to Atomic and Nuclear Physics

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to Atomic and Nuclear Physics
Welcome!
All these slide presentations are at
http//www.hep.shef.ac.uk/Phil/PHY008.htm and
also on FY website
Phil Lightfoot, E47, (24533) p.k.lightfoot_at_shef.ac
.uk
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Most Important Thing !!!!!!
Im always available to help with any aspect of
the course Stop me if youre confused My contact
details are on the top of your lecture notes
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How to pass Physics !!!!!!
Read 3 or 4 pages ahead in the notes so you are
familiar with content of the next lecture and in
the lecture ask about the bits you didnt
understand.
In the lecture have a go at answering the
questions theres no such thing as a stupid
answer.
Have a go at the questions from the problem
classes. These are very similar to the exam
questions and if you can do these then youll be
fine.
Just because your first homework is set for 27th
April doesnt mean you cant get loads of
feedback on how youre doing. Ask questions, do
problem class questions, have a go at questions
in lectures.
Come to see me whenever you dont understand
something. Dont wait until just before the exam.
Email me saying what you want to talk about and
when youre free and well go somewhere until
youre happy.
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Review of last lecture The Photoelectric effect
Scientist noted that when some metals were
illuminated by a source of light, electrons were
emitted from the metal with a certain kinetic
energy.
But the scientists were confused. The kinetic
energy of the ejected photoelectrons was found to
depend on the frequency of light and the type of
metal used and not, as expected, on the
brightness of the light source.
Einstein offered an explanation in 1905 by
proposing that light was not simply a wave but
rather made up of tiny quanta or packets of
energy, the energy of a single photon
proportional to the frequency of light.
They had a big problem thinking of it as a
particle like a ball since then it would have to
have weight and processes like interference and
diffraction couldnt be explained.
What Einstein was saying was that we should
actually think of light as a bunch of tiny
packets of energy each with a small amount of
energy.
He even went further and said that each photon
(packet of energy) would have an energy of
Where f is the frequency of the light and h is
Planks constant (6.63 10-34 Js)
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The Photoelectric effect
Einstein proposed that there was a minimum energy
E0 required to release a photoelectron from a
metal.
He called E0 the work function and suggested that
this value was a constant for a particular metal,
but was different for different metals.
When a photon is absorbed within a metal, some of
the photons energy will be used up in freeing a
photoelectron from the metal, and if there is any
energy remaining, then this will appear as
kinetic energy of the ejected photoelectron.
Where f is frequency of light and h is Planks
constant (6.63 10-34 Js), E0 is the work
function in joules.
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The Compton effect
In the last section we saw how Einstein imagined
waves of light to be in fact tiny particles
called photons in order to explain the
photoelectric effect, and how he linked the
energy of the photon to the frequency of the wave
with the equation
Where f is frequency of light and h is Planks
constant (6.63 10-34 Js).
But this isnt Einsteins most well known
equation !!! What is ???
Where m is the mass of the photon and c is the
speed of light
In this equation he is defining a relationship
between the energy of the photon of light and its
mass imagining it to a ball-like packet.
So now we have two expressions for the energy of
a photon. Lets equate them to each other
Momentum p is always equal to its mass multiplied
by its velocity i.e.
So
and since therefore
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The Compton effect
Where p is the momentum in kgms-1, h is
6.6310-34 Js, and ? is the wavelength in m. This
is called the de Broglie equation after the
person that did the maths based on Einsteins
equations.
This is a very important equation because it
links momentum of a photon to its wavelength.
Many scientists questioned this interpretation.
What was needed was an experiment to demonstrate
the particle nature of photons of light.
In 1923 Arthur Compton did this by setting up a
collision between X-ray photons and electrons.
relationship.
http//faculty.gvsu.edu/majumdak/public_html/Onlin
eMaterials/ModPhys/QM/Compton/compton.html
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The Compton effect
The experiment showed that the X-ray photons and
electrons behaved exactly like ball bearings
colliding on a table top.
Because the electron was scattered, the photon
must have transferred both momentum and kinetic
energy to it. This can only be explained by
assuming that photons have momentum.
But he observed something else. Before the
collision the photon had one wavelength and after
the collision its wavelength had increased.
Remember that since
if the photon loses energy then ? decreases
Clearly the electron had been given energy,
conservation of energy indicating that the
scattered photon must therefore have lower energy
than prior to the collision.
The Compton effect is important because it
demonstrates that light cannot be explained
purely as a wave phenomenon, the classical theory
of an electromagnetic wave scattered by charged
particles unable to explain any shift in
wavelength.
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The Compton effect an example
Lets imagine that we collide a photon of green
light (? 530 nm) with a stationary electron. If
after the collision the momentum is shared
equally between the photon and the electron, what
is the final wavelength of the photon and the
velocity of the electron?
What is initial momentum of a photon of green
light?
What is the mass of an electron?
9.1 10-31 kg
What is the momentum of a stationary electron?
zero
If momentum is shared equally, how much does the
photon end up with ?
0.625 10-27 kgms-1
What is the final wavelength of the photon ?
What is the final velocity of the electron ?
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de Broglie and matter waves
So de Broglie had said that light, previously
thought of as a wave could actually be thought of
as a particle with momentum like a ball.
The next obvious question of course was.
Could particles previously thought of as behaving
like balls actually behave like waves in some
situations???
This became known as wave-particle duality.
To test this concept de Broglie needed to try to
get matter to demonstrate wave-like properties
such as diffraction or interference for example
using a double slit apparatus.
Interference
Wave-like behaviour
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de Broglie and matter waves
What would you expect if you fired two Uzis
through two holes in a wall ?
Would the bullets act like particles or waves?
Lets repeat the experiment but instead of using
bullets lets use electrons
As expected the bullets from different guns dont
interfere with each other and you just get bullet
holes in two specific places. Big heavy bullets
are acting like particles
If they had instead acted like waves what would
the bullet hole distribution on the wall have
looked like ??
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de Broglie and matter waves
When we repeat this experiment using electrons we
find amazingly that we get interference patterns
on the far wall even though we think of electrons
as little balls!!
The only way that this can be explained is if the
electrons are acting like waves (called matter
waves) and constructively and destructively
interfering.
http//chaos.nus.edu.sg/simulations/Modern20Physi
cs/Interference/interference.html
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de Broglie and matter waves
Lets try to work out why electrons can act like
waves but bullets do not .
Earlier we derived the de Broglie equation where
p is the momentum in kgms-1, h is 6.6310-34 Js,
and ? is the wavelength in m.
Momentum of a particle is defined as its velocity
multiplied by its mass where m is the mass and u
is the velocity.
The kinetic energy of a particle is given as E so

So by rearranging we find
And therefore
.
Finally we have the expression
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de Broglie and matter waves
p is the momentum in kgms-1, m is in kg, E is in
joules, h is 6.6310-34 Js, and ? is the
wavelength in m.
From double slit experiments using light we know
that we only see clear evidence of interference
(i.e. those pretty interference fringes) if the
spacing of the slits is about the same as the
wavelength ? of the light.
It is therefore sensible to select matter waves
of similar wavelength to the slit spacing. This
caused de Broglie a great deal of difficultly.
Lets see why using an example.
What is the momentum of an electron of mass
9.1?10-31 kg moving with velocity u 4.68?107 m
s-1 ??
What is its wavelength according to the de
Broglie equation ??
.
This is really really small. He couldnt find or
make a slit that narrow so he couldnt
demonstrate interference.
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de Broglie and matter waves
The best he could do was select a very thin
crystal of carbon atoms close to this spacing. He
then shone an electron beam at it, the regular
spacing of atoms forming a kind of diffraction
grating producing rings on a distant screen.
This diffraction pattern is of very similar
appearance to that seen from a light source of
similar wavelength, as shown in the figure to the
right.
.
This was proof that electrons were acting like
waves.
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de Broglie and matter waves
p is the momentum in kgms-1, m is in kg, E is in
joules, h is 6.6310-34 Js, and ? is the
wavelength in m.
So why dont we see diffraction effects when a
person collides with a metal grating?!
People are heavier than electrons and therefore
they have a higher momentum. Looking at the
equation above a big momentum p means that ? will
be tiny.
Remember that in order to see diffraction effects
the wavelength should be approximately the same
as the slit separation. But the wavelength ? of
the matter wave is incredibly small and certainly
far smaller than the smallest slit separations.
Question How slow does a 2000 kg elephant have
to move in order to have the same momentum as an
electron travelling at 2107 ms-1 ??
.
This also explains why we see fringes in the
double slit experiment using electrons but not
bullets !!!
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Rutherford The gold foil scattering experiment
In 1904, Thomson considered atomic structure to
be represented by a plum pudding in so far as
the atom was made up of negatively charged
electrons surrounded by a soup of positive charge
to keep the overall charge neutral.
To find out for sure, in 1909 Rutherford fired
positively charged helium ions (sometimes called
alpha particles) with high energy at gold atoms
in a thin gold film.
Since it was believed that positive and negative
charges were spread evenly within the atom and
that therefore only weak electric forces would be
exerted on the helium ions passing through the
thin foil, he expected to find that most ions
travelled straight through the foil with no
deviation.
.
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Rutherford The gold foil scattering experiment
What he found, to great surprise, was that whilst
most passed straight through the foil, a small
percentage (about 1 in 10000) were deflected at
very large angles and some even bounced back
toward the ion source.
What did this say about the atomic structure of
gold ??
Because helium ions are about 8000 times the mass
of an electron and impacted the foil at very high
velocities, it was clear that very strong forces
were necessary to deflect these particles.
(Imagine firing bullets at soup. Even if just one
ricocheted back it would be surprising!!)
http//serc.carleton.edu/sp/compadre/interactive/e
xamples/19267.html
.
http//webphysics.davidson.edu/Applets/pqp_preview
/contents/pqp_errata/cd_errata_fixes/section4_7.ht
ml
http//galileoandeinstein.physics.virginia.edu/mor
e_stuff/Applets/rutherford/rutherford.html
http//galileoandeinstein.physics.virginia.edu/mor
e_stuff/Applets/rutherford/rutherford2.html
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Rutherford The gold foil scattering experiment
The positive helium ions had clearly been
repelled by an incredibly large positive charge
within the atom, this charge concentrated in a
dense region also containing most of the mass.
This work led in 1913 to Rutherford declaring the
atom to contain a very small nucleus of high
positive charge (equal to the number of electrons
in order to maintain neutrality) and to be
similar to the solar-system-like model, in
which a positively charged nucleus is surrounded
by an equal number of electrons in orbital shells.
We now know that the diameter of the nucleus
ranges from 2 to 9 10-15 m from hydrogen to
uranium, and the diameter of the nucleus to be
almost 10-5 the diameter of the atom.
.