quantum and nuclear physics
DESCRIPTION
Quantum and Nuclear Physics. The Photoelectric effect. Waves or Particles?. The Photoelectric effect. How are the electrons released?. Powerful red laser. No electrons released. …and some energy is given to the electron as kinetic energy. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/1.jpg)
Quantum and Nuclear Physics
The Photoelectric effect
![Page 2: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/2.jpg)
Waves or Particles?
![Page 3: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/3.jpg)
The Photoelectric effect
![Page 4: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/4.jpg)
How are the electrons released?
Powerful red laser
No electrons released
![Page 5: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/5.jpg)
Photoelectron EnergyPhoton
-Some energy is needed to release the electron (the work function φ)…
…and some energy is given to the electron as kinetic energy.
Photon Energy = work function + kinetic energy of electron
![Page 6: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/6.jpg)
Determining Planck’s constant
• Add different filters under the light source
![Page 7: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/7.jpg)
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
![Page 8: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/8.jpg)
Photoelectric Effect: Vstop vs. Frequency
stopeV hf
min0stopV hf
Slope = h = Planck’s constanthfmin
![Page 9: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/9.jpg)
Determining “h” from the graph
maxKhf E
![Page 10: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/10.jpg)
Photoelectric Effect: IV Curve Dependence
Intensity I dependence
Frequency f dependence
Vstop= Constant
Vstop f
f1 > f2 > f3
f1
f3
f2
![Page 11: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/11.jpg)
Is light a wave or a particle?
• http://www.schoolphysics.co.uk/age16-19/Quantum%20physics/text/Photoelectric_effect_animation/index.html
E max=
=Φ
V= Stopping voltage
![Page 12: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/12.jpg)
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
![Page 13: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/13.jpg)
Questions
Tsokos page 396 q’s 1-7.
![Page 14: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/14.jpg)
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.
![Page 15: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/15.jpg)
Davisson and Germer -- VERY clean nickel crystal. Interference is electron scattering off Ni atoms.
ee
ee
e
e
e ee
e e
scatter off atoms
e det.
move detector around,see what angle electrons coming offNi
![Page 16: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/16.jpg)
![Page 17: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/17.jpg)
ee
ee ee
e
e
e det.
Ni
Observe pattern of scattering electrons off atomsLooks like …. Wave!
# e’s
scatt. angle 5000
See peak!!
so probability of angle where detectelectron determined by interferenceof deBroglie waves!
![Page 18: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/18.jpg)
Electron diffraction
Diffraction rings
![Page 19: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/19.jpg)
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!
![Page 20: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/20.jpg)
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.
![Page 21: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/21.jpg)
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.)
![Page 22: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/22.jpg)
Questions
Tsokos page 396 q’s 8-10
![Page 23: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/23.jpg)
History of Quantum MechanicsMax 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.
![Page 24: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/24.jpg)
Participants of the 5th Solvay Congress, Brussels, October 1927
A. Einstein
M Curie
M. Planck
N. Bohr
L.V. de Broglie
W. HeisenbergW. PauliE. Schrödinger
![Page 25: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/25.jpg)
• 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!!
Models of the Atom–
–
––
–
+
+
+ –
![Page 26: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/26.jpg)
Different view of atomsThe Bohr Atom
The Schrodinger Atom
Electrons are only allowed to have discrete energy values and these correspond to changes in orbit.
Electrons behave like stationary 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
Amplitude
![Page 27: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/27.jpg)
SpectraConsider a ball in a hole:
When the ball is here it has its lowest gravitational potential energy.
We can give it potential energy by lifting it up:
If it falls down again it will lose this gpe:
20J
5J
5J
30J
![Page 28: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/28.jpg)
SpectraA 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”
![Page 29: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/29.jpg)
SpectraIf we illuminate the atom we can excite the electron:
0eV
-0.85eV
-1.5eV
-3.4eV
-13.6eV
Q. What wavelength of light would be needed to excite this electron to ionise it?
Light
Energy change = 3.4eV = 5.44x10-
19J.Using E=hc/λ wavelength = 3.66x10-7m(In other words, ultra violet light)
![Page 30: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/30.jpg)
Example questions1) 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.
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?
0eV
-0.85eV
-1.5eV
-3.4eV
-13.6eV
![Page 31: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/31.jpg)
SpectraContinuous spectrum
Absorption spectrum
Emission spectrum
![Page 32: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/32.jpg)
Emission SpectraHydrogen
Helium
Sodium
![Page 33: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/33.jpg)
Observing the Spectra
Light source
GasCollimator
Diffraction grating(to separate the colours)
Microscope
(to observe the spectrum)
![Page 34: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/34.jpg)
Questions
Tsokos page 405 q’s 1-7.
![Page 35: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/35.jpg)
• 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!!
Models of the Atom–
–
––
–
+
+
+ –
![Page 36: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/36.jpg)
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.
Schrödinger model
![Page 37: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/37.jpg)
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.
![Page 38: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/38.jpg)
Electron in a box model
Electrons will form standing waves of wavelength 2L/n
![Page 39: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/39.jpg)
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
![Page 40: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/40.jpg)
![Page 41: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/41.jpg)
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!
![Page 42: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/42.jpg)
Heisenberg uncertainty principle
![Page 43: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/43.jpg)
Nuclear physics
![Page 44: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/44.jpg)
Determining the size of the nucleus
![Page 45: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/45.jpg)
25
20
15
10
5
0
0 2 4 6 8 10
distance from nucleus / 10–14 m
Approach of alpha particle to nucleus
Z = 79 (gold)
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?
![Page 46: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/46.jpg)
1. Substituting values gives
.MeV7.2210m 100.1m J C 1085.84
C 106.1792 = 6
1411212
19
P
E
2. Halves, because the potential energy is proportional to 1/r.3. About 4.6x10–14 m, where the graph reaches 5 MeV.
![Page 47: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/47.jpg)
Circular pathsRecall:
++ -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, orr =mv
Bq
![Page 48: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/48.jpg)
Bainbridge mass spectrometerIons are formed at D and pass through the cathode C and then through a slit S1
A particle with a charge q 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
Bqv = Mv2/r
Therefore
r = Mv/(Bq)
Hyperlink
![Page 49: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/49.jpg)
![Page 50: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/50.jpg)
Nuclear energy levels
There are 2 distinct length of tracksin this Alpha decay
Therefore, the energy levels in the nucleus are discrete
![Page 51: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/51.jpg)
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
![Page 52: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/52.jpg)
Changes in Mass and Proton Number
11
5
0
+1C
11
6B β+
90
39Sr
90
38Y β
0
-1+
Beta - decay:
Beta + decay:
“positron”
![Page 53: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/53.jpg)
Radioactive Decay Law
dN/N = -λdt which when integrated, gives
Taking antilogs of both sides gives:
![Page 54: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/54.jpg)
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:
![Page 55: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/55.jpg)
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.
![Page 56: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/56.jpg)
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).
![Page 57: Quantum and Nuclear Physics](https://reader035.vdocument.in/reader035/viewer/2022062315/568152d6550346895dc0f14e/html5/thumbnails/57.jpg)
Questions
Tsokos page 412 q’s 1-20