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WileyPLUS Assignment 3
Chapters 22, 24 - 27
Due Friday, March 27 at noon
Q24.21: Intensity = Power/AreaQ25.10: Concave mirror - not covered in course, do not attemptQ26.72: Farsighted person - incorrectly marked sometimes?
Assignment 4 to follow, due last day of term
1Friday, March 27, 2009
Week of March 30 - April 3
Tutorial and Test 4: ch. 27, 28
PHYS 1030 Final Exam
Wednesday, April 15
9:00 - 12:00
Frank Kennedy Gold Gym
30 multiple choice questionsFormula sheet provided
2Friday, March 27, 2009
Then, !y !py ! h
Heisenberg Uncertainty Principle
sin!= "/W
Diffraction minima:W sin!= m"
So, !py !"
W× px !
"
W× h
"
!py !h
W
m = 1
m = 1
or, W!py ! h
Heisenberg: sending the particle through the aperture is a way of measuring the particle’s position in the y-direction to a precision: Δy = W.
For small angles, !! "
W! #py
px
m = 1
! = h/px
3Friday, March 27, 2009
Heisenberg Uncertainty Principle – Alternative View
The position of a particle is measured by shining light on it. Because the photons carry momentum, the act of observing the particle perturbs its motion so that its momentum becomes uncertain.
The ultimate resolution in measuring the position of the particle is limited by the wavelength of the light, so "y ! !.
The photon carries momentum and disturbs the motion of the particle in bouncing off it. So the momentum of the particle after observing it is uncertain to "py ! p = h/!.
Then, "y !py ! λ " h/! = h
vy
Photon,
p = h/!
Reflected
photon
Particle
4Friday, March 27, 2009
Because of the wavelike behaviour of matter it is not possible to measure both the position and momentum of an object to arbitrarily high precision.
A more complete analysis shows that the best possible measurements of position and momentum follow the Heisenberg uncertainty relation:
!y !py ≥h
4"
A similar relation holds for the uncertainty of the energy of a particle and the time during which the energy is measured or during which the particle occupies an energy state:
!E !t ≥ h
4"
The Heisenberg uncertainty relations apply to any measurements.
Heisenberg Uncertainty Principle
Also, ∆x ∆px ≥h
4π, ∆z ∆pz ≥
h4π
5Friday, March 27, 2009
A source of uncertainty...
Edition 5 of Cutnell defines the uncertainty relation as:
!y !py ≥h
2"
We use the version in editions 6 and 7:
!E !t ≥ h
4"!y !py ≥
h
4"
The formula sheet uses this version
6Friday, March 27, 2009
Prob. 29.32: An electron is trapped within a sphere whose diameter is 6"10-15 m (~size of the nucleus of an oxygen atom).
What is the minimum uncertainty in the electron’s momentum?
7Friday, March 27, 2009
Prob. 29.-/36: Suppose the minimum uncertainty in the position of a particle is equal to its de Broglie wavelength.
If the particle has an average speed of 4.5"105 m/s, what is the minimum uncertainty in its speed?
8Friday, March 27, 2009
Summary, Wave-Particle Duality
• Light has properties of particles: – energy is carried in packets of radiation (photons) of energy hf = hc/! (photoelectric effect, blackbody radiation) – Photons carry momentum p = E/c = h/! (verified by Compton effect, not covered)
• Particles have properties of waves: – diffraction according to a de Broglie wavelength ! = h/p
• Energy does not take on just any value, but is quantized
" (Planck’s theory of blackbody radiation). Further evidence in quantization of energy levels of atoms (next chapter).
• There is a fundamental limit to how well both position and momentum can be known (Heisenberg Uncertainty Principle).
9Friday, March 27, 2009
Chapter 30: The Nature of the Atom
• Rutherford scattering and the nuclear atom - –#the nucleus is really small!
• Atomic line spectra – wavelengths characteristic of each element
• Bohr model of the hydrogen atom –# - quantization of angular momentum ! quantization of energy
• x-rays
• The laser
• Omit 30.5, 6 – quantum mechanical picture of H-atom, Pauli Exclusion Principle, periodic table
10Friday, March 27, 2009
Early model of the atom
• About 10-10 m in size.
• Positively charged (“pudding”), with negatively charged electrons (“plums”) embedded in the pudding – “plum pudding model”
Stability of the plum pudding was questionable.
Where would the characteristic spectral lines come from?
#
11Friday, March 27, 2009
Geiger+Marsden: Fall of the plum puddingAlpha (#) particles – nuclei of 4He atom, emitted by some radioactive nuclei – were scattered from a thin gold foil and observed on a screen as flashes of light.
Far more were scattered at large angle than would be possible with the weak electric field inside a “plum pudding” atom.
Rutherford: realized that the positive charges of the atom had to be contained in a very small volume – the nucleus –#electric field very strong close to it.
$ planetary model of the atom with electrons in orbit around nucleus
12Friday, March 27, 2009
Very schematic picture of an atom
Size of the nucleus ! 10-15 – 10-14 m
Size of the atom ! 10-10 m
Equal amounts of + and – charges in the neutral atom.
Electrons are in orbit around the nucleus in a planetary model of the atom.
(charge +Ze)
13Friday, March 27, 2009
Problem with a Planetary Model of the Atom
• The electrons suffer centripetal acceleration in their orbital motion.
• Accelerated charges should radiate electromagnetic energy.
$ The electrons should lose energy and spiral into the nucleus in very little time.
$ A planetary atom should not be stable!
$ Classical theory does not explain the structure of the atom.
$ Small systems, such as atoms, must behave differently from
large.
14Friday, March 27, 2009
Fraunhofer absorption lines from the sun
http://www.harmsy.freeuk.com/fraunhofer.html
1 Å (Angstrom) = 0.1 nm
• Fraunhofer lines – due to absorption of sunlight by elements in the atmosphere of the sun •chemical elements emit and absorb light at the same wavelengths• Models of the atom need to explain this
The sun: blackbody radiation for T ! 6000 K
15Friday, March 27, 2009
Line spectrum of the hydrogen atom
InfraredUltraviolet Visible
m Series
1 Lyman ultraviolet
2 Balmer visible
3 Paschen infrared
4 Brackett infrared
Balmer found by trial and error a simple formula to calculate the wavelength of all lines of the hydrogen atom:
R = Rydberg constant = 1.097 " 107 m-1
1
!= R
(1
m2− 1
n2
)m= 1,2,3 . . .
n= m+1,m+2,m+3 . . .
n = 4n = 5
n = 6
n = $
series limit
n = 3
n = 4n = 2
m = 1 m = 2 m = 3+ other series
16Friday, March 27, 2009
1
!= 1.097×107
(1
12− 1"
)= 1.097×107
! = 91.2 nm (series limit)
Spectrum of H-atom
Example: Lyman series, m = 1:
For n = $: “series limit”
For n = 2:
1
!= 1.097×107
(1
12− 1
22
)= 0.823×107
! = 121.5 nm
R = Rydberg constant = 1.097 " 107 m-1
1
!= R
(1
m2− 1
n2
)m= 1,2,3 . . .
n= m+1,m+2,m+3 . . .
(series limit, shortest wavelength in Lyman series)
(longest wavelength in Lyman series)
m = 1
n = 2n = $
m–1
m–1
17Friday, March 27, 2009
Bohr model of the hydrogen atom
Now superseded by more modern quantum mechanical ideas, but gives the correct answers.
Assume:• A planetary model with electron in orbit around the nucleus.• There are certain electron orbits that are stable (“stationary states”). This is contrary to classical theory and is not explained in this model.• Light is emitted or absorbed when an electron changes state.• Energy is conserved, so the energy of the photon is the difference in energy between the initial and final states:
E = hf = Ei – Ef " – what are the allowed energies?
E = hf
“Stationary” states
= Ei – Ef
18Friday, March 27, 2009
The electron is kept in orbit by the attractive Coulomb force between it and the nucleus.
F =kZe
2
r2=mv
2
r
mv2 =
kZe2
r
The potential energy is:
PE=−kZe2
r
And so the total mechanical energy, E = KE + PE, is:
But, what are the “good” values of r corresponding to the stationary states?
= 2 KE (A)
E =kZe
2
2r− kZe
2
r=−kZe
2
2r(B)
Energies of hydrogen atom in the Bohr model
A hydrogen-like atom or ion with just one electron in orbit
m
19Friday, March 27, 2009
Combine with (A): mv2r = kZe
2
So, v2 =
kZe2
mr=
[nh
2!mr
]2
r =1
kZe2m
[nh
2!
]2= (5.29×10−11 m)
n2
Z
En =−[2!2mk2e4
h2
]Z2
n2, n= 1,2,3 . . .
Ln = mvr =nh
2!n = 1, 2, 3...
Energies of hydrogen atom in the Bohr model
And then, substituting into (B), the total mechanical energy is:[E =−kZe
2
2r
]
En =−13.6 Z2
n2eV
Assertion (Bohr): the angular momentum of the electron around the nucleus can have only certain, quantized, values:
m
n = “quantum number”
20Friday, March 27, 2009
Emission of a photon
Emission and absorption occur at the same wavelengths (as seen in the Fraunhofer absorption lines of sunlight).
E = hfEi = Ef +h f
Absorption of a photon
E = hf
Ef +h f = Ei
21Friday, March 27, 2009
Spectrum of hydrogen-like atoms and ions
Energy levels: En =−13.6 Z2
n2eV
E = hf
ni
nf
(Just one electron in orbit around the
nucleus)
In agreement with Balmer’s formula[R=
(13.6×1.602×10−19 J)hc
= 1.097×107 m−1]
= Ei – E
f
Energy of photons: E = Ei − Ef =hc
λ= 13.6Z2
[1n2
f
− 1n2
i
]eV
Z = 1 for hydrogen
22Friday, March 27, 2009
Energy levels of the hydrogen atom – Bohr
En
erg
y %
En
erg
y %
Origin of the lines of the hydrogen atom spectrum
En =−13.6n2
eV
Ground state
Free electron
(Z = 1)
23Friday, March 27, 2009
Prob. 30.-/7: It is possible to use electromagnetic radiation to ionize atoms. To do so, the atoms must absorb the radiation, the photons of which must have enough energy to remove an electron from an atom.
What is the longest radiation wavelength that can be used to ionize the ground state of the hydrogen atom? En
ergy
→
Free electron
The least energy to remove a ground state electron is E$ – E1.
Ground state,n = 1
24Friday, March 27, 2009
Ener
gy →
Ground state
Free electron
First excited state, n = 2
Prob. 30.7/8: The electron in a hydrogen atom is in the first excited state, when the electron acquires an additional 2.86 eV of energy.
What is the quantum number, n, of the state into which the electron moves?
25Friday, March 27, 2009