic - so splitting cannot be understood.It also
fails totally for atoms such as helium with two electrons where the
motion of both electrons needs to be treated.
These problems were overcome by:
• treating the electrons as waves
• adding in an extra interaction between the electron’s magnetic moment
-which is proportional to the electron’s spin s = 1/2 - and the magnetic
field created by the electrons motion - which is proportional to the
orbital angular momentum of the electron, l.
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When this was done it was found that instead of the two quantum
numbers
n = 1,2,3 … ( or nr = 0,1,2,…)
k = 1,2,3, … n
to describe energy levels,and a selection rule
k = 1
being needed to describe the number of spectral lines,
with the wave model four quantum numbers:
n = 1,2,3,…
- the principal quantum number
l= 0,1,2... n-1
- the orbital angular momentum quantum number
ml = -l, - (l-1), -(l -2), …-2,-1, 0, 1, 2, … (l -2), (l -1), l - the orientation
of angular momentum quantum number
ms = 1/2
-the orientation of spin quantum number
are needed to describe the energy levels. A selection rule l = 1 is again
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needed to describe the number of spectral lines.
The wave model explains the level schemes of atoms in a more
coherent way than the Bohr model AND predicts reaction rates something the Bohr model cannot do. It also together with a hugely
important principle:
The Pauli Exclusion Principle
No two electrons in an atom can have the same four quantum
numbers
explains the Periodic Table of the elements.
Hydrogen atom - ground state
1 electron n=1, l =0, ml = 0, ms = 1/2
Helium atom - ground state
2 electrons n = 1, l = 0, ml = 0, ms = +1/2,ms= -1/2
Lithium atom - ground state
3 electrons n=1, l =0, ml = 0, ms = +1/2,ms= -1/2
n=2, l =0, ml = 0, ms = 1/2
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First the l-shells are filled - the sub-shells
Then the n-shells filled - the shells.
We often give the n- and l- values specific letters rather than numbers spectroscopic notation:
n = 1 K-shell
l= 0
s
n=2 L-shell
l=1
p
n=3 M =shell
l=2
d
etc
etc
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X-Ray Spectra
If we excite atoms by bombardment with high energy electrons - say 10
keV or so - then as they de-excite the light they emit is of much higher
frequency than the visible range. We call this light, of such high
frequency - and hence low wavelength ( of order Angstroms) - that it
can penetrate matter, X-radiation or X-rays.
Electrons
H.T.
V kV
-
X-ray detector
+
X-rays
Anode
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These spectra can be divided into a continuum background plus line
spectra. The continuum has a short wavelength cut-off point which is
dependent on the initial bombarding electron eneergy.
Eelectron= eV = h = hc/min
i.e. min = hc/eV
This, in fact, provides a very accurate method of measuring h - Planck’s
constant. All the electron’s kinetic energy is converted to electromagnetic
energy.
Unfortunately the rest of the continuum does not have such a simple type
of interpretation. As the incident electron decelerates in the anode - by
collision with tungsten atomic electrons - EM radiation is emitted ( i.e. Xrays). This is called bremsstrahlung - brake radiation.
An advanced theory - Quantum electrodynamics (QED) does though
predict the shape of the continuum for a thick anode.
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Line spectra
With the line spectra we can deal! As you might have guessed these are
due to a bombarding electron knocking out an inner shell electron from
an atom of the anode. Electrons in higher shells then drop down
sequentially to fill first the hole created in that shell, then the
subsequent holes created in higher shells.
Example
Let the electron in an anode atom that the bombarding electron knocks
out be in the n=1 state.
Then electrons in higher shells can de-excite :
n=2  n=1.
n=3  n=2 or n=1
n=4  n=3 or n=2 or n=1
Hence we get the X-ray line spectra.
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Moseley observed a regularity in line spectra from different anode
elements.
Using spectroscopic notation:
n
1
2
3
4
5….
K
L
M
N
O…..
Moseley obtained data for two prominent lines known as the K, L lines
for different anode elements.
K L  K transition
L M  L transition
K M K transition
L N  L transition
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He noticed that the wavelengths of these line could be fit by the simple
formulas ( Moseley’s Laws):
For K
1/ ~ C K (Z-1)2
For L
1/ ~ C L (Z-7.4)2
Now Bohr model - for a single electron atom with nucleus of charge
Ze+ - predicts for the energy of a shell:
En = - R hcZ2/n2 = -13.6Z2/n2 eV
Therefore the energy of X-rays emitted for a transition between states
n1 and n2 is:
E 12 = R hcZ2(1/n22 -1/n12)
Now E = hc/ 
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So the Bohr Model predicts :
For the K line: 1/ = [R (1/12 - 1/22)]Z2
For the L line: 1/ = [R (1/22 - 1/32)]Z2
In both cases bracketed quantities gave good agreement with C K and
C L . The factors (Z-1)2 and (Z-7.4)2 rather than Z2 were explained as
being due to the ‘screening effect’ of electrons between the nucleus and
the electron interacting with the bombarding electron. This effect of
course is not dealt with by the Bohr model.
Modern calculations can calculate this screening effect with high
accuracy.
Hence Moseley’s work was a direct confirmation of the predictions of
the Bohr model.
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Auger Electron Emission
An alternative to X-ray emission as the atom de-excites is the emission
of Auger electrons:
Auger
Electron
K
L
Say an electron in the atom is in the L-shell. If a hole has been
created by electron bombardment in the K-shell, then the L shell
electron de-excites to the K-shell, losing energy NOT by emitting
an X-ray but by giving its excess energy to another electron in the
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L-shell.This is then emitted as the Auger electron.
The kinetic energy of the Auger electron, TAuger = (IK - IL) - IL
where IK, IL are the energies required to remove an electron from
an energy level.
I = - E ( the energy of the level caculated by Bohr model).
Fluorescent Yield
This is a measure of the probability of X-ray emission rather than
Auger emission for different elements. It is the number of X-rays
emitted per vacancy in a given shell. For instance if there is a
vacancy in the K-shell - created by electron bombardment say then the fluorescent yield is K.
K = KL + KM + KN ...
the sum of X-rays emitted from higher shells per vacancy in the42
K-shell.
1
K
0
20
40
60
80
Z
So light elements preferentially decay by Auger electron emission, heavy
elements by X-ray emission.
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