Download 31.1 Nuclear Structure

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.
• Recall that E0 = mc2
• If the separated nucleons are also at rest, that
energy is in the form of greater mass.
• 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 = 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 ). 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
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
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.
α DECAY
The general form for α decay is:
A
Z
P 
A 4
Z 2
D 
2
2
He
where P is the parent nucleus, D is the daughter nucleus and
The process whereby one element becomes another is called
transmutation.
This reaction occurs when the mass of the parent nucleus is
greater than the mass of the daughter nucleus plus the
mass of an alpha particle. This is the case for some high-Z
nuclei located beyond Bi on the binding energy curve.
It is energetically favorable for these nuclei to eject an alpha
particle, because the daughter is bound more tightly than
the parent.
α DECAY of Uranium
A
Z
P 
A 4
Z 2
D 
2
2
He
α DECAY and the release of energy
The atomic mass of 238U is 238.0508 u, the atomic mass of 234Th is
234.0436 u and the atomic mass of a helium atom is 4.0026.
Determine the energy released when uranium 238 undergoes α
decay.
238.0508u
{234.0436u
+ 4.0026u}
238.0462
Since energy is released, the combined mass of thorium and the
alpha particle should be less than that of the uranium .
31.4 Radioactivity
α DECAY of Uranium
238.0508u
{234.0436u
+ 4.0026u}
238.0462
∆E = ∆mc2 = (.0046u)(931.5)MeV/u = 4.3 MeV.
Most of this energy is in the form of kinetic energy of the recoiling
particles.
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.
In this case a neutron within the nucleus decays into a proton and a
negative particle that is indistinguishable from an orbital electron. 14C
becomes 14N due to this mechanism.
β plus decay also occurs. In this case a proton decays into a neutron and
positron, which has the same mass as an electron, but a positive charge.
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
31.4 Radioactivity
Gamma knife – 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
decay process is:
A.Alpha decay.
B.Beta-plus decay.
C.Beta-minus decay.
D.Gamma decay.
Stop to think
The cobalt isotope 60Co (Z = 27) decays
to the nickel isotope 60Ni (Z = 28). The
decay process is:
A.Alpha decay.
B.Beta-plus decay.
C.Beta-minus decay.
D.Gamma decay.
31.5 The Neutrino
During beta decay, energy is released. However, it is found that
most beta particles do not have enough kinetic energy to account for
all of the energy released.
The additional energy is carried away by a neutrino.
Th 
234
90
Pa 
234
91
0
1
e  
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
remaining vs. number of tosses
would result in an exponential graph,
like the one on the right.
Tossing a single coin is a random
process, but doing so repeatedly
shows a definite pattern.
Radioactive decay shows this same
pattern.
How often does alpha, beta, or gamma 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. 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.
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
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 activity of 1.85 x 105 Bq?
b. What is the activity of the source 10
years later?
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 activity of 1.85 x 105 Bq?
A = λ N and we are looking for N. We
can use the relationship between T1/2
and λ :
T1 2 
ln 2

Cesium activity
First, express T1/2 in terms of seconds:
T1/2 = 30.0 y(3.15 x 107 s/y) = 9.45 x 108 s
λ = (ln 2)/T1/2
λ = .693/9.45 x 108 s
λ = 7.33 x 10-10 s-1
Thus the number of 137 Cs is:
A/ λ = N
1.85 x 105 Bq/ 7.33 x 10-10 s-1 = 2.5 x 1014
atoms.
Cesium activity
What is the activity of the source 10 years later?
A  Ao e   t
A= 1.47 x 105 Bq
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.
•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.
•T1/2 = 5730 years for 14 C. It
undergoes beta decay to 14 N.
•Other isotopes with longer
half-lives are used to date
geological materials.
•. In particular Uranium was used to obtain
an date for the age of the Earth (5.4 billion
years)
Carbon dating
• Archeologists excavating a site have found a
piece of charcoal from a fireplace. Lab
measurements find the 14 C activity 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 = 1.807 x 1011 s
λ = 1.209 x 10-4 y-1
 t
A  Ao e
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
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
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
31.1 Nuclear Structure


r  1.2 1015 m A1 3
31.1 Nuclear Structure
Conceptual Example 1 Nuclear Density
It is well known that lead and oxygen contain different atoms and
that the density of solid lead is much greater than gaseous oxygen.
Using the equation, decide whether the density of the nucleus in a
lead atom is greater than, approximately equal to, or less than that
in an oxygen atom.

r  1.2 10
15

13
mA