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WileyPLUS Assignment 5 is available
Chapters 28, 29, 30
Due Friday, April 9 at 11 pm
Friday, April 9
Review - send questions!
PHYS 1030 Final Exam
Friday, April 23, 1:30-4:30 pm
Frank Kennedy Brown Gym,
30 questions, the whole course, formula sheet provided
Monday, April 5, 2010
The Laser
1
Metastable state (~10-3 s)
Photon:
E = Ei – Ef
Population inversion
Light amplification occurs when:
• there is a “population inversion”,
• the excited state is long-lived – a “metastable state”,
• a photon of energy Ei - Ef triggers an electron in state Ei to fall to
state Ef, emitting a second photon of energy Ei - Ef : (stimulated
emission).
The result is a build up of photons of the same wavelength that are in
phase – coherent light.
If there is no population inversion, the initial photon is more likely to be
absorbed by an electron that moves from Ef to Ei.
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2
Lasers
Many uses:
• playback/recording of CDs, DVDs
• checkout scanners
• welding metal parts
• accurate distance measurement from time of travel of light pulses
• telecommunications
• study of molecular structure
• medicine
– selective removal of tissue, example, shaping the cornea of the eye
– removal of “port wine stain” birth marks
– destroying tumours with light-activated drugs
– reattaching detached retinas
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3
Laser Altimetry of Mars
Volcano
Impact crater
~7 km deep!
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Red: high
Purple: low
4
Photorefractive keratectomy (PRK)
Offers an alternative treatment for nearsightedness and
farsightedness that does not rely on lenses
UV laser light
absorbed
by cornea,
causing
vaporization
of material
Correcting nearsighted
Correcting farsighted
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Holography
A laser beam is split into two parts. The first, a
reference beam, travels directly to the film.
The second reflects from the object and
recombines with the reference beam at the
film, generating interference fringes there.
Hologram
(interference
fringes, not
an image)
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6
http://hyperphysics.phy-astr.gsu.edu/Hbase/optmod/holfilm.html#c1
A Hologram
Image encoded
in the hologram
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Viewing the image by
shining laser light on
the hologram
d
The fringes act as slits
of varying spacing
O
" for first-order
sin ! = maximum (the
d
brightest)
d = spacing of the fringes
P
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Holograms and Parallax
The images were taken through a transmission hologram. The hologram was
illuminated from behind by a helium-neon laser which has been passed through
a diverging lens to spread the beam over the hologram.
This recorded interference pattern actually contains much more information
than a focused image, and enables the viewer to view a true three-dimensional
image which exhibits parallax. That is, the image will change its appearance if
you look at it from a different angle, just as if you were looking at a real 3D
object.
http://hyperphysics.phy-astr.gsu.edu/Hbase/optmod/holog.html
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Summary of Chapter 30
• The positive charge of the the atom is concentrated in a nucleus
that is ~10-15 - 10-14 m in radius.
• The angular momentum of the electron in an atom is quantized,
leading to quantized energy levels in hydrogen-like atoms:
En = –13.6 Z2/n2 eV
• Photons are emitted and absorbed by atoms at the same wavelength
! identification of elements in the “atmosphere” of stars,
discovery of helium.
• The energies of K" x-rays can be calculated by replacing Z by Z–1
! identification of some chemical elements for the first time.
• Stimulated emission, the laser.
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Chapter 31: Nuclear Physics & Radioactivity
• Nuclear structure, nuclear size
• The strong nuclear force, nuclear stability, binding energy
• Radioactive decay, activity, half life
• The neutrino
• Radioactive age measurement
• Decay series
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The Nucleus
Protons and neutrons (“nucleons”) are closely packed together in
nuclei that are roughly spherical in shape.
Proton: q = +e
Neutron: q = 0
neutrons and protons have almost the same mass
Number of protons, Z = atomic number
Number of neutrons, N = neutron number
Total number of nucleons, A = mass number, or nucleon number
chemical symbol of
the element
A=Z+N
Nuclei are specified by:
Example, 146C
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A
ZX
Z is sometimes omitted, as the chemical symbol
gives the same information
12
Isotopes: Nuclei of the same chemical element (same atomic number,
Z), but different A and N.
Example:
12
C, 13C, 14C.
Only 12C, 13C are stable.
The radius of a nucleus of mass number A is:
r = r0A1/3, r0 = 1.2 × 10−15 m
That means that nuclei have the same
density:
Density =
p
n
mass
AmN
�4 3
volume 3 !r
(mN = average mass of a nucleon)
AmN
3mN
3 × 1.67 × 10−27 kg
17
3
Density = 4 3 =
3 = 4!(1.2 × 10−15 m)3 = 2.3 × 10 kg/m !!!
4!r0
3 !r0 A
Comparable with the supposed density of a black hole or a neutron star.
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One isotope (X) contains an equal number of neutrons and protons.
Another isotope (Y) of the same element has twice the number of
neutrons as the first.
Determine the ratio rY/rX of the nuclear radii of the isotopes.
1.145
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The Strong Nuclear Force
Nuclei are held together by the
strong nuclear force.
– gravity is much too weak
– the Coulomb force between proton
charges is repulsive and decreases
nuclear stability.
The “valley of stability”
Stable nuclei
Effect of Coulomb
repulsion between
protons
The strong nuclear force is:
– attractive
– extends over only ~10-15 m
(a short-range, nearest-neighbour
force)
The repulsive Coulomb force
between protons favours nuclei with
slightly more neutrons than protons.
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Binding energy, mass defect
Z protons
+ N neutrons
Z protons, N neutrons
Mass
Mass defect:
Zmp + Nmn
Δm = mass defect
mnucleus
!m = (Zm p + Nmn) − mnucleus
Binding energy: B = !m c2
mnucleus = (Zm p + Nmn) − B/c2
Binding energy = energy to break up the nucleus into its constituent nucleons.
Mass tables give the mass of neutral atoms with electrons in orbit.
Then, Δm = [(ZmH + Nmn) – matom], mH = mass of H-atom
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Atomic and Nuclear Mass
Atomic mass unit (u): 12 u = mass of 12C atom (including the 6 electrons)
1 u is equivalent to a mass energy, mc2, of 931.5 MeV.
Mass
Particle
Electric
Charge
Kilograms
Atomic Mass
Units (u)
Electron
-e
9.109382!10-31
5.485799!10-4
Proton
+e
1.672622!10-27
1.007276
Neutron
0
1.674927!10-27
1.008665
Hydrogen atom
0
1.673534!10-27
1.007825
1 u = 1.66054x10-27 kg
Atomic mass (including Z electrons) = nuclear mass + mass of Z electrons.
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Example
The mass defect is:
!m = 2 ! (mass of H atom + mass of neutron) – (mass of 4He atom)
= 4.0330 – 4.0026 u
!m = 0.0304 u
Using the energy equivalent of the atomic mass unit, the binding energy
is: B = (0.0304 u) ! 931.5 MeV/u = 28.4 MeV.
B 28.4 × 106 × 1.6 × 10−19 J
Mass defect, !m = 2 =
= 5 × 10−29 kg
8
2
c
(3 × 10 )
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Binding energy per nucleon, B/A
Fission
energy release
Unstable
beyond 209Bi
Peak value,
8.7 MeV
per nucleon
Sharp fall due to small
number of nearest
neighbour nucleons –
short range nuclear force
Decrease due to
Coulomb repulsion
between protons
Why are certain nuclei unstable?
Because neighbouring nuclei have lower mass energy. Decay is possible
to the lower mass nuclei while releasing kinetic energy.
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Prob. 31.13/47: Mercury 202Hg (Z = 80) has an atomic mass of
201.970617 u. Obtain the binding energy per nucleon.
• Work out the mass defect knowing the mass of the atom.
• Convert the mass defect to a binding energy.
Mass
Particle
Electric
Charge
Kilograms
Atomic Mass
Units (u)
Electron
-e
9.109382!10-31
5.485799!10-4
Proton
+e
1.672622!10-27
1.007276
Neutron
0
1.674927!10-27
1.008665
Hydrogen atom
0
1.673534!10-27
1.007825
7.90 MeV/nucleon
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Radioactivity
Three forms:
• Alpha (") – the nucleus of a 4He atom is emitted from the “parent” nucleus
• Beta (#)
– an electron (+ or – charge) is emitted
• Gamma ($) – a nucleus falls from one energy level to another and emits a
gamma ray photon
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Conserved quantities in radioactive decay
Conserved quantities:
• number of nucleons (nucleon number)
• charge
• energy
• linear momentum
• angular momentum
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Alpha Decay
Nucleon number
Charge
A
ZX
→
Parent
Nucleon number: A = (A – 4) + 4
Charge:
Z = (Z – 2) + 2
A−4
Z−2Y
+ 42He
"-particle
Daughter
!
!
238
92 U
→
234
90 Th
+
4
2 He
Daughter and "-particle: greater binding energy, lower combined
mass than parent energy is released in the decay.
Energy released = [mX – (mY + m")] ! 931.5 MeV, if masses in atomic
mass units (u).
The "-particles have a kinetic energy of typically a few MeV.
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Alpha Decay in a Smoke Detector
Alpha particles from a weak source
collide with air molecules and ionize
them, which allows a current to
flow between the plates.
In the presence of smoke, ions
colliding with smoke are generally
neutralized (i.e. neutral atoms are
formed), so that the current
decreases and the alarm is tripped.
When the battery is low, the current is low, which also trips the alarm!
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Prob. 31.23/20: Find the energy (in MeV) released when alphadecay converts radium 226Ra (Z = 88, atomic mass = 226.02540 u)
into radon 222Rn (Z = 86, atomic mass = 222.01757 u).
The atomic mass of an alpha particle is 4.002603 u.
4.87 MeV
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Beta (β–) Decay
Nucleon number
Charge
A
ZX
→
A
Z+1Y
+ e−
(or #–)
Nucleon number: A = A + 0
Charge: #
Z = (Z + 1) + (–1)
234
90 Th
Positron (β+) Decay
Nucleon number
Charge
A
ZX
→
A
Z−1Y
→
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Z = (Z – 1) + 1
+ e−
+ e+
(or #+)
Nucleon number: A = A + 0
Charge: #
234
91 Pa
22
11 Na
→
22
10 Ne
+ e+
26
Beta (β–) Decay
234
90 Th
Nucleus
→
234
91 Pa
+ e−
234
90 Th
Z = 90
234
91 Pa
Neutron into proton
Z = 91
Beta decay
90 electrons
(protactinium)
e–
90 electrons
Neutral atom
has 91 electrons
To calculate the energy released using tabulated masses of neutral
Neutral atom
atoms, the e– that is generated in the beta decay is lumped in with the
90 existing atomic electrons to form a neutral Pa atom and then,
!E = [mT h − mPa] × 931.5 MeV
atomic masses, in atomic mass units
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Beta (β+) Decay
22
11 Na
Nucleus
→
22
10 Ne
+ e+
22
10 Ne
22
11 Na
Z = 11
11 electrons
Proton into neutron
Beta decay
Z = 10
e+
10 electrons
1 electron, e–
Neutral atom
Neutral atom
has 10 electrons
Using tabulated atomic masses, the energy released in the decay is,
%
&E = [mNa - (mNe + 2me)] x 931.5 MeV
atomic masses, in atomic mass units
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