Transcript Motion in a Straight Line

to Atomic and Nuclear
Physics
All these slide presentations are at:
http://www.hep.shef.ac.uk/Phil/PHY008.htm
and also on FY website
Phil Lightfoot, E47, (24533) [email protected]
Most Important Thing !!!!!!
I’m always available to help with
any aspect of the course
Stop me if you’re confused
My contact details are on the top of
your lecture notes
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 didn’t understand.
In the lecture have a go at answering the questions – there’s 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 you’ll be fine.
Just because your first homework is set for 27th April doesn’t mean you
can’t get loads of feedback on how you’re doing. Ask questions, do
problem class questions, have a go at questions in lectures.
Come to see me whenever you don’t understand something. Don’t wait
until just before the exam. Email me saying what you want to talk about
and when you’re free and we’ll go somewhere until you’re happy.
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 couldn’t 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
Plank’s constant (6.63 × 10-34 Js)
photon
E
hf
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 photon’s 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.
Ekineticenergy  h f  E0
Where f is frequency of light and h is Plank’s constant
(6.63 × 10-34 Js), E0 is the work function in joules.
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
E photon  h f
Where f is frequency of light and h is Plank’s constant
(6.63 × 10-34 Js).
But this isn’t Einstein’s most well known equation !!! What is ???
E photon  m c
2
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. Let’s equate them to
each other
E photon  h f  m c 2
Momentum p is always equal to its mass multiplied by its velocity i.e.
So
E photon  h f  p c
and since
c f 
therefore
p  mc
p
h

The Compton effect
p
h

Where p is the momentum in kgms-1, h is 6.63×10-34 Js, and λ is
the wavelength in m. This is called the de Broglie equation after
the person that did the maths based on Einstein’s 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.
http://faculty.gvsu.edu/m
ajumdak/public_html/On
lineMaterials/ModPhys/
QM/Compton/compton.
html
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 E photon  h f 
hc

if the photon loses energy then λ 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.
The Compton effect : an example
Let’s imagine that we collide a photon of green light (λ = 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?
6.631034
 27
1
p 

1.25

10
kgms
 530109
What is the mass of an electron?
h
9.1 × 10-31 kg
What is the momentum of a stationary electron?
If momentum is shared equally, how
much does the photon end up with ?
What is the final wavelength
of the photon ?
What is the final velocity
of the electron ?
zero
0.625 × 10-27 kgms-1
h
6.631034
 
 1061nm
 27
p 0.62510
m om entum 0.6251027
1
velocity 


687
ms
m ass
9.11031
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
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?
As expected the bullets from
different guns don’t interfere
with each other and you just
get bullet holes in two specific
places. Big heavy bullets are
acting like particles
Let’s repeat the
experiment but
instead of using
bullets lets use
electrons
If they had instead acted like
waves what would the bullet
hole distribution on the wall
have looked like ??
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/Modern%
20Physics/Interference/interference.html
de Broglie and matter waves
Let’s try to work out why electrons can act like waves but bullets do not ….
p
h

Earlier we derived the de Broglie equation where p is the momentum in
kgms-1, h is 6.63×10-34 Js, and λ 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
2E
u
m
p  mu
mu2
E
2
 2E 

And therefore p  m u  m
 m   2m E


.
Finally we have the expression :
p  2mE 
h

de Broglie and matter waves
p  2mE 
h

p is the momentum in kgms-1, m is in kg, E is in joules, h is
6.63×10-34 Js, and λ 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. Let’s 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 ??
31
7
23
p  mu  9.110
 4.6810  4.2710
kgms1
What is its wavelength according to the de Broglie equation ??
.
h 6.626 1034
11
 

1.55

10
m
 23
p 4.27 10
This is really really small. He couldn’t
find or make a slit that narrow so he
couldn’t demonstrate interference.
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.
de Broglie and matter waves
p  2mE 
h

p is the momentum in kgms-1, m is in kg, E is in joules, h is
6.63×10-34 Js, and λ is the wavelength in m.
So why don’t 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 λ 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 λ 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 2×107 ms-1 ??
.m
u
elephant elephant
uelephant 
 meue
31
meue
9.110  2 10

 9.11027 ms1
melephant
2000
7
This also explains why we see
fringes in the double slit experiment
using electrons but not bullets !!!
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.
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/examples/19267.html
.
http://webphysics.davidson.edu/Applets/pqp_preview/contents/pqp_errata/cd_errata_fixes/section4_7.html
http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/rutherford/rutherford.html
http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/rutherford/rutherford2.html
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.