Download Chapter 31 - Radioactivity

Chapter 31- Nuclear Physics
31.1 Nuclear Structure
The atomic nucleus consists
of positively charged protons
and neutral neutrons.
31.1 Nuclear Structure
atomic mass
number
atomic
number
A
Z
N


 




Number of protons
and neutrons
Number of
protons
Number of
neutrons
31.1 Nuclear Structure
Nuclei that contain the same number of protons but a different
number of neutrons are known as isotopes.
The Strong Force
The strong force has four
important properties:
1. It is an attractive force
between any two nucleons.
2. It does not act on electrons.
3. It is a short-range force,
acting only over nuclear
distances.
4. Over the range where it
acts, it is stronger than the
electrostatic force that tries
to push two protons apart.
As the number of protons increase, the number
of neutrons must increase even more for stability
All elements with >83 protons are unstable
Radioactivity – the spontaneous disintegration or
rearrangement of internal nuclear structure.
• A nucleus is a bound system. You need to
supply energy to separate a stable nucleus into
separated nucleons.
•This energy is called the binding energy.
• Experimental evidence shows that the mass
of any nucleus is less than the mass of the
same number and type of nucleons when they
are separated, i.e. at rest and out of the range
of the forces from other nucleons. Why is this?
•According to Einstein, mass and energy are
related. The energy of an object at rest is:
E0 = mc2
• The binding energy is transformed into the
greater mass of the separated nucleons.
• Binding Energy (B) = mseparated c2 - mnuc c2
31.3 The Mass Deficit of the Nucleus and Nuclear Binding Energy
B  Mass defect c 2  mc 2
Where Δm, the mass defect, is the difference
between the nuclear mass and the mass of the
separated nucleons.
31.3 The Mass Deficit of the Nucleus and Nuclear Binding Energy
B  Mass defect c 2  mc 2
• Values for binding energy are very small, when
using SI units of Joules.
• Often the unit used is called the electron volt,
where:
1 eV = 1.60 x 10-19 J
1 MeV = 1 x 106 eV
31.3 The Mass Deficit of the Nucleus and Nuclear Binding Energy
1u of mass is equivalent to 931.5 MeV/c2
c2 = 931.5 MeV/u
B = mseparated c2 - matom c2
31.3 The Mass Deficit of the Nucleus and Nuclear Binding Energy
B = mseparated c2 - matom c2
•Use atomic mass units (amu) for the mass of the
atom (matom ) when calculating the resting energy
of the atom. Values in periodic table.
•Use amu value for neutron (1.0087 u) for all
neutrons (Nmn ).
•Use atomic mass of 1hydrogen (1.0078 u) for all
protons (ZmH ). This accounts for the mass of the
electrons. Note that this is slightly different from
the 1.00794 value given in the periodic table!
31.3 The Mass Deficit of the Nucleus and Nuclear Binding Energy
Example 3 The Binding Energy of the Helium Nucleus Revisited
The atomic mass of helium is 4.0026u and the atomic mass of 1hydrogen
isotope is 1.0078u. Using atomic mass units, instead of kilograms,
obtain the binding energy of the helium nucleus.
31.3 The Mass Deficit of the Nucleus and Nuclear Binding Energy
m  4.0330 u  4.0026 u  0.0304 u
c 2  931.5 MeV/u
B  mc2  (0.03304u)(931.5MeV / u)  28.3 MeV
QUESTION
The atomic mass of this particular isotope of iron is
55.9349u.
Recall that the neutron has a mass of 1.0087u and
the 1hydrogen atom a mass of 1.0078u
Binding Energy per Nucleon Curve
31.4 Radioactivity
When radioactive material disintegrates spontaneously, certain kinds of
particles and/or high energy photons are released.
These are called, respectively alpha (α) rays, beta (β) rays, and gamma(γ)
rays.
A magnetic field can separate these three types of particles emitted by
radioactive nuclei.
Alpha (α) DECAY
The general form for α decay is:
A
Z
P 
A 4
Z 2
D 
2
2
He
• P is the parent nucleus,
• D is the daughter nucleus and
• He is an alpha particle, which is a helium nucleus
(2 protons and 2 neutrons)
• Since the daughter product has 2 less protons than
the parent nucleus, it becomes another element
• The process whereby one element becomes
another is called transmutation.
α DECAY
A
Z
P 
A 4
Z 2
D 
2
2
He
• When alpha decay occurs, the daughter product
and the alpha particle have less mass than the
parent nucleus.
• This is in marked contrast to the separation of a
stable nucleus, where the separated nucleons has
a greater mass.
• The fact that the daughter nucleus plus alpha
particle have less mass implies that alpha decay
releases energy.
• Alpha decay occurs in some high-Z nuclei located
beyond bismuth (Bi) on the binding energy curve.
α DECAY and the release of energy
239
Plutonium 94 Pu has a mass of 239.052 u, and
235
U with a mass of
undergoes α decay to become 92
235.043 u. The mass of a helium atom (used for the
mass of the alpha particle) is 4.0026. Determine the
energy released (in MeV) when Plutonium 239
undergoes α decay.
239
94
Pu 
235
92
U 
4
2
He  energy
239.052u  (235.043u + 4.0026u + E)
E = .0064 u (931.5 MeV/u) = 5.96 MeV
This energy manifests itself as kinetic
energy of the alpha particle and daughter
particle.
Beta Decay and Quarks
• Nucleons are composed of elementary particles
called quarks.
• There are 6 known types of quarks, but we will
only consider 2 of these types:
– Up quarks have a charge of +2/3 e
– Down quarks have a charge of - 1/3 e
Beta Decay and Quarks
• In the most common type of beta decay, ( known
as β- decay), one of the down quarks in the
neutron decays into an up quark, by emitting a
beta particle with a total charge of -1. This
particle is indistinguishable from an orbital
electron.
Beta Decay and Quarks
• In the most common type of beta decay, ( known as βdecay), one of the down quarks in the neutron decays
into an up quark, by emitting a beta particle with a total
charge of -1. This particle is indistinguishable from an
orbital electron.
• In the process, an extremely small particle (0.0004% of
the mass of the electron) called an antineutrino( ) is
emitted.
β- DECAY
The general form for β- decay is:
A
Z
P 
A
Z 1
D 
0
1
e 
β- decay, like α decay, results in transmutation of the parent
nucleus into a different element and releases energy.
This type of radioactivity occurs in unstable nuclei with more
neutrons than protons.
14C decays into 14N due to this mechanism.
β- DECAY
The radioactive isotope of carbon, Carbon 14,
has an atomic mass of 14.003241 u. It is
converted by β- decay into Nitrogen 14
(atomic mass= 14.003074 u).What is the
energy (in MeV) released in this process?
β- DECAY
The radioactive isotope of carbon, Carbon 14,
has an atomic mass of 14.003241 u. It is
converted by β- decay into Nitrogen 14
(atomic mass= 14.003074 u).
C 14 N 14 + E
E = (14.003241-14.003074) u x 931.5 MeV/u
E = .1556 MeV
Note that the mass of the β- particle and the
antineutrino is included in the atomic mass
of the nitrogen.
β+ DECAY
β plus decay also occurs, but is a much less common
phenomenon. In this case a proton decays into a
neutron, a neutrino, and positron, which has the
same mass as an electron, but a positive charge.
A
Z
P 
A
Z 1
D 
0
1
e 
β+ DECAY
β plus decay requires energy, it does not release it.
Therefore it does not happen spontaneously
A
Z
P 
A
Z 1
D 
0
1
e 
31.4 Radioactivity
γ DECAY
γ decay occurs when a nucleus in a higher energy state
spontaneously jumps to a lower energy state. The
result is emission of a very high frequency photon.
A
Z
P

excited energy
state

A
Z
P  
lower energy
state
Note that γ decay does not result in transmutation.
31.4 Radioactivity
Gamma knife surgery – I hope I never see you here!
Stop to think
The cobalt isotope 60Co (Z = 27) decays
to the nickel isotope 60Ni (Z = 28). The
number in the rh corner is A, atomic
mass. The decay process is:
A.Alpha decay.
B.Beta-plus decay.
C.Beta-minus decay.
D.Gamma decay.
31.6 Radioactive Decay and Activity
Imagine tossing a coin. Will it
be heads or tails?
If you tossed 1000 coins,
you’d very likely find 500
heads and 500 tails.
If you tossed the 500 heads,
you’d get about 250 heads,
250 tails. Tossing the 250
heads you’d get about 125
heads and so on.
A graph of number of heads
(N) remaining vs. number of
tosses (T) would result in an
exponential graph, like the one
on the right.
31.6 Radioactive Decay and Activity
Tossing a single coin is a
random process, but tossing
many coins results in a
definite pattern.
Each time you toss, about ½
the coins will be heads, and ½
will be tails.
Radioactive decay shows this
exact same pattern.
How often does the radioactive decay occur within a
sample of radioactive material?
•We can define λ as the decay constant, the probability that a nucleus
will decay in the next second. λ has units of inverse seconds (s-1)
•For example if λ=.01 s-1, it means that a nucleus has a 1% chance of
decay in the next second.
•The probability that a nucleus will decay in a small time period, Δt, is:
λ Δt
•If there are N independent nuclei in a sample, the number of nuclei
expected to decay in the time period, Δt:
Δ N = -N λ Δt
•And therefore
Δ N/ Δt = - λN
•The negative sign indicates that the ΔN is a loss of radioactive
nuclei.
How often does alpha, beta, or gamma decay occur within
a sample of radioactive material?
•The activity (A) of a radioactive
sample is defined as the number
of disintegrations per second:
A = |Δ N/ Δt|
We have shown that the activity is
equal to λ N.
•The SI unit of activity is the
becquerel (one Bq = 1
disintegration per second)
•A non-SI unit of activity often
used in the medical industry is the
Curie, where 1 Ci = 3.7 x 1010
Bq.
•Activity can be determined by
direct measurement.
This is a graph of N, the number of radioactive nuclei in
a sample, vs. time.
It can be shown that the exponential graph, shown on
the left, has the following equation:
N  N oe t
where N is the number of radioactive or parent nuclei at
a given time t, N0 is the number of parent nuclei at time
t = 0, and λ is the decay constant.
Since A = λN we can also say:
A  Ao e   t
Half- Life, T1/2
It is often convenient to know how long
it will take for one-half the sample to
decay.
 t
N  N oe
N0
 N o e  T1 / 2
2
1
 e  T1 / 2
2
Taking the natural log of both sides:
ln 2  T1 2
T1 2 
ln 2

where T1/2 the half-life, is defined as
the time in which ½
of the radioactive nuclei
disintegrate.
31.6 Radioactive Decay and Activity
A sample starts with 1000 radioactive
atoms. How many half-lives have
elapsed when 750 atoms have
decayed?
A.2.5
B.2.0
C.1.5
D.0.25
Cesium activity
• The isotope 137Cs is a standard source
of gamma rays. The half-life is 30.0
years.
a. How many 137Cs atoms are in a source
that has an current activity of 1.85 x
105 Bq?
b. What is the activity of the source 10
years later?
Cesium activity, part A
• The isotope 137Cs is a standard source of
gamma rays. The half-life is 30.0 years.
a. How many 137Cs atoms are in a source that
has a current activity of 1.85 x 105 Bq?
Known
Find
T1/2 = 30.0 years
N, number of Cs atoms
A = λ N = 1.85 x 105 Bq
To find λ, the decay
:
constant, we can use the
relationship between it and
half-life
T1 2 
ln 2

Cesium activity, part A
a. How many 137Cs atoms are in a source
that has a current activity of 1.85 x 105 Bq?
A = λ N = 1.85 x 105 Bq
T1 2 
ln 2

λ = 7.33 x 10-10 s-1
N = A/ λ = 2.53 x 1014 atoms
Cesium activity
b. What is the activity of the source 10 years later?
A  Ao e   t
A0 = 1.85 x 105 Bq
λ = 7.33 x 10-10 s-1
t = 10 yrs converted to seconds
t= 3.15 x 108 s
A= 1.47 x 105 Bq
Activity decreases over time.
Cesium activity
1. Express T1/2 in terms of seconds:
T1/2 = 30.0 y(3.15 x 107 s/y) = 9.45 x 108 s
2. Find decay constant, λ:
λ = (ln 2)/T1/2
λ = .693/9.45 x 108 s
λ = 7.33 x 10-10 s-1
3. Thus the number of 137 Cs atoms is:
A/ λ = N
1.85 x 105 Bq/ 7.33 x 10-10 s-1 = 2.5 x 1014 atoms.
Formation of C-14
• Carbon 14 is produced in the upper atmosphere
when energized neutrons form from interactions
between high energy cosmic radiation and
atmospheric gases.
• The energetic neutron is absorbed by a nitrogen 14
nucleus and a proton is ejected.
• Transmutation occurs and an unstable isotope of
carbon (C-14) is created.
31.7 RadiocarbonDating – visiting with Oetzi
In 1949, Willard Libby developed a method of
using the radioactive isotope 14 C to determine
the age of organic materials up to about 50,000
years old. Libby won a Nobel Prize for his
work.
.
31.7 RadiocarbonDating – visiting with Oetzi
•The concentration of 14 C in the is about 1 part per trillion (1
atom of 14 C for 8.3 x 1011 atoms of 12 C. This seems small,
but is measurable by modern chemical techniques.
•Living organisms have an activity of 0.23 Bq per gram
carbon. After death, this activity decreases.
•materials.
31.7 RadiocarbonDating – visiting with Oetzi
•T1/2 = 5730 years
for 14 C. It
undergoes beta
decay to 14 N.
•Other isotopes
with longer halflives are used to
date geological
materials.
•Uranium-Lead
isotope dating was
used to obtain a
value for the age of
the Earth (4.5 billion
years).
Carbon dating
• Archeologists excavating a site have found a
piece of charcoal from a fireplace. Lab
measurements find the 14 C activity of the
charcoal to be .07 Bq per gram. What is the
radiocarbon age of the charcoal?
Known
Find
A= .07Bq/g
t
A0 = .23 Bq/g
T1/2 = 5730 years
λ = (ln 2)/T1/2
 t
o
A Ae
Carbon dating
• First, find λ, the decay constant
Known
Find
A= .07Bq/g
t
A0 = .23 Bq/g
T1/2 = 5730 years
Tip: If we leave λ in terms of the T1/2 , we can
find t in years : λ = (ln 2)/T1/2 = .693/5730
λ = 1.209 x 10-4 y-1
Carbon dating
Known
Find
A= .07Bq/g
t
A0 = .23 Bq/g
T1/2 = 5730 years λ = 1.209 x 10-4 y-1
A  Ao e
 t
Ln (A/A0) = - λt
Ln (A/A0)/- λ = t
T = 9839 or 9800 years
Conceptual Example 12 Dating a
Bottle of Wine
A bottle of red wine is thought to have
been sealed about 5 years ago. The
wine contains a number of different
atoms, including carbon,oxygen, and
hydrogen. The radioactive isotope of
carbon is the familiar C-14 with ½ life
5730 yr. The radioactive isotope of
oxygen is O-15 with a ½ life of 122.2 s.
The radioactive isotope of hydrogen is
called tritium and has a ½ life of 12.33 yr.
The activity of each of these isotopes is
known at the time the bottle was sealed.
However,
only one of the isotopes is useful for
determining the age of the wine. Which
is it?
A. C-14
B. O-15
C. H-3
EOC 47
The practical limit to ages that can be
determined by radiocarbon dating is about
41,000 years. In a sample of this age,
what percentage of C-14 remains?
EOC 47
The practical limit to ages that can be
determined by radiocarbon dating is about
41,000 years. In as sample of this age,
whatj percentage of C-14 remains?
0.70%
EOC #50
When a sample from a meteorite is
analyzed, it is determined that 93.8% of
the original mass of a certain radioactive
isotope is still present. The age of the
meteorite is calculated to be 4.51 x 109
yrs. What is T1/2 in years of this isotope?
EOC #50
When a sample from a meteorite is
analyzed, it is determined that 93.8% of
the original mass of a certain radioactive
isotope is still present. What is T1/2 in
years of this isotope?
Answer:4.88 x 1010 years
31.8 Radioactive Decay Series
The sequential decay of one nucleus after another is
called a radioactive decay series.
31.8 Radioactive Decay Series
31.9 Radiation Detectors
A Geiger counter
31.9 Radiation Detectors
A scintillation
counter