5/1/2016
Chapter 28 Lecture
Nuclear
Physics
Becquerel and the emissions from uranyl
crystals
 Becquerel found that uranium crystals that had
not been exposed to sunlight formed images on
photographic plates.
 The uranium emitted radiation without an
external source of energy.
Rutherford and experiments investigating
the charge of emitted particles
 Rutherford covered a uranium sample with thin
aluminum sheets to investigate how metal layers
affected the amount of radiation.
Pierre and Marie Curie and the particles
responsible for Becquerel's rays
 Chemical changes or changes in the amount of
light shining on a sample did not lead to
changes in the amount of radiation produced by
uranium salts.
 These findings suggested that the electrons
in the atoms were not responsible for the
rays.
 Marie and Pierre Curie concluded that the
Becquerel rays must come from the nuclei of
atoms.
Rutherford’s Testing experiment
 Radiation emanating from radioactive sample
passes through a B field pointing into the page.
A scintillating screen (which glows when hit with
radiation) is in the plane perpendicular to the
beam.
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Rutherford’s Testing experiment
 If the radiation contains positively charged
particles, then according to the right-hand rule,
the magnetic field should deflect upward with
respect to the original direction.
Rutherford’s Testing experiment
 The scree glowed in three places: one straight
down, one deflected up, and one deflected
down. The location of the downward-deflected
radiation was much farther from the original
beam than for the upward-deflected radiation.
The early model of the nucleus
 The nucleus of an atom is made of positively
charged alpha particles and negatively charged
electrons.
 When a nucleus contains a large number of
alpha particles, they start repelling each other
more strongly than the electrons can attract
them, and the alpha particles leave the
nucleus.
 This leaves behind electrons that repel each
other; thus beta rays are emitted.
 The nucleus is left in an excited state and
emits a high-energy photon, a gamma ray.
Rutherford’s Testing experiment
 If the radiation contains negatively charged
particles, then according to the right-hand rule,
the magnetic field should deflect downward with
respect to the original direction.
Alpha particles, beta rays, and gamma rays
 Rutherford found positively charged particles
with a mass-to-charge ratio twice that of a
hydrogen ion; they were called alpha rays or
alpha particles.
 Negatively charged particles had the same
mass-to-charge ratio as that of the electron; they
were called beta rays.
 Neutral radiation was thought to consist of highenergy electromagnetic waves, called gamma
rays.
Problem with the early model of the nucleus
 A hydrogen atom is lighter than an alpha
particle.
 What, then, is the composition of its nucleus?
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Size of the nucleus: Too small for an
electron
 We can use the uncertainty principle and the
size of the nucleus to show that our current
models would result in atoms that rapidly lose
their electrons.
The search for a neutral particle
 An alpha particle has the charge of two protons
but four times the mass of a proton; thus it
cannot be made of two protons.
 In 1920, Rutherford suggested a neutral
particle with the approximate mass of a
proton.
 In 1928, Bothe and Becker took the initial
step in this search.
The neutral radiation is not a gamma-ray
photon
 By comparing the energies and momenta of the
particles knocked out of different atoms,
Chadwick determined that the particles were
uncharged particles with a mass approximately
equal to that of the proton.
ISOTOPES
 Carbon ions are accelerated
to high speeds by letting
them pass through a region
with high potential difference
and then pass them through
a magnetic field, observing
the paths the take. Atoms of
a single chemical element
take different paths.
Revising ideas of the structure of the
nucleus
• A new model of nuclear
constituents evolved
that involved protons
and neutrons.
– The protons
accounted for the
electric charge of the
nucleus.
– The uncharged
neutrons accounted
for the extra mass.
Centripetal Force = Magnetic Force
FC  FB
m  v2
 Bvq
R
v
BqR
m
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Change in Energy = Kinetic Energy
Whiteboard: Write an expression for “R”
BqR
m
v
U E  q  V
KE  q  V
v
2  q  V
m
R
2  m  V
B2  q
m  v2
 q  V
2
v
2  q  V
m
ISOTOPES
R
Isotopes
2  m  V
B2  q
Radius of the circle does
depend on the mass
 Atoms of a single chemical
element with the same
atomic number have different
masses.
Thus,
different
atoms must have a different
number of neutrons in the
nucleus.
Representing the nucleus
symbolically
Total number of nucleons
(protons + neutrons)
(A = Z + N)
 Atoms of a particular element with different
numbers of neutrons are called isotopes of that
element.
 The electronic structure of an element's
isotopes is the same, which means their
chemical behaviors are almost identical.
 However, the nuclei behave quite differently.
 Isotopes of carbon are:
12
13
6
6
C
C
14
6
C
WHITEBOARD
 Determine the number of protons and neutrons
in each of the following nuclei:
U
238
92
Element symbol
Number of protons
(Z)
6
3
Li
16
8
O
27
13
Al
Z = 92
Z=3
Z=8
Z = 13
N = 146
N=3
N=8
N = 14
N = 238 – 92 = 146
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WHITEBOARD
Nuclear and Quantum Physics
 Determine the number of protons and neutrons
in each of the following nuclei:
56
26
Fe
64
30
Zn
107
47
Question the very structure and behavior of matter itself.
Although sought after for thousands of years, this knowledge has
just started to present itself to the human race through modern
physics experiments of the past 120 years.
Ag
Z = 26
Z = 30
Z = 47
N = 30
N = 34
N = 60
Structure of the Nucleus
Coulomb repulsion is not the whole story!
There are four fundamental forces that we know
of that govern the interactions of the universe.
Gravitational Force
A concentrated sphere of protons and neutrons
that makes up 99.9% of an atom’s mass.
Although consistent with experiments throughout the early 1900s,
this model was met with much skepticism.
Electromagnetic Force
Weak Nuclear Force
Strong Nuclear Force
With all of the electric repulsion between protons,
how could this arrangement possibly be stable?
The strong nuclear force
Attractive force between protons and protons, protons and
neutrons, neutrons and neutrons.
Nuclear force and binding energy
 How can protons stay bound together when they
repel each other so strongly?
Very strong (100 times stronger than Coulomb force), but only
at very short range (< 1 × 10-15 m).
Basically disappears at distances greater than 3 × 10-15 m.
Cannot be expressed algebraically –
only with multivariable differential equations.
Study more physics to learn how it works in detail!
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How are nuclei born?
Nuclear force
 Some attractive force must balance this
electrical repulsive force and must attract
neutrons as well; it has to be an attractive force
for both protons and neutrons.
 We call this attractive force a nuclear force.
 The nuclear force must weaken to nearly zero
extremely rapidly with increasing distance
between nucleons.
 If it didn't, then nuclei of nearby atoms would
be attracted to each other, clumping together
into ever-larger nuclei.
At a distance, the strong nuclear force is negligible,
so Coulomb repulsion dominates proton-proton interaction.
d >> 1 × 10-15 m
However, if the protons are forced to get close enough to one
another by some immense external pressure (like it in a star)…
d < 1 × 10-15 m
Felectric << Fnuclear
now the strong nuclear force dominates, and a nucleus is born.
(Usually some neutrons will also be there to increase stability)
Nuclear binding energy
 The binding energy of the nucleus is the energy
that must be added to the nucleus to separate it
into its component protons and neutrons.
Felectric >> Fnuclear
Wacky Units!
In nuclear and atomic physics, we use some seemingly
strange units for mass and energy.
Don’t be afraid! Just know what they are and how/when to convert them!
Mass
 The nucleus is a bound system, so its nuclear
potential energy plus electric potential energy
plus kinetic energy of the protons and neutrons
must be negative.
1 atomic mass unit (amu) = 1 u = 1.660539 × 10-27 kg
Proton mass = 1.007 u
Neutron mass = 1.009 u
Energy
1 electron-volt (eV) is the amount of energy gained by an
electron that moves through a potential difference of 1 Volt.
Ue = qΔV
1 eV = 1.6 × 10-19 J
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One atomic mass unit (u)
Total number of nucleons
(protons + neutrons)
(A = Z + N)
12
6
 What is the rest energy (in J, eV, and MeV) of 1
atomic mass unit (u) ?
C
E = mc2
One atomic mass unit equals one-twelfth of the
mass of a carbon atom with six protons and six
neutrons, including the six electrons in the atom.
In terms of kilograms.
EXPERIMENT
211H
+
1 atomic mass unit (amu) = 1 u = 1.660539x10-27 kg
Element symbol
Number of protons
(Z)
+
WHITEBOARD
E = 934.1 MeV
E = 931.5 MeV
MASS DEFECT (m)
 The mass defect of a certain type of nucleus is
the difference in mass between the constituents
(as Hydrogen atoms and neutrons) and the
mass of the atom itself
=
201n
E = 1.4945x10-10 J
E = 9.3409x108 eV
=
4
2
He
m = [Zmhydrogen atom + (A - Z)mneutron] – matom
mHydrogen = 2(1.007825 u) = 2.01565 u
mNeutron = 2(1.008665 u) = 2.01733 u
mHelium = 4.03298 u
mHelium = 4.002602 u
Binding energy (BE)
 The binding energy of a nucleus is the rest
energy equivalent of its mass defect:
Mass-Energy Equivalence
Matter is energy.
BE  mc 2
BE = mc2
 If mass is “lost” in a nuclear reaction, then its energy must be
have been converted to another form. This is almost always in the
form of kinetic energy and electromagnetic waves (light energy).
 The binding energy represents the total energy
needed to separate the nucleus into its
component nucleons.
 During a nuclear explosion, the tremendous and rapid conversion
of mass energy to kinetic energy can cause surrounding particles
to recoil with high velocities.
 Also, the light produced by this conversion of matter to energy is
often of high-frequency (UV, X-ray and Gamma), making it very
dangerous to surrounding organisms.
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Binding Energy
When nuclei form, they are bound together by the strong force.
In order to break the nucleus apart, you
need to put energy into the system.
This means that the system has negative potential energy.
Similar to the negative energy between two particles that attract
electrically, but now the attraction is caused by the strong nuclear force.
Example: The nucleus shown above (a Helium nucleus)
has a binding energy of -931.5 MeV.
This means that you would need to put 931.5 MeV of energy into the
nucleus to separate it into two free protons and two free neutrons.
Binding energy per Nucleon
 The binding energy per nucleon is the binding
energy BE of the atom divided by the number of
nucleons A in the atom’s nucleus:
Binding energy per nucleon = BE/A
How to use
BE  mc 2
Speed of light in a vacuum (3 × 108 m/s)
Law of Conservation of Energy:
Total energy before = Total energy after
Just apply this using E = mc2 to determine how much
mass-energy is converted to other forms, or vice versa.
Always use mass in kilograms when using c!
Representing nuclear reactions
 The advantage of writing nuclear reactions as
shown here is that atomic masses (found in
atomic mass tables) can be used to analyze the
energy transformations that occur during the
reactions:
 The binding energy per nucleon is a good
indicator of the stability of a nucleus; the larger
the binding energy per nucleon the more stable
is the nucleus.
Rules for nuclear reactions (1/3)
Rules for nuclear reactions (2/3)
RULE 1: ELECTRIC CHARGE CONSERVATION.
RULE 2: NUCLEON NUMBER CONSERVATION.
 The total electric charge of the reacting nuclei
and particles equals the total charge of the
nuclei and particles produced by the reaction.
This condition is satisfied if the total atomic
number Z of the reactants equals the total
atomic number of the products including Z=-1 for
free electrons
 The total number of nucleons A (protons +
neutrons) of the reactants always equals the
total number for the products.
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Solution
WHITEBOARD
In the alpha decay of a Radium nucleus, the following is known.
226
88
During an alpha decay, the total number of nucleons
must add up before and after, and the total number of
protons must add up before and after.
Ra ® 24 He +?
226
88
Ra ® ? + 24 He
226 = X + 4
88 = A + 2
What is the missing product?
WHITEBOARD
WHITEBOARD
 Determine the missing products in the following
reactions:
1
1
1
1
H
1
0
n
94
38 Sr
137
56
Ba
WHITEBOARD SOLUTION
1
1
H  37Li  24He 24He  kinetic energy
a.
b.
c.
d.
Find
Find
Find
Find
the
the
the
the
mass of the reactants.
mass of the products
mass defect (m)
rest energy of the mass defect (in MeV)
mReactants = 8.023828 u
mproducts = 8.005204 u
m = 0.018624 u
Q = 17.35 MeV
H  37Li  24He 24He  kinetic energy
 
m Li   7.016003 u
m He   4.002602 u
m 11 H  1.007825 u
7
3
4
2
Rules for nuclear reactions (3/3)
RULE 3: ENERGY CONSERVATION
 The rest energy of the reactants equals the rest
energy of the products plus any change in
kinetic energy of the system.
 (mc2)reactants =  (mc2)products + Q
 Q, the reaction energy can be either positive (if
the products’ rest energy is less that that of the
reactants) or negative (if the products’ rest
energy is greater)
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Binding energy and energy release
 The higher the binding energy per nucleon, the
more energy needed to split the nucleus into its
constituent protons and neutrons.
Fusion and chemical elements
 Two or more small nuclei fuse together, forming
a larger nucleus and releasing huge amounts of
energy. Takes place in stars like the Sun.
 Fusion occurs naturally in stars; it requires high
heat and high pressure to overcome the
repulsion from the electric charges.
Binding energy and energy release
 The graph predicts:
 When two small nuclei combine, energy should be
released.
 When a large nucleus breaks apart, energy should be
released.
Fusion and chemical elements
 Supernova explosions contribute to the chemical
composition of the universe.
 The elements lighter than iron that are
produced in stars' cores before the explosion
are ejected into space.
 The elements heavier than iron that are
produced during the explosion are then
ejected into space.
WHITEBOARD
Fission and nuclear energy
 The energy released by the Sun comes from
several sources, including the proton-proton
chain of fusion reactions:
 Determine the energy released in this chain of
reactions in MeV. Use the masses
,
, and
to determine the rest energy
converted to other forms.
Q = 28.29711 MeV
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Fission and nuclear energy
Fission and nuclear energy
Fission and nuclear energy
Fission and nuclear energy
Nuclear Reactions
Fission - Chain Reaction!
Nuclear fission – a large nucleus splits into two or more smaller
nuclei. Takes place in nuclear power plants and bombs.
Nuclear fusion – two or more small nuclei fuse together,
forming a larger nucleus and releasing huge amounts of energy.
Takes place in stars like the Sun.
A neutron is absorbed by a heavy
nucleus, making it unstable.
The unstable nucleus splits into two
smaller nuclei, releasing some energy
and more neutrons in the process.
These neutrons go on to get absorbed
by other nuclei, sparking a chain
reaction that grows exponentially.
This is a nuclear explosion.
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In a nuclear power plant,
these are absorbed by water
to prevent meltdown.
In a nuclear power plant, most of the released neutrons are
absorbed by water molecules to control the rate of fission and
prevent the reaction from “going critical”
(becoming a nuclear bomb)
WHITEBOARD
 Determine the energy in the fission reaction.
 One neutron collides with and joins momentarily with the
U-235atomic nucleus to form an excited U-236, which
quickly splits into Ba-141, Kr-92 an 3 neutrons.
1
0
n1.008665u 
U 235.043924u 
235
92
92
36
141
56
Kr 91.926156u 
Ba140.914411u 
Bohr's liquid drop model of the nucleus
 Frisch and Meitner decided that Bohr's liquid
drop model of a nucleus could explain the
observation of fission.
 Surface tension holds a water drop together;
likewise, the nuclear forces hold the nucleons
together.
 The protons repel each other and overwhelm
the effect of the "surface tension."
 The nucleus can then stretch itself and divide
into two smaller pieces.
Q = 173 MeV
The Three Types of Nuclear Decays
Alpha decay, Beta decay, and Gamma
decay.
Alpha decay
 When one of the alpha particles leaves, this
emission reduces the number of protons in the
original nucleus as well as the electric repulsion
between the remaining protons.
All three of these decays obey the
three conservation laws of the universe.
Conservation of Energy
Conservation of Momentum
Conservation of Charge
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α-decay
WHITEBOARD
A parent nucleus releases an alpha particle (Helium nucleus),
and recoils in the process.
4
2
He
Alpha particle – two protons
and two neutrons bound
together (a Helium nucleus)
To analyze an alpha decay, you must use
momentum conservation and energy conservation.
Beta decay
 During beta decay, a particular element is
transformed into an element with a Z number
that is larger by 1.
 Determine the kinetic energy of the product
nuclei when polonium-212 undergoes alpha
decay.
 The masses of the nuclei involved in the decay
are
m = 0.0097 u
Q = 9.03555 MeV
β-decay
Within the nucleus, a neutron decays into a proton,
while emitting an electron (beta particle).
The most important part about beta decay is that it
obeys the Law of Conservation of Charge.
The Law of Conservation of Charge
In all chemical and nuclear reactions,
electric charge must be conserved.
Beta minus
decay:
Beta plus decay
1
0
n11p  10e 
Neutral  Positive + Negative
1
1
Problems with beta decay
 The problem with spin conservation:
 In beta decay, the spin quantum number was
not conserved.
 The problem with energy conservation:
 The total energy of the products was always
observed to be less than the total energy of
the reactants.
p 01n  10e
Positive  Neutral + positive
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Beta decay
Beta-minus and beta-plus decay
 Wolfgang Pauli proposed an explanation for beta
decay that did not require abandoning energy
conservation or spin number conservation.
 He hypothesized that some unknown particle
carried away the missing energy and accounted
for the discrepancy in spin number.
 This particle had zero electric charge, zero mass,
and a spin number of either +1/2 or
–1/2.
 Enrico Fermi called the particle a neutrino,
meaning "little neutral one."
Gamma decay
γ-decay
• After alpha or beta
decay, the nucleus can
be left in an excited state
from which it then emits
one or more photons to
return to its ground state.
An excited nucleus emits
a photon (gamma ray),
and decreases its state of
vibration in the process.
WHITEBOARD
 The following nuclei undergo different types of
radioactive decay. Determine the daughter
nucleus for each and write an equation
representing each decay reaction.

WHITEBOARD
 The following nuclei undergo different types of
radioactive decay. Determine the daughter
nucleus for each and write an equation
representing each decay reaction.

beta-minus decay
alpha decay
239
94
4
Pu  235
92 U  2 He  energy
144
58
0
Ce144
59 Pr  -1 e  v  energy
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Whiteboard!
WHITEBOARD
 The following nuclei undergo different types of
radioactive decay. Determine the daughter
nucleus for each and write an equation
representing each decay reaction.

65
30
beta-plus decay (produces a positron)
Zn Cu  e  v  energy
65
29
0
1
The following nuclear decay occurs spontaneously.
U
238
92
Th
234
90
4
2
3.6 x 105 m/s
He
?
m = 4.00 u
m = 234.04 u
m = 238.05 u
a) What type of decay is this?
b) What is the speed of the recoiling Helium nucleus?
c) What is the combined kinetic energy (MeV) of the two
nuclei after the decay? (Without using your answer from b)
1 u = 1.660539 × 10-27 kg
U
238
92
Th
234
90
3.6 x
105
m/s
4
2
He
U
238
92
Th
234
90
4
2
He
pi = 0
m = 4.00 u
m = 234.04 u
m = 238.05 u
Conservation of Momentum
Spi = Sp f
vα = 2.11 x
Conservation of Energy
BE  mc 2
A typical momentum problem!
0 = mThvTh + ma va
107
m = 4.00 u
m = 234.04 u
m = 238.05 u
m/s
The alpha particle has much less
mass, so it will recoil with much
greater velocity!
(Note: Don’t need to convert to kg in
this part, as you are working with the
same units on both sides of the eqn.)
Half-life
 Using a particle detector such as a Geiger
counter, we can measure the number of nuclei
that decay in a short time interval and determine
the number N of radioactive nuclei that remain
in the sample as a function of time.
First, convert the masses into kilograms.
Then subtract to find out how much mass is “lost” (converted to KE).
Finally, use BE = mc2 to find that the energy released in the decay is
m = 0.01 u
Q = 9.315 MeV
Observational experiment
Number N of
radioactive nuclei
Clock reading
t [min]
N0
0
0.5 N0
1
0.25 N0
2
0.125 N0
3
0.0625 N0
4
0.03125 N0
5
0.015625 N0
6
Undetectable
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Observational experiment
Number N of
radioactive
nuclei
Clock reading
t [min]
102400
0
51200
1
25600
2
12800
3
6400
4
3200
5
1600
6
800
10
N/N0
Half-life
(N/N0)
In terms of 1/2
 The half-life T of a particular type of radioactive
nucleus is the time interval during which the
number of nuclei in a given sample is reduced
by one-half.
 After n-lives, the fraction of radioactive nuclei
that remain is
N
1
 n
N0 2
Ratio of
decays
t  nT
Time
(expressed in half-lives)
Half-life
Half-life
Half-life
y  102400  e 0.693 x
N vs. Time
N
100000
90000
80000
N
1
 n
N0 2
t  nT
N  N 0  e  t
70000
60000
y = 102400e-0.693x
50000
40000
30000
N  102400  e 0.693t
20000
10000
0
0
2
3
4
5
6
7
8
9
10
11
12
t (min)
N
Number of radioactive nuclei
N0
Number of radioactive nuclei at t = 0 s
(N0 = 102400)
Decay constant ( = - 0.693 s-1)
Euler number

e
Half-lives and decay constants of some
common nuclei
1
Euler number
n 
1
1
1
1
1
 n!  0!  1!  2!  3!  ...
n 0
T

n 
1
 n!  2.71828182845905
n 0
n 
1
 n!  e
n 0
e  2.71828182845905
e  2.71
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EXPONENTIAL RELATIONSHIP
N  N0  e
EXPONENTIAL RELATIONSHIP
 t
N vs. Time


Ln N   Ln N   Lne 
Ln N   Ln N 0  e  t
 t
0
90000
80000
70000
60000
y = 102400e-0.693x
50000
40000
30000
20000
10000
0
0
Ln N   Ln N 0   t  Lne 
How do we
solve for t?
Ln N   Ln N 0   t
1
2
3
4
5
6
7
8
9
10
11
12
11
10
9
8
7
6
5
4
3
2
1
0
12
y = -0.6931x + 11.537
0
1
2
3
4
5
6
7
8
9
10
11
12
y  0.6931x  11.537
y  102400  e 0.693 x
N  N0  e
Ln N   t  Ln N 0 
Ln(N) vs. Time
100000
Ln N   Ln N 0    t 
 t
LnN   Ln102400   0.6931t 
LnN   0.6931 t  11.537
N  102400  e 0.693t
EXPONENTIAL RELATIONSHIP
N vs. Time
105000
100000
Ln(N) vs. Time
12
11
10
9
8
7
6
5
4
3
2
1
0
95000
90000
y = -0.6931x + 11.537
85000
Number of
radioactive
nuclei
First half-life
80000
75000
70000
65000
Decay
constant
60000
55000
0
1
2
3
4
5
6
7
8
9
10
11
Second half-life
50000
12
45000
Ln N   Ln N 0    t 
t
 Ln N   Ln N 0 

1
2n
N  N 0  e  t
N0 
Ln N 0   Ln N 
30000

1
 N 0  e  T
2
1
 e  T
2
N  N 0  e  t
25000
20000
 N 

 Ln
 N0 
t
15000
10000
5000
0
0
1
2
3
4
5
6
7
8
9
10
11
12


Decay rate and half-life


 At time t = T (one half-life), the number N of
radioactive nuclei remaining is one-half the
number N0 at time zero:
1
Ln   Ln e T
2
t  nT
n 1
t T
y = 102400e-0.693x
35000

N
1
 n
N0 2
N  N0 
t
40000
1
Ln   Ln e T
2
1
Ln     T
2
1
Ln 
2
T   

T
0.693

 If the decay constant is large, then the material
decays rapidly and consequently has a short halflife T.
17
5/1/2016
WHITEBOARD
WHITEBOARD (SOLUTION)
 Radium-226, a waste product of nuclear reactors,
has a half-life of 1600 years.
 Find the decay constant of the () Radium-226
 Use two different methods to determine the
fraction of radium-226 remaining in a reactor
fuel rod 8000 years after it is removed from the
reactor.
t  nT
N
 e  t
N0
N
1
 n
N0 2
t  nT
n
0.693

0.693
T
t
T
8000
1600
n
T
  4.33125 x10 4 yr 1
N
 e  t
N0
n5
WHITEBOARD (SOLUTION)
  4.33125 x10 4 yr 1
N
1
 n
N0 2
N
 e  t
N0
N
1
 5
N0 2
N
 0.03125
N0
4
1
N
 e  4.33125 x10 yr 8000 yr
N0
N
 0.03127
N0
DECAY RATE AND HALF LIFE
T
0.693

0.693
T

N  N 0  e  t
N
 e  t
N0
 N 
  Ln e t
Ln
 N0 
 
 N 
  t
Ln
 N0 
N
1
 n
N0 2

WHITEBOARD
 Strontium-90, has a half-life of 28.9 years.
 Find the decay constant of the () Strontium-90
 Use two different methods to determine the
fraction of Strontium-90 remaining after 101.15
years.
n

t
n  3.5
T
0.693
T
  0.023979 yr 1
N
1
 n
N0 2
N
 0.08839
N0
N
N
 0.08843
 e  t
N0
N0
Radioactive dating
 Archeologists and geologists are interested in
determining the age of a radioactive sample
from the known fraction N/N0 of radioactive
nuclei that remain in the sample:
 N 
0.693
  
Ln
t
T
 N0 
 N  T
 
t   Ln
 N 0  0.693
 In this equation, T is the half-life of the
radioactive material and t is the sample's age
when N radioactive nuclei remain.
18
5/1/2016
Carbon dating
 Any plant or animal that metabolizes carbon
incorporates about one carbon-14 atom into its
structure for every 1012 carbon-12 atoms it
metabolizes.
 Carbon is no longer metabolized by the
organism after death, so the carbon-14 starts
to transform into nitrogen-14.
 After
5730
years,
the
carbon-14
concentration decreases by one-half.
 A measurement of the current carbon-14
concentration indicates the age of the
remains.
WHITEBOARD
WHITEBOARD
 The ratio of carbon-14 in an old bone compared
to the number in a fresh bone of the same mass
is 0.050.
 Determine the age of the bone.
 N  T
 
t   Ln
 N 0  0.693
t   Ln0.05 
5730 years
0.693
Tcarbon = 5730 years
t  24769.9 years
Radioactive decay series
 A bone found by an archeologist contains a
small amount of radioactive carbon-14. The
radioactive emissions from the bone produce a
measured decay rate of 3.3 decays/s. The same
mass of fresh cow bone produces 30.8
decays/s. Estimate the age of the sample.
 N  T
 
t   Ln
 N 0  0.693
 3.3  5730 years
t   Ln

0.693
 30.8 
Tcarbon = 5730 years
t  18468 years
19