20/08/2010
Modern understanding: the ``onion’’ picture
SPH4UI Physics
Nuclear Binding, Radioactivity
Protons
Nucleus
Atom and neutrons
Let’s see what’s inside!
Introduction: Development of Nuclear
Physics

1896 – the birth of nuclear physics

Rutherford showed the radiation had three types


Nice Try








3

Alpha (He nucleus)
Beta (electrons)
Gamma (high-energy photons)
1911 Rutherford, Geiger and Marsden performed scattering experiments


Becquerel discovered radioactivity in uranium compounds
Established the point mass nature of the nucleus
Nuclear force was a new type of force
1919 Rutherford and coworkers first observed nuclear reactions in which
naturally occurring alpha particles bombarded nitrogen nuclei to produce
oxygen
1932 Cockcroft and Walton first used artificially accelerated protons to
produce nuclear reactions
1932 Chadwick discovered the neutron
1933 the Curies discovered artificial radioactivity
1938 Hahn and Strassman discovered nuclear fission
1942 Fermi achieved the first controlled nuclear fission reactor
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Some Properties of Nuclei

All nuclei are composed of protons and neutrons

The atomic number (charge), Z, equals the number of protons in the
nucleus
The neutron number, N, is the number of neutrons in the nucleus
The mass number, A, is the number of nucleons in the nucleus






Exception is ordinary hydrogen with just a proton
A=Z+N
Nucleon is a generic term used to refer to either a proton or a neutron
The mass number is not the same as the mass
A
Z
X

Notation

27
Example:
13



Charge:

The electron has a single negative charge, -e (e = 1.60217733 x 10-19 C)
The proton has a single positive charge, +e

The neutron has no charge



Makes it difficult to detect
It is convenient to use atomic mass units, u, to express masses


Mass number is 27
Atomic number is 13
Contains 13 protons
Contains 14 (27 – 13) neutrons
Thus, charge of a nucleus is equal to Ze
Mass:

Al


where X is the chemical symbol of the element
Charge and mass

1 u = 1.660559 x 10-27 kg
Based on definition that the mass of one atom of C-12 is exactly 12 u
Mass can also be expressed in MeV/c2


From ER = m c2
1 u = 931.494 MeV/c2
The Z may be omitted since the element can be used to determine Z
Quick problem: protons in your
body
Summary of Masses
Masses
Particle
kg
u
MeV/c2
Proton
1.6726 x 10-27
1.007276
938.28
Neutron
1.6750 x 10-27
1.008665
939.57
Electron
9.101 x 10-31
5.486x10-4
0.511
What is the order of magnitude of the number of protons in your body?
Of the number of neutrons? Of the number of electrons? Take your
mass approximately equal to 70 kg.
An iron nucleus (in hemoglobin) has a few more neutrons
than protons, but in a typical water molecule there are
eight neutrons and ten protons. So protons and neutrons
are nearly equally numerous in your body, each
contributing 35 kg out of a total body mass of 70 kg.
 1 nucleon 
28
N  35kg 
  10 protons
27
 1.67 10 kg 
Same amount of neutrons and electrons.
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The Size of the Nucleus



First investigated by Rutherford in
scattering experiments
He found an expression for how
close an alpha particle moving
toward the nucleus can come
before being turned around by the
Coulomb force
The KE of the particle must be
completely converted to PE
 2e  Ze 
qq
1 2
mv  ke 1 2  ke
2
r
d

Size of Nucleus

Since the time of Rutherford,
many other experiments have
concluded the following


Most nuclei are approximately
spherical
Average radius is
r  ro A
or
d
4ke Ze
mv 2
1
3
2
For gold: d = 3.2 x 10-14 m, for silver: d = 2 x 10-14 m
Such small lengths are often expressed in femtometers where 1 fm = 10-15 m
(also called a fermi)

A
27
13
Example:
Al
15
has radius r  3.6 10 m
Z
Density of Nuclei
The volume of the nucleus (assumed to be
spherical) is directly proportional to the total
number of nucleons
 This suggests that all nuclei have nearly the
same density
 Nucleons combine to form a nucleus as
though they were tightly packed spheres
ro = 1.2 x 10-15 m
Nuclear Stability


There are very large repulsive electrostatic forces between protons

The nuclei are stable because of the presence of another, shortrange force, called the nuclear (or strong) force



These forces should cause the nucleus to fly apart
This is an attractive force that acts between all nuclear particles
The nuclear attractive force is stronger than the Coulomb repulsive
force at the short ranges within the nucleus
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20/08/2010
Nuclear Stability chart


Light nuclei are most stable if N = Z
Heavy nuclei are most stable when
N>Z


Isotopes


As the number of protons increase,
the Coulomb force increases and so
more nucleons are needed to keep
the nucleus stable
The nuclei of all atoms of a particular element must contain
the same number of protons
They may contain varying numbers of neutrons
 Isotopes of an element have the same Z but differing N
and A values
No nuclei are stable when Z > 83
A
Z

Example: 11C 12C
6
6
13
6
C
14
6
C
Binding Energy Plot
Binding Energy
Iron (Fe) is most binding energy/nucleon. Lighter have too
few nucleons, heavier have too many.
The total energy of the bound
system (the nucleus) is less than
the combined energy of the
separated nucleons
 This difference in energy is
called the binding energy of
the nucleus
 It can be thought of as the
amount of energy you
need to add to the
nucleus to break it apart
into separated protons
and neutrons
Binding Energy per Nucleon
BINDING ENERGY in MeV/nucleon

atomic number (charge), Z
neutron number, N
nucleon number, A,
X
Fission
Fission = Breaking large atoms into small
Fusion = Combining small atoms into large
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Question
Binding Energy Notes


Except for light nuclei, the binding energy is about 8 MeV per
nucleon
The curve peaks in the vicinity of A = 60


Nuclei with mass numbers greater than or less than 60 are not as
strongly bound as those near the middle of the periodic table
The curve is slowly varying at A > 40


Where does the energy released in the nuclear
reactions of the sun come from?
This suggests that the nuclear force saturates
A particular nucleon can interact with only a limited number of other
nucleons
(1) covalent bonds between atoms
(2) binding energy of electrons to the nucleus
(3) binding energy of nucleons
Question
Question
Which element has the highest binding energy/nucleon?
•
Neon (Z=10)
•
Iron (Z=26)
•
Iodine (Z=53)
Which of the following is most correct for the total
binding energy of an Iron atom (Z=26)?
9 MeV
For Fe, B.E./nucleon  9MeV
234 MeV
56
26
270 MeV
Total B.E  56x9=504 MeV
Fe has 56 nucleons
504 Mev
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20/08/2010
Binding Energy
Big Problem: binding energy
E = m c2
Einstein’s famous equation
Proton: mc2 = 938.3MeV
Adding these, get
1877.8MeV
Neutron: mc2= 939.5MeV
93
Calculate the average binding energy per nucleon of 41 Nb
Using atomic mass units
Difference is
Binding energy,
2.2MeV
Deuteron: mc2 =1875.6MeV
MDeuteron = MProton + MNeutron – |Binding Energy|
Calculate the average binding energy per nucleon of
93
41
Nb
Given:
In order to compute binding energy, let’s first find the
mass difference between the total mass of all protons and
neutrons in Nb and subtract mass of the Nb:
m p= 1.007276u
m n= 1.008665u
Number of protons:
m  41mp  52mn  mNb
 411.007276u   52 1.008665u    92.9063768u 
Find:
Eb = ?
 0.8425192u
Thus, binding energy is
Eb 
 m  c2   0.842519u  931.5 MeV u   8.44 MeV
A
93



N p  41
Number of neutrons: Nn  93  41  52
Mass difference:
1u=931.5 MeV
Radioactivity
Radioactivity is the spontaneous emission of radiation
Experiments suggested that radioactivity was the result of the
decay, or disintegration, of unstable nuclei
Three types of radiation can be emitted
 Alpha particles
 The particles are 4He nuclei
 Beta particles
 The particles are either electrons or positrons
 A positron is the antiparticle of the electron
 It is similar to the electron except its charge is +e
 Gamma rays
 The “rays” are high energy photons
nucleon
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20/08/2010
The Decay Processes –
General Rules
Types of Radioactivity
B field into
screen



Radioactive
sources
a particles:
b
4
2

detector
He nuclei
particles: electrons
When one element changes into another element, the process
is called spontaneous decay or transmutation
The sum of the mass numbers, A, must be the same on both
sides of the equation
The sum of the atomic numbers, Z, must be the same on both
sides of the equation
Conservation of mass-energy and conservation of momentum
must hold
Barely penetrate a piece of paper
Can penetrate a few mm of aluminum
g photons (more energetic than x-rays)
Can penetrate
several cm of
lead
Alpha Decay -- Example
Alpha Decay


When a nucleus emits an alpha particle it loses two protons and two
neutrons
 N decreases by 2
 Z decreases by 2
 A decreases by 4
Symbolically
A
Z


X
A 4
Z 2
Y  He
4
2
X is called the parent nucleus
Y is called the daughter nucleus

Decay of 226Ra
226
88



4
Ra  222
86 Rn  2 He
Half life for this decay is 1600
years
Excess mass is converted into
kinetic energy
Momentum of the two particles
is equal and opposite
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Beta Decay – Electron Energy
Beta Decay




During beta decay, the daughter nucleus has the same number of
nucleons as the parent, but the atomic number is one less
In addition, an electron (positron) was observed
The emission of the electron is from the nucleus
 The nucleus contains protons and neutrons
 The process occurs when a neutron is transformed into a proton
and an electron
 Energy must be conserved




The energy released in the decay
process should almost all go to
kinetic energy of the electron
Experiments showed that few
electrons had this amount of
kinetic energy
To account for this “missing”
energy, in 1930 Pauli proposed the
existence of another particle
Enrico Fermi later named this
particle the neutrino
Properties of the neutrino




Zero electrical charge
Mass much smaller than the
electron, probably not zero
Spin of ½
Very weak interaction with matter
Gamma Decay
Beta Decay


Symbolically


Gamma rays are given off when an excited nucleus “falls” to a lower
energy state

Y  e 
A
Z
X
A
Z 1
A
Z
X
A
Z 1
Y  e  
 is the symbol for the neutrino
 is the symbol for the antineutrino



The excited nuclear states result from “jumps” made by a proton or neutron
The excited nuclear states may be the result of violent collision or more
likely of an alpha or beta emission
Example of a decay sequence



Similar to the process of electron “jumps” to lower energy states and giving off
photons
To summarize, in beta decay, the following pairs of particles are
emitted
 An electron and an antineutrino
 A positron and a neutrino
The first decay is a beta emission
The second step is a gamma emission
12
5
B  126 C *  e 
C*  126 C  g
12
6


The C* indicates the Carbon nucleus is in an excited state
Gamma emission doesn’t change either A or Z
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20/08/2010
Decay Rules
Practice
A nucleus undergoes a decay. Which of the following is FALSE?
1) Nucleon Number is conserved.
2) Atomic Number (charge) is conserved.
3) Energy and momentum are conserved.
1. Nucleon number decreases by 4
2. Neutron number decreases by 2
a: example
U  234
90Th  a
238
92
1) 238 = 234 + 4
2) 92 = 90 + 2
recall 2 He  a
Nucleon number conserved
Charge conserved
4
b: example
1
0
n  p e  
g:
A
Z
P*  ZA P  00g
example
0 
1
1
1
3. Charge on nucleus increases by 2
4
a decay is the emission of 2 He  a
A decreases by 4
0
0
Neutrino needed to
conserve energy and
momentum.
Ex.
238
92
Z decreases by 2
(charge decreases!)
Practice
The nucleus
234
90
Th
Decay
undergoes b decay.

Which of the following is true?
1. The number of protons in the daughter
nucleus increases by one.
2. The number of neutrons in the daughter
nucleus increases by one.
b  decay is accompanied by the emission of an
electron: creation of a charge -e.
Th  X
Pa ee
234
90
In fact, n  p  e
neutrino “escape.”
???
234
??
91

 e
U  23490Th  24 He
0 0  0 0
11
00
Which of the following decays is NOT allowed?
1
238 = 234 + 4
92 = 90 + 2
U  234
90Th  a
238
92
4
Po  210
82 Pb  2 He
214 = 210 + 4
C  N g
14 = 14+0
2
214
84
3
14
6
4
40
19
14
7
6 <> 7+0
K  p e  
40
20
0 
1
84 = 82 + 2
0
0
40 = 40+0+0
19 = 20-1+0
inside the nucleus, and the electron and
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20/08/2010
Decay
The Decay Constant
Which of the following are possible reactions?

The number of particles that decay in a given time is proportional to
the total number of particles in a radioactive sample
N   N  t 


λ is called the decay constant and determines the rate at which the
material will decay
The decay rate or activity, R, of a sample is defined as the number
of decays per second
R
(a) and (b). Reactions (a) and (b) both conserve total charge and total
mass number as required. Reaction (c) violates conservation of mass
number with the sum of the mass numbers being 240 before reaction
and being only 223 after reaction.
Decay Curve
Radioactivity

Decays per second,
or “activity”
N
  N
t
No. of nuclei
present
decay constant
Start with 16
14C


How many will be left after another 6000 years?
2) 4
3) 8
The decay curve follows the
equation
N  N 0 e  t
atoms.
After 6000 years, there are only 8 left.
1) 0
N
 N
t
The half-life is also a useful
parameter
The half-life is defined as the
time it takes for half of any
given number of radioactive
nuclei to decay
T1 2 
ln 2 0.693



Every 6000 years ½ of atoms decay
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20/08/2010
Units

The unit of activity, R, is the Curie, Ci


1 Ci = 3.7 x 1010 decays/second
The SI unit of activity is the Becquerel, Bq

t
 1 T1/2
N (t )  N 0e t  N 0   
2
1 Bq = 1 decay / second
 Therefore,

Decay Function
1 Ci = 3.7 x 1010 Bq
The most commonly used units of activity
are the mCi and the µCi
time
Practice
The half-life for beta-decay of 14C is ~6,000 years. You
test a fossil and find that only 25% of its 14C is un-decayed.
How old is the fossil?
3,000 years
At 0 years: 100% remains
6,000 years
12,000 years
At 6,000 years: 50% remains
Radioactivity Quantitatively
Decays per second,
or “activity”
N
  N
t
Survival:
N (t )  N0e t
decay constant
No. of nuclei present
at time t
No. we started
with at t=0
Instead of base e we can use base 1/2:
t
At 12,000 years: 25% remains
No. of nuclei
present
 1  T1/2
e  t   
where
2
Half life
T1/2 
0.693

t
Then we can write
 1 T1/2
N (t )  N 0e t  N 0   
2
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20/08/2010
Radioactivity Example
The half-life for beta-decay of 14C is 5730 years. If you start
with 1000 carbon-14 nuclei, how many will be around in
22920 years?
t
 1 T1/2
N (t )  N 0e t  N 0   
2
t
 1  T1/2
N (t )  N 0   
2
1
 1000  
2
 62.5
22920
5730
T1/2 
T1/ 2 
0.693

Uses of Radioactivity









 t
 1000e
1.209104  22920)
Ionization type smoke detectors use a radioactive source to ionize the
air in a chamber
A voltage and current are maintained
When smoke enters the chamber, the current is decreased and the
alarm sounds
Radon pollution

N  t   N0e
Beta decay of 14C is used to date organic samples
The ratio of 14C to 12C is used
Smoke detectors

0.693
0.693

5730
 1.209 104
Carbon Dating

Radon is an inert, gaseous element associated with the decay of radium
It is present in uranium mines and in certain types of rocks, bricks, etc
that may be used in home building
May also come from the ground itself
 62.5
Binding Energy
Which system “weighs” more?
1) Two balls attached by a relaxed spring.
2) Two balls attached by a stretched spring.
3) They have the same weight.
M1 = Mballs + Mspring
M2 = Mballs + Mspring + Espring/c2
M2 – M1 = Espring/c2 ≈ 10-16 Kg
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Q Values (nice to know)
Strong Nuclear Force


Acts on Protons and Neutrons

Strong enough to overcome Coulomb
repulsion

Acts over very short distances

Energy must also be conserved in nuclear reactions
The energy required to balance a nuclear reaction is called the Q
value of the reaction


Two atoms don’t feel force
Problem: nuclear reactions
An exothermic reaction
 There is a mass “loss” in the reaction
 There is a release of energy
 Q is positive
An endothermic reaction
 There is a “gain” of mass in the reaction
 Energy is needed, in the form of kinetic energy of the incoming
particles
 Q is negative
Determine the product of the reaction
What is the Q value of the reaction?
Given:
Determine the product of the reaction
What is the Q value of the reaction?
7
3
Li  24 He  ?  n
reaction
7
3
Li  24 He  YX ?  01n
In order to balance the reaction, the total amount
of nucleons (sum of A-numbers) must be the
same on both sides. Same for the Z-number.
Number of nucleons (A): 7  4  X  1  X  10
Number of protons (Z): 3  2  Y  0  Y  5
Thus, it is B, i.e.
7
3
Li  24 He  105 B  01n
Find:
The Q-value is then


Q=?
Q   m  c 2  m7 Li  m 4 He  m10 B  mn c 2
It is easier to
use atomic
mass units
rather than kg.
  7.016005u  4.002603u  10.012938u  1.008665u  931.5Mev / u 
 2.79MeV
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20/08/2010
Summary

Nuclear Reactions






Nucleon number conserved
Charge conserved
Energy/Momentum conserved
a particles = 24 He nucleii
b - particles = electrons
g particles = high-energy photons
Survival:

N (t )  N0e t
T1/2 
0.693

Decays

Half-Life is time for ½ of atoms to decay
14