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!
1Friday, April 3, 2009
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
2Friday, April 3, 2009
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
3Friday, April 3, 2009
The Nucleus
Protons and neutrons (“nucleons”) are closely packed together in nuclei that are roughly spherical in shape.
Proton: q = +eNeutron: q = 0
Number of protons, Z = atomic numberNumber of neutrons, N = neutron number
Total number of nucleons, A = mass number, or nucleon number
A = Z + N
Nuclei are specified by:A
ZX
chemical symbol of
the element
Example, 146CZ is sometimes omitted, as the chemical symbol
gives the same information
neutrons and protons have almost the same mass
4Friday, April 3, 2009
That means that nuclei have the same
density:
Density =mass
volume! AmN
43!r3
(mN = average mass of a nucleon)
Density =AmN
43!r30A
=3mN
4!r30
The radius of a nucleus of mass number A is:
r = r0A1/3, r0 = 1.2×10−15 m
pn
=3×1.67×10−27 kg4!(1.2×10−15 m)3
= 2.3×1017 kg/m3 !!!
Comparable with the supposed density of a black hole or a neutron star.
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.
5Friday, April 3, 2009
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.
6Friday, April 3, 2009
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.
Stable nuclei
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.
Effect of Coulomb
repulsion between
protons
The “valley of stability”
7Friday, April 3, 2009
Binding energy, mass defect
MassZm
p + Nm
n
mnucleus
Δm = mass defect
Z protons, N neutrons
!m= (Zmp+Nmn)−mnucleus
Mass defect:
Binding energy = energy to break up the nucleus into its constituent nucleons.
Binding energy: B= !m c2
Alternatively, a neutral atom with Z electrons is broken up into N
neutrons and Z hydrogen atoms. Then, Δm = [(ZmH + Nmn) – matom].
Z protons
+ N neutrons
mnucleus = (Zmp +Nmn)−B/c2
8Friday, April 3, 2009
Atomic mass unit (u): 12 u = mass of 12C atom (including the 6 electrons)
Atomic and Nuclear Mass
MassMass
ParticleElectric
ChargeKilograms
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 is equivalent to a mass energy, mc2, of 931.5 MeV.
Atomic mass (including Z electrons) = nuclear mass + mass of Z electrons.
9Friday, April 3, 2009
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
Mass defect, !m=B
c2=28.4×106×1.6×10−19 J
(3×108)2 = 5×10−29 kg
Using the energy equivalent of the atomic mass unit, the binding energy
is: B = (0.0304 u) ! 931.5 MeV/u = 28.4 MeV.
10Friday, April 3, 2009
Binding energy per nucleon, B/A
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
Unstable
beyond 209Bi
Why are certain nuclei unstable?
Because neighbouring nuclei have lower mass energy. Decay is possible
to the lower mass nuclei while releasing kinetic energy.
Fission ⇒ energy release
11Friday, April 3, 2009
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.
MassMass
ParticleElectric
ChargeKilograms
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
12Friday, April 3, 2009
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
13Friday, April 3, 2009
Conserved quantities in radioactive decay
Conserved quantities:
• number of nucleons (nucleon number)
• charge
• energy
• linear momentum
• angular momentum
14Friday, April 3, 2009
Alpha Decay
A
ZX → A−4
Z−2Y + 4
2He
Parent Daughter (!)
Nucleon number: A = (A – 4) + 4 !
Charge: Z = (Z – 2) + 2 !
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.
Nucleon number
Charge
23892U → 234
90Th + 42He
15Friday, April 3, 2009
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!
16Friday, April 3, 2009
Prob. 31.20/50: Find the energy (in MeV) released when alpha-
decay 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.
17Friday, April 3, 2009
Beta (β–) Decay
Nucleon number: A = A + 0
Charge: # Z = (Z + 1) + (–1)
Nucleon number
Charge
A
ZX → A
Z+1Y + e−
Nucleon number: A = A + 0
Charge: # Z = (Z – 1) + 1
(or "–)
Positron (β+) Decay
A
ZX → A
Z−1Y + e+Nucleon number
Charge
(or "+)
234
90Th → 234
91Pa + e
−
2211Na → 22
10Ne + e+
18Friday, April 3, 2009
Beta (β–) Decay
Z = 90
90 electrons
Nucleus
Neutral atom Neutral atom
has 91 electronsTo calculate the energy released using tabulated masses of neutral
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 = [mTh−mPa]×931.5 MeV
Beta decay
234
90Th → 234
91Pa + e
−
234
90Th
Z = 91
90 electrons
234
91Pa
e–
(protactinium)
atomic masses, in atomic mass units
Neutron into proton
19Friday, April 3, 2009
Beta (β+) Decay
Z = 11
11 electrons
Nucleus
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
Beta decayZ = 10
10 electrons
e+
atomic masses, in atomic mass units
2211Na → 22
10Ne + e+
1 electron, e–
2211Na 22
10Ne
Proton into neutron
20Friday, April 3, 2009
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
21Friday, April 3, 2009
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html
Beta-decay – a problem
X (parent, at rest)
Y (daughter)
e– or e+
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?! No, energy is conserved...
Expected energy of the e+
Beta-decay
X → Y + e
22Friday, April 3, 2009
The Neutrino
6429Cu → 64
28Ni + e+ + ν
A third particle, a neutrino, is also emitted in the decay, so that
the released energy is shared between three particles instead
of two:
Expected energy of the e+
The neutrino is very difficult to detect.
23Friday, April 3, 2009
Neutrino detector in Japan
24Friday, April 3, 2009
Neutrino detector in Japan - X-Files version
25Friday, April 3, 2009
Gamma (γ) Decay
A
ZX∗ → A
ZX+ !
Excited state
of the nucleus
Nuclear
Energy
# ray photon
X*
X
Gamma rays are produced in the decay
(de-excitation) of a nuclear state.
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.
26Friday, April 3, 2009
Gamma Knife – to zap a tumour
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.
60Co gamma ray source:
60
28Ni∗ → 60
28Ni+ ! (# 1.2 MeV)
60
27Co → 60
28Ni∗+ e
−+ !̄ Tumour
Half of the 60Co decays away in 5.3 years, so has to be replaced...
27Friday, April 3, 2009
Winnipeg Free Press,
April 4, 2004
28Friday, April 3, 2009
Winnipeg Free Press, March 19, 2008
29Friday, April 3, 2009