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Atomic PhysicsThe 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)

A = area of detector in cm2.

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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 second.

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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 kas 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) cm-2

1.296 (2Z/T)2 cosec4(/2) 10-27 cm2/sr A/d2 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?

The kinetic energy of the beam is: T = 1/2 Mv2

The potential energy is :V = 2Ze2/r

The Total Energy E = 1/2 Mv2 +2Ze2/r

Nucleus

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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=1/2Mv2

Therefore as E is a constant,2Ze2/d = 1/2Mv2

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

82Pb containing 208 protons and neutrons, 82 protons and 120 neutrons only has a radius of ~ 7 fm.

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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

m

c m/sec

(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 = n2h2/42me2 (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 is continuous at all r we can replace this with an INTEGRAL

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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 = n2h2/42me2 we get

Total Energy of electron in the nth orbit En is = - 22me4/n2h2

-13.6 eV

-3.39 eV

n=1

n=2

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 = 22me4/h3{1/nf2 - 1/ni

2}

and k = 1/ = /c = 22me4/h3c{1/nf2 - 1/ni

2}

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 = 22me4/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 considering the motion of both.

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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.

Proton Electron

x r-x

r

Centre-of-mass

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 = 22 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 ellipse shapes - including a circle).

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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= -22 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= -22 e4/n2h2{ 1 + Z22(n/k -3/4)/n2}

where = 2e2/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=1

n=2

n=3

k=1

k=2 k=1

k=3 k=2 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.}

n=1

n=2

n=3

k=1

k=2 k=1

k=3 k=2 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 needed to describe the number of spectral lines.

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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 PrincipleNo two electrons in an atom can have the same four quantum numbersexplains the Periodic Table of the elements.Hydrogen atom - ground state1 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 state3 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.

X-rays

X-ray detector H.T.

V kV

-

+

Electrons

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. X-rays). 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 12 = R hcZ2(1/n2

2 -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:

K

L

Auger Electron

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 L-shell.This is then emitted as the Auger electron.

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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 the K-shell.

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1

0Z

20 40 60 80

So light elements preferentially decay by Auger electron emission, heavy elements by X-ray emission.

K