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WileyPLUS Assignment 4
Chapters 28 - 31
Due Thursday, April 9 at 11 pm
Review of Course Next Week
Monday, after completing chapter 31,
and Wednesday
Send questions!
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1
PHYS 1030 Final Exam
Wednesday, April 15
9:00 - 12:00
Frank Kennedy Gold Gym, seats 1 - 290
30 multiple choice questions
Formula sheet provided
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2
Chapter 31: Nuclear Physics & Radioactivity
• Nuclear structure, nuclear size
• The strong nuclear force, nuclear stability, binding energy
• Radioactive decay, activity
• 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
4
Isotopes: Nuclei of the same chemical element (same atomic number,
Z), but different A and N.
Example: 12C, 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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Prob. 31.-/6: One isotope (X) contains an equal number of neutrons
and 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.
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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.
Alternatively, a neutral atom with Z electrons is broken up into N
neutrons and Z hydrogen atoms. Then, Δm = [(ZmH + Nmn) – matom].
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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
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. 31.47/10: 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
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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
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. 31.20/50: 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.
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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+
18
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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Prob. 31.27: Find the energy released when "+ decay converts 22Na
(Z = 11, atomic mass = 21.994434 u). Notice that the atomic mass
for 22Na includes the mass of 11 electrons, whereas the atomic mass
for 22Ne (Z = 10, atomic mass = 21.991383 u) includes the mass of
only 10 electrons.
Using tabulated atomic masses, the energy released in the decay is,
$
%E = [mNa - (mNe + 2me)] x 931.5 MeV
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Beta-decay – a problem
Beta-decay
X→Y+e
Y (daughter)
Expected energy of the e+
X (parent, at rest)
e– or e+
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html
Kinematics: if energy and momentum are conserved, the electron (e– or e+)
should have a well-determined kinetic energy following the beta-decay.
But, the electron does not have a well-determined energy, as seen above.
Is energy not conserved?!
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No, energy is conserved...
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The Neutrino
A third particle, a neutrino, is also emitted in the decay, so that
the released energy is shared between three particles instead
of two:
64
29 Cu
→
64
28 Ni
+ e+ + ν
Expected energy of the e+
The neutrino is very difficult to detect.
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Neutrino detector in Japan
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Neutrino detector in Japan - X-Files version
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Gamma (γ) Decay
A ∗
ZX
→
A
ZX
+!
Excited state
of the nucleus
Nuclear
Energy
X*
Gamma rays are produced in the decay
(de-excitation) of a nuclear state.
# ray photon
X
This is similar to the production of a
photon by an atom, except that the energy
levels are associated with the nucleus
itself, not with electrons in the atom.
Gamma rays are generally of higher energy and are even more
penetrating than x-rays.
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Gamma Knife – to zap a tumour
60
Co gamma ray source:
60
27 Co
→
60 ∗
−
¯
28 Ni + e + !
60 ∗
28 Ni
→
60
28 Ni + !
Tumour
(# 1.2 MeV)
Gamma rays from 60Co sources are
channeled through collimators in a
metal helmet.
Gamma rays are concentrated at the
site of the tumour, to selectively
destroy the malignant tissue.
Half of the 60Co decays away in 5.3 years, so has to be replaced...
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Winnipeg Free Press,
April 4, 2004
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Winnipeg Free Press, March 19, 2008
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