### Ch 30 Nuclear Physics

```Nuclear Physics
The Atom
 All matter is composed of atoms.
 An atom is composed of three subatomic
particles: electrons (-), protons (+), and
neutrons (0)
 The nucleus of the atom contains the protons
and the neutrons (also called nucleons.)
 The electrons surround (orbit) the nucleus.
 Electrons and protons have equal but
opposite charges.
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The Atom
 Protons and neutrons have nearly the same mass and
are 2000 times more massive than an electron.
 Electron discovered by J.J. Thomson in 1897
 Proton discovered by Ernest Rutherford in 1918
 Neutron discovered by James Chadwick in 1932
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Rutherford's Alpha-Scattering Experiment
 J.J. Thomson’s “plum pudding” model predicted the
alpha particles would pass through the evenly
distributed positive charges in the gold atoms.
a particle = helium nucleus
4
The Atom
 Diameter of nucleus = about 10-14 m
 Electron orbit diameter = about 10-10 m
 Atomic Mass is concentrated in the nucleus
(>99.97%)
 Therefore the volume (or size) of an atom is
determined by the orbiting electrons.
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Nucleus
 Nucleus composed of neutrons and protons
 Protons are made of two “up” quarks and one “down” quark.
 Neutrons are made of one “up” quark and two “down” quarks.
u
u
+ 2/3 + 2/3
d
1/3
u
+ 2/3
d
1/3
d
1/3
Proton Charge:
2/3 + 2/3 -1/3 = +1
Neutron Charge:
2/3 - 1/3 -1/3 = 0
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Definitions
 Atomic Number (Z) – the # of protons in the
nucleus (“defines” the element – the # of
protons is always the same for a given
element)
 Atomic Number also designates the number
of electrons in an element.
 If an element either gains or loses electrons,
the resulting particle is called an ion.
 For example, if a sodium atom (Na) loses an
electron it becomes a sodium ion (Na+)
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More Atomic Designations
 Atomic Number (Z) – the # of protons
 Mass Number (A) – protons + neutrons, or the total
number of nucleons
 Isotope – when the number of neutrons vary in the
nucleus of a given element (always same number of
protons)
 Only 112 elements are known, but the total number
mass number
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Isotopes
 Some elements have several isotopes
(like carbon – 12C, 13C, 14C)
 Isotopes of a single element have the ‘same’
chemical properties (due to same number of
electrons), but they have different masses
(due to varying number of neutrons.)
 Due to their various masses isotopes behave
slightly different during reactions.
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Carbon Isotopes
 Neutron number = mass number – atomic number
N=A-Z
Symbol
Protons Neutrons Mass #
(Z)
(N)
(A)
12C
6
6
12
13C
6
7
13
14C
6
8
14
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Three Isotopes of Hydrogen
In naturally occurring Hydrogen - 1 atom in 6000 is
deuterium and 1 in 10,000,000 is tritium. Heavy water = D2O
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Example - Determining the Composition of an Atom
 Determine the number of protons, electrons, and
neutrons in the fluorine atom 19F






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Atomic Number (Z) = 9
so protons = 9 and electrons = 9
Mass Number (A) = 19
A = N + Z {N = Neutron Number}
so N = A – Z = 19 – 9 = 10
neutrons = 10
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Review of Atom
 Protons & Neutrons – in nucleus
 Electrons – orbit around nucleus
 Mass Number (A) = protons + neutrons
 Atomic Number (Z) = # of protons
 Neutron Number (N) = # of neutrons
 Isotope – an element with different # of
neutrons (same # of protons)
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Atomic Mass
 The weighted average mass of an atom of
the element in a naturally occurring sample
 The Atomic Mass is measured in unified
atomic mass units (u) – basically the weight
of a proton or neutron.
 The 12C atom is used as the standard, and is
assigned the Atomic Mass of exactly 12 u.
 The weighted average mass of all carbon is
slightly higher than 12 (12.011) because
some is 13C and 14C.
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Schematic Drawing of a Mass Spectrometer
The ion with the greatest mass is deflected the least, the ion
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with the least mass is deflected the most.
Calculating an Element’s Atomic Mass
 Naturally occurring chlorine is a mixture
consisting of 75.77% 35Cl (atomic mass =
34.97 u) and 24.23% 37Cl (atomic mass =
36.97 u). Calculate the atomic mass for
the element chlorine.
 Calculate the contribution of each Cl isotope.
 0.7577 x 34.97 u = 26.50 u (35Cl)
 0.2423 x 36.97 u = 8.96 u (37Cl)
 Atomic Mass for Cl = 35.46 u = Total
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Strong Nuclear Force
 Two fundamental forces of nature –
gravitational and electromagnetic.
 The electromagnetic force between a proton
(+) and an electron (-) is 1039 greater than the
gravitational forces between the two particles.
m1
r
e-
m1
p+
re
 Strong nuclear force between nucleons
p+
p+
p+
n
n
n
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Strong Nuclear Force
 According to Coulomb’s Law like charges repel each




other. Therefore the repulsive forces in a nucleus are
huge and the nucleus should fly apart.
There must exist a third fundamental force that
somehow holds the nucleus together.
That 3rd force is the strong nuclear force. It is much
stronger than the electromagnetic force.
But strong only at short range! (< 10-14 m)
As nucleus gets larger, the strong nuclear force is
much weaker and is overcome by the electromagnetic
force. The nucleus decays or disintegrates.
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spontaneous process of nuclei undergoing a
change by emitting particles or rays
 Nuclide – a specific type of nucleus 238U or 14C
undergo spontaneous decay (disintegration)
 Substances that give off such radiation are
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 Parent nucleus – the original nucleus before
decay
 Daughter nucleus (or daughter product) – the
resulting nucleus after decay
 Radioactive nuclei can decay (disintegrate) in
three common ways



Alpha decay
Beta decay
Gamma decay
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 Alpha decay – disintegration of a nucleus into a nucleus
of another element, with the emission of an alpha particle
(α) - a helium nucleus (42He)
A
B + 4He
2
 Beta decay – a neutron is transformed into a proton, with
the emission of a beta particle (β) – an electron ( -10e)
A
B + 0e
-1
 Gamma decay – occurs when a nucleus emits a gamma
ray (γ) and becomes a less energetic form of the same
nucleus
A*
B+γ
The * means nucleus is in excited state.
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Alpha(a), Beta(b), Gamma(g)
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Nuclear Decay Equations - Examples
• Alpha decay =
232Th
90

228Ra
88
+ 24He
A = 228+4=232
Z = 88+2 = 90
14N + 0e
• Beta decay = 14
C

6
7
-1
A = 14 + 0 = 14
Z=7-1=6
204Pb + g
• Gamma decay = 204
Pb*

82
82
A = 204 + 0 = 204
Z = 82 + 0 = 82
 In a nuclear decay equation, the sums of the
mass numbers (A) and the sums of the atomic
numbers (Z) will be the equivalent on each
side
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The Products of Alpha Decay – Example
•
238 U
92
•
238 U
92

A
X
Z
•
238 U
92

234X
90
undergoes alpha decay. Write the equation
for the process
• 238
U?
92
+
4He
2
238 = A + 4 so A = 234
92 = Z + 2 so Z = 90
+ 4He
2
 Must determine the mass number (A), the
atomic number (Z), and the chemical symbol
for the daughter product
4He
• 238 U  234
Th
+
90
92
2
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Decay Series of Uranium-238 to Lead-206
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 Pattern emerges:


Most stable nuclides have an even number of
both protons and neutrons (even-even nuclides)
Most unstable nuclides have an odd number of
both protons and neutrons (odd-odd nuclides)
 A nuclide will be radioactive if:
- Its atomic number (Z) is greater than 83
– n < p (except for 11H and 32He)
14N)
– It is an odd-odd nuclide (except for 21H, 36Li, 10
B,
5
7
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 Identify the radionuclide in each pair, and
222Rn
a) 208
Pb
and
82
86
• Z above 83
19Ne and 20Ne
b) 10
10
• fewer n than p
63Cu and 64Cu
c) 29
29
• odd-odd
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 Half-Life – the time it takes for half of the nuclei
of a given sample to decay
 In other words – after one half-life has expired,
only one-half of the original amount of
 After 2 half-lives only one-quarter (½ of ½) of
the original amount of the radionuclide remains
undecayed
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Decay of Thorium-234
 Decay of Thorium-234 over Two Half-Lives
 Thorium-234 has a Half-Life of 24 days
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 Curve can be applied
Number
of Half Lives
Number
of Nuclei
0
1 No
1
½ No
2
¼ No
3
1/8 No
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Half-Life
 Carbon-14 dating can be used for 10s of
thousands of years.
 Uranium-238 dating can be used for 100s of
millions of years
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Example - Finding the Number of Half-Lives
 What fraction and mass of a 40 mg sample of iodine-
131 (half-live = 8d) will remain in 24d?
• Step 1 – find the number of half-lives that have
passed in 24 days: 24 days = 3 half-lives
8d/half-life
 Step 2 – Start with the given amount No = 40mg, and
half it 3 times (3 half-lives)



Once  No/2 = 20 mg (after 8 days)
Twice  No/4 = 10 mg (after 16 days)
Thrice  No/8 = 5 mg (after 24 days)
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Example - Finding the Elapsed Time
 How long would it take a sample of
14
C to decay
to one-fourth its original activity? (half-live of 14C
is 5730 years)
 Solution: No  No/2  No/4
So 14C would need to decay for two half-lives in
order to be reduced to ¼ its original activity.
 (2 half-lives)(5730 y / half-life) = 11,460 years
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Determining the Half-Live
 In order to determine the half-life of a
particular radionuclide, we must monitor the
activity of a known amount in the laboratory
 Activity – the rate of emission of the decay
particles (usually in counts per minute, cpm)
 When (time) the initial activity rate has fallen
to one-half – we have reached One Half-Life
 Measured with a Geiger Counter

Geiger developed this in 1913 helping
Rutherford
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Nuclear Reaction
 We know that radioactive nuclei can spontaneously
change into nuclei of other elements, a process
called transmutation.
 Particle (proton or neutron) can also be added to a
nucleus to change it into another element.
 This process is called a nuclear reaction.
a + A
B + b
 a is particle that bombards nucleus A
to form nucleus B and an emitted particle b
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Gold from Mercury!
 Particle accelerators can create gold (Au) by
bombarding Mercury with Hydrogen.
197Au + 4He
• 11H + 200
Hg

80
79
2
 But very expensive! \$1,000,000 per ounce!
 Elements with atomic numbers greater than
92 have been created this way.
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Example - Equation for a Nuclear Reaction
 Complete the equation for the proton bombardment of
lithium-7.
• 11H + 73Li  ???? + 01n
 Note, the sum of the mass #’s on left = 8.
 The mass # on the right must also = 8, therefore the
missing particle must have a mass # = 7.
 The sum of the atomic #’s on left = 4
 Therefore the sum of the atomic #’s on right must also
equal 4.
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Example - Equation for a Nuclear Reaction
• 11H + 73Li  ???? + 01n
 The missing particle must have an atomic
number = 4
 Therefore the missing particle has a mass
number of 7 and an atomic number of 4.
 This element is 7Be (beryllium.)
4
 Completed equation 
• 11H + 73Li  74Be + 10n
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Smoke Detector
• A weak radioactive source ionizes the air and
sets up a small current. If smoke particles
enter, the current is reduced, causing an
alarm.
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Units Used for Mass and Energy
 Mass in atomic mass units
 Energy in mega electron volts (MeV)
1 MeV = 1.60 x 10-13 J
 1 u has energy of 931 MeV
 1 u = 1.66 x 10-27 kg ≈ mass of proton
 1 eV = qe (1 V) = 1.6 x 10-19 C V (1 J = C V)
 E = mc2 = (1.66 x 10-27 kg)(3.0x108m/s)2
= 1.49 x 10-10 J / (1.60 x 10-13 J / MeV)
 931 MeV / u
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 Alpha: Emits a helium nucleus (heavy, charged particle)
 Beta: Emits an electron (light, charged particle)
 Gamma: Emits very high energy EM wave (uncharged)
 The figure below shows penetration of these three types
 Only Gamma radiation harmful to humans externally.
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 Radiometric dating – the determination of age
by the measurement of the rate of decay of
material
 Recall that an atomic nuclei is said to be
radioactive when it will naturally decay.
 The product of decay is generally called the
daughter nuclei or daughter product.

Daughter products may themselves be stable
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 Half-life – the length of time taken for half of
the radionuclide in a sample to decay
 This rate of decay has been found to always
be constant.

Unaffected by temperature, pressure, and
chemical environment
 The older the rock the less parent and the
more daughter product is present.
 Different radioactive parents may have
drastically different half-lives.
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 As the parent nuclide decays the proportion
of the parent decreases and the proportion
of the daughter increases.
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Rock “Clocks” – Condition #1
 Radioactive decay can serve as a “clock” for
dating rocks, if the following conditions are
met
1

Over the lifetime of the rock, no daughter or
parent has been added or subtracted.


This condition requires that there has been no
contamination of the rock.
If either parent or daughter nuclides are added or
subtracted by metamorphism or fluid movement,
the date obtained is not valid.
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Rock “Clocks” – Condition #2
2
 The age of the rock is reasonably close to the



If too many half-lives transpire it may become
impossible to measure the amount of the
remaining parent nuclide.
If only a small portion of one half-life transpires
then it may be impossible to measure the
amount daughter product present.
In either case, a valid date cannot be
obtained.
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Rock “Clocks” - Condition #3
3
 No daughter product was present when the rock
initially formed.



If daughter product was present when the rock
formed, later analysis of the rock will result in an
inaccurate parent to daughter ratio.
Sometimes it may be possible to determine the
amount of daughter nuclide initially present.
In order to use radiometric dating techniques at all,
the rocks must actually contain the appropriate
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Condition #3 – Sometimes a Problem
 In the case of uranium-lead dating, there are
 For example, we know that primordial lead
 We also know that lead-204 is never created
 Therefore if any lead-204 is present we know
that the other three lead isotopes are also
present and we know their ratios.
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 Since lead-204 is present, we know how
much of the other isotopes are primordial and
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 Note that since the half-lives vary the range of
ages also varies.
 Not all rocks can be radiometrically dated, only
those with the appropriate mineral present.
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 Note that since the half-lives vary the range of ages also
varies.
 Not all rocks can be radiometrically dated, only those with
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the appropriate mineral present.
 The ratio of U-235 to its daughter, Pb-207, is 1
to 3 in a certain rock. That is, only 25% of the
original U-235 remains. (The half-life of U-235
is 704 x 106 years.) How old is the rock?
 To decay from 100% to 25% takes 2 half-lives.
 100%  50%  25%
 (2) x (704 x 106 years) = 1408 x 106 years
 or 1.41 billion years = age of the rock
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Carbon Dating
 Developed in 1950 by American, Willard Libby
 Carbon-14 (14C) dating is the only radiometric
dating technique that can be used to date onceliving organisms.
 14C is a radionuclide with a half-life of 5730
years.
 The age of an ancient organic remain is
measured by comparing the amount of 14C in
the ancient sample compared to the amount of
14
C in modern organic matter.
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Carbon Dating





14
C is a natural product
formed in the atmosphere.
About one in a trillion C
atoms in plants is 14C.
14
C is incorporated into all
living organisms.
Living matter has an
counts/minute/gram C.
At death the 14C present
begins to decay.
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Carbon Dating – Modern Methods
 In the newest carbon dating techniques, the
14
12
amounts of both C and C are measured.
 The ratio of these two isotopes in the ancient
sample is compared to the ratio in living
matter.
 Using this method only very small samples
are needed and specimens as old as 75,000
years can be accurately dated.
 Beyond 75,000 years, the amount of
14
C still
not decaying is too small to measure.
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Carbon Dating - Limitations
 Carbon dating techniques assumes that the
14
amount of C in the atmosphere (and
therefore in living organisms) has been
constant for the past 75,000 years.
 We now know that the amount of
14
C in the
atmosphere has varied by (+) or (-) 5%.
 These variations in
14
C levels have been due
to changes in solar activity and changes in
the Earth’s magnetic field.
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Carbon Dating - Limitations
 These slight variations in
14
C abundance
have been corrected by careful analyses of
California’s 5,000 year-old bristlecone pines.
 An extremely accurate calibration curve has
been developed for 14C dates back to about
5000 B.C.
 Carbon dating is widely used in archaeology,
and has been used to date bones and other
organic remains, charcoal from fires, beams
in pyramids, the Dead Sea Scrolls, and the
Shroud of Turin.
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Equations
 N=A–Z
A
X
Z
N = number of neutrons
 Alpha Decay:
A
B + 42He
 Beta Decay:
A
B + 0e
 Gamma Decay:
A*
B+γ
-1
 Half Life = Time / (time/half-life)
 E = mc2
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