Transcript Quantum and Nuclear Physics

Quantum and Nuclear Physics
The Photoelectric effect
Waves or Particles?
The Photoelectric effect
How are the electrons released?
Powerful red laser
No electrons released
Photoelectron Energy
…and some energy is
given to the electron as
kinetic energy.
-
Some energy is needed
to release the electron
(the work function φ)…
Photon Energy = work function + kinetic energy of electron
Determining Planck’s constant
• Add different filters under the light source
Photoelectric experiment
• Take measurements of stopping potential
and wavelength to determine Planck’s
constant and the threshold frequency
Plot a graph of stopping
potential versus frequency
Photoelectric Effect: Vstop vs. Frequency
eVstop  hf  
Vstop  0 
hfmin

hfmin  
Slope = h = Planck’s constant
Determining “h” from the graph
hf  EK max  
Photoelectric Effect: IV Curve Dependence
Intensity I
dependence
Vstop= Constant
f1 > f2 > f3
Frequency f
dependence
f1
f2
Vstop f
f3
Is light a wave or a particle?
• http://www.schoolphysics.co.uk/age1619/Quantum%20physics/text/Photoelectric
_effect_animation/index.html
=Φ
E max=
V= Stopping voltage
1. The work function for lithium is 4.6 x 10-19 J.
(a) Calculate the lowest frequency of light that will cause photoelectric emission. (6.9 x
1014 Hz )
(b) What is the maximum energy of the electrons emitted when light of 7.3 x 1014 Hz is
used? (0.24 x 10-19 J )
2. A frequency of 2.4 x 1015 Hz is used on magnesium with work function of 3.7 eV.
(a) What is energy transferred by each photon?
(b) Calculate the maximum KE of the ejected electrons.
(c) The maximum speed of the electrons.
(d) The stopping potential for the electrons.
(a)
1.6 x 10-18 J
(b)
1.0x 10-18 J
(c) v = 1.5 x 106 m s-1
(d)
Vs = 6.3 V
Questions
Tsokos page 396 q’s 1-7.
Review of Bohr and deBroglie
• Background:
– Balmer found equation for Hydrogen spectrum but
didn’t know what it meant.
– Rutherford found that atoms had a nucleus, but didn’t
know why electrons didn’t spiral in.
• Bohr postulates quantized energy levels for no good
reason, and predicts Balmer’s equation.
• deBroglie postulates that electrons are waves, and
predicts Bohr’s quantized energy levels.
• Note: no experimental difference between Bohr
model and deBroglie model, but deBroglie is a lot
more satisfying.
Davisson and Germer -- VERY clean nickel crystal.
Interference is electron scattering off Ni atoms.
e
e
e e
e
e
Ni
e e det. e
e
scatter off atoms
e
e
move detector around,
see what angle electrons coming off
See peak!!
so probability of angle where detect
electron determined by interference
of deBroglie waves!
# e’s
0
e
e
e e
500
scatt. angle 
e e det. e
e
e
Ni
Observe pattern of scattering
electrons off atoms
Looks like ….
Wave!
Electron diffraction
Diffraction rings
Calculating the De Broglie λ
λ = h/p
(= h/(2Ekm)1/2 )
h = Planck’s constant p = Momentum
In 1923, French Prince Louis de
Broglie, generalised Einstein's work
from the specific case of light to cover
all other types of particles. This work
was presented in his doctoral thesis
when he was 31. His thesis was
greeted with consternation by his
examining committee. Luckily, Einstein
had received a copy in advance and
vouched for de Broglie. He passed!
de Broglie questions
• Calculate the wavelengths of the
“deBroglie” waves associated with
• a)a 1kg mass moving at 50ms-1
• b)an electron which has been accelerated
by a p.d. of 500V.
a)Discuss briefly deBroglie’s hypothesis and
mention one experiment which gives evidence to
support it.
b)Calculate the wavelength of the “deBroglie
wave” associated with an electron in the lowest
energy Bohr orbit. (The radius of the lowest
energy orbit according to the Bohr theory is
5·3×10-11m.)
Questions
Tsokos page 396 q’s 8-10
History of Quantum Mechanics
Max Planck's work on the 'Black Body' problem started the quantum
revolution in 1900. He showed that energy cannot take any value but is
arranged in discrete lumps – later called photons by Einstein.
In 1913, Niels Bohr proposed a model of the atom with quantised
electron orbits. Although a great step forward, quantum physics was still
in its infancy and was not yet a consistent theory. It was more like a
collection of classical theories with quantum ideas applied.
Starting in 1925 a true 'quantum mechanics' – a set of mathematically
and conceptual 'tools' – was born. At first, three different incantations of
the same theory were proposed independently and were then shown to
be consistent. Quantum mechanics reached its final form (essentially
unchanged from today) in 1928.
Participants of the 5th Solvay
Congress, Brussels, October 1927
E. Schrödinger W. Pauli
W. Heisenberg
N. Bohr
M. Planck
M Curie
A. Einstein
L.V. de Broglie
Models of the Atom
• Thomson – Plum Pudding
–
–
–
–
–
– Why? Known that negative charges can be removed from atom.
– Problem: just a random guess
• Rutherford – Solar System
– Why? Scattering showed hard core.
– Problem: electrons should spiral into nucleus in ~10-11 sec.
+
• Bohr – fixed energy levels
– Why? Explains spectral lines.
– Problem: No reason for fixed energy levels
+
• deBroglie – electron standing waves
– Why? Explains fixed energy levels
– Problem: still only works for Hydrogen.
• Schrodinger – will save the day!!
+
–
Different
view
of
atoms
The Bohr Atom
Electrons are only allowed to
have discrete energy values and
these correspond to changes in
orbit.
The Schrodinger Atom
Electrons behave like stationary Amplitude
waves. Only certain types of
wave fit the atom, and these
correspond to fixed energy
states. The square of the
amplitude gives the probability
of finding the electron at that
point
+
0eV
Consider a ball in a hole:
When the ball is
here it has its
lowest gravitational
potential energy.
Spectra
5J
We can give it
potential energy by
lifting it up:
If it falls down again it
will lose this gpe:
5J
30J
20J
Spectra
A similar thing happens to electrons. We can “excite” them
and raise their energy level:
0eV
-0.85eV
-1.5eV
-3.4eV
-13.6eV
An electron at this energy
level would be “free” – it’s
been “ionised”.
These energy levels are
negative because an electron
here would have less energy
than if its ionised.
This is called “The ground
state”
Spectra
If we illuminate the atom we can excite the electron:
Q. What wavelength of light
would be needed to excite
this electron to ionise it?
0eV
-0.85eV
-1.5eV
-3.4eV
Light
Energy change = 3.4eV = 5.44x10-19J.
Using E=hc/λ wavelength = 3.66x10-7m
-13.6eV
(In other words, ultra violet light)
Example questions
1) State the ionisation energy of this
atom in eV.
2) Calculate this ionisation energy in
joules.
3) Calculate the wavelength of light
needed to ionise the atom.
0eV
-0.85eV
-1.5eV
-3.4eV
4) An electron falls from the -1.5eV to
the -3.4eV level. What wavelength
of light does it emit and what is the
colour?
5) Light of frequency 1x1014Hz is
incident upon the atom. Will it be
able to ionise the atom?
-13.6eV
Spectra
Continuous spectrum
Absorption spectrum
Emission spectrum
Emission Spectra
Hydrogen
Helium
Sodium
Observing the Spectra
Microscope
(to observe the spectrum)
Light source
Collimator
Gas
Diffraction grating
(to separate the colours)
Questions
Tsokos page 405 q’s 1-7.
Models of the Atom
• Thomson – Plum Pudding
–
–
–
–
–
– Why? Known that negative charges can be removed from atom.
– Problem: just a random guess
• Rutherford – Solar System
– Why? Scattering showed hard core.
– Problem: electrons should spiral into nucleus in ~10-11 sec.
+
• Bohr – fixed energy levels
– Why? Explains spectral lines.
– Problem: No reason for fixed energy levels
+
• deBroglie – electron standing waves
– Why? Explains fixed energy levels
– Problem: still only works for Hydrogen.
• Schrodinger – will save the day!!
+
–
Schrödinger model
Schrödinger set out to develop an alternate formulation of quantum
mechanics based on matter waves, à la de Broglie. At 36, he was
somewhat older than his contemporaries but still succeeded in
deriving the now famous 'Schrödinger Wave Equation.' The solution of
the equation is known as a wave function and describes the behavior
of a quantum mechanical object, like an electron.
At first, it was unclear what the wave function actually represented.
How was the wave function related to the electron? At first,
Schrödinger said that the wave function represented a 'shadow wave'
which somehow described the position of the electron. Then he
changed his mind and said that it described the electric charge
density of the electron. He struggled to interpret his new work until
Max Born came to his rescue and suggested that the wave function
represented a probability – more precisely, the square of the absolute
magnitude of the wavefunction is proportional to the probability
that the electron appears in a particular position. So,
Schrödinger's theory gave no exact answers… just the chance for
something to happen. Even identical measurements on the same
system would not necessarily yield the same results! Born's key role
in deciphering the meaning of the theory won him the Nobel Prize in
Physics in 1954.
Quantum Mechanical tunneling
In the classical world the positively charged alpha particle needs enough
energy to overcome the positive potential barrier which originates from
protons in the nucleus. In the quantum world an alpha particle with less
energy can tunnel through the potential barrier and escape the nucleus.
Electron in a box model
Electrons will form standing waves of wavelength 2L/n
Kinetic Energy of an electron in a
box
• When the momentum expression for the
particle in a box :
•
• is used to calculate the energy associated
with the particle
Heisenberg uncertainty principle
Heisenberg made one fundamental and long-lasting contribution to the
quantum world – the uncertainty principle. He showed that quantum
mechanics implied that there was a fundamental limitation on the accuracy to
which pairs of variables, such as (position and momentum) and (energy and
time) could be determined.
If a 'large' object with a mass of, say, 1g has its position measured to an
accuracy of 1 , then the uncertainty on the object's velocity is a minute 10-25
m/s. The uncertainty principle simply does not concern us in everyday life. In
the quantum world the story is completely different. If we try to localize an
electron within an atom of diameter 10-10 m the resulting uncertainty on its
velocity is 106 m/s!
Heisenberg uncertainty principle
Nuclear physics
Determining the size of the nucleus
Approach of alpha particle to nucleus
Z = 79 (gold)
25
20
15
10
5
0
0
2
4
6
distance from nucleus /
8
10–14
10
m
1. Make an arithmetical check to show that at distance r = 1.0x10–14 m, the
electrical potential energy, is between 20 MeV and 25 MeV, as shown by the
graph.
2.How does the electrical potential energy change if the distance r is doubled?
3.From the graph, at what distance r, will an alpha particle with initial kinetic
energy 5 MeV colliding head-on with the nucleus, come to rest momentarily?
1.
Substituting values gives
EP =
2.
3.
2  79  1.6  1019 C
4  8.85 1012 C 2 J 1 m 1  1.0  1014 m
 106  22.7 MeV.
Halves, because the potential energy is proportional to 1/r.
About 4.6x10–14 m, where the graph reaches 5 MeV.
Recall:
+
+
Circular paths
2 protons, 2 neutrons,
therefore charge = +2
-
1 electron, therefore
charge = -1
Because of this charge, they will be deflected by magnetic
fields:









These paths are circular, so Bqv =
mv2/r,
or
r =mv
Bq
Bainbridge mass spectrometer
Ions are formed at D and pass through the cathode C and then through a slit S1
A particle with a charge q
Hyperlink
and velocity v will only pass
through the next slit S2 if the
resultant force on it is zero –
that is it is traveling in a
straight line. That is if:
Therefore
In the region of the Mag field
r = Mv/(Bq)
Bqv = Mv2/r
Therefore
Nuclear energy levels
There are 2 distinct
length of tracks
in this Alpha decay
Therefore, the energy
levels in the nucleus
are discrete
The existence of Neutrinos
How can a 2 body
system create a
spectrum of
energies?
There must be a
3rd particle
The Neutrino
was postulated
A 2 body system only
has one solution
A 3 body system
has many solutions
Changes in Mass and Proton Number
Beta - decay:
90
Sr
38
90
Y
39
+
0
β
-1
“positron”
Beta + decay:
11
6
C
11
B
5
+
0
+1
β
Radioactive Decay Law
dN/N = -λdt which when integrated, gives
Taking antilogs of both sides gives:
Half life and the radioactive decay
constant
When N = No/2 the number of radioactive nuclei will have halved
Therefore when t = T1/2
N = No/2 = Noe-λT1/2 and so 1/2 = e-λT1/2 .
Taking the inverse gives 2 = eλT1/2 and so:
Measuring long half lives
• If the half life is very long, then the activity (A) is
constant
• Analysis of a decay curve cannot give the half
life.
• If the mass of the substance is measured, then
• A = -λN, so a measurement of the activity
enables Measuring long half lives to be
calculated (N from mass).
• T1/2 can be calculated from λ.
Measuring short half lives
• Each decay can cause an ionisation
• This can generate an electric current
• If the current is displayed on an
oscilloscope, then
• The limit is the response time of the
oscilloscope (typically µs).
Questions
Tsokos page 412 q’s 1-20