Particle Nature of
The Nucleus
• -All nuclei have mass that is multiple of
specific number.
• -t.f. nucleus made of smaller particles:
protons, neutrons.
1
Nuclides – elements same p+
Isotopes – specific types of nuclides
• Mass number
A
•
X
• atomic proton# Z
• Proton Z number identifies Z type of element.
• Ex: all H has one proton. Oxygen has 8 protons
in nucleus. Proton number is equal to number of
2
e- in neutral element.
Mass or Nucleon Number
• Nucleons live in nucleus -protons and neutrons.
• Mass number (A) is equal to the total all
neutrons and protons in the nucleus.
• If p+ and no considered as atomic weight 1, the
mass number gives the atomic weight in amu u.
3
Isotopes
• Elements come in different isotopes.
• Two isotopes of the same substance will have equal
proton numbers, but different numbers of
neutrons.
• For the same element, the Z number must always
be the same but the A number may vary.
• FYI for the most part dif isotopes of the same
element behave the same way chemically.
4
Nuclides
• Nuclide is a specifically defined isotope.
• Nuclides are defined by many different aspects,
such as half life, mode of decay, percent
abundance, and so on.
• More specific than isotope. Isotopes are sets of
nuclides having the same number of p+, but
different number of no - . Each individual isotope
is a separate nuclide.
5
Calculations of nucleons:
• For a given isotope:
protons (Z) + neutrons = mass number (A)
• Its easy to find the number of no in isotope.
A – Z is total neutrons.
• Find the number of no in
37 Cl
17
6
Mass Spectrometer
separates isotopes by mass
•
•
•
•
•
Ionization (thermal)
Accelerate ions E field
Velocity detector
Magnetic Deflector
Detector.
7
Velocity Selector
Crossed E and B fields
• Velocity selector
• Felc = Fmag
• Eq = qvB
v = E/B
8
Entering Magnetic Field
Heavier ions are deflected less than lighter ones.
Fmag = Fc.
qBr = mv
qvB = mv2/r.
r=
qB = mv/r.
mv
qB
Detectors Count How many land at each point.
10
• Read Hamper 7.2 and the Mass
Spectrometer pg 255-256
• Do Hwk worksheet w reading.
11
Nuclear Force
12
The strong nuclear force holds nucleons together.
Heating metals or shining EM radiation can strip
e- from atom. Not so to pull apart nucleus.
13
Binding E (BE) = work (J or eV) needed to pull apart
nucleus.
• When work is added to nucleus to split it up, where
does the E go?
• If they are not in motion no KE, or PE chm or PE elc
what happens to work put in?
• Einstein -separate nucleons have greater mass
than when they are bound in nucleus.
E from W to split atom was
converted to mass!
• E = mc2.
14
The higher the BE between nucleons, the more
work needed to split it up, the more stable the
nucleus – the less likely to decay (fall apart)
15
Binding E per Nucleon. Total E of Nucleus
Num. Nucleons.
16
Higher binding E per nucleon = stable elements.
17
Mass Defect/Deficit
• The difference between the mass of the atom
and the mass of the individual constituents.
• Since the nucleus has less mass than the sum of
its parts, the difference is called mass defect
and it equals the BE.
• Look at the table, which are the most stable?
18
What happens when nucleus in
not stable?
• Strive for stability.
19
Mass Defect
20
20
Because E and mass are equivalent it can
be shown that 1 u = 931.5 MeV.
• We can use the above relationship to
calculate the BE per nucleon for each
element.
21
Calculation of BE per nucleon
1. Calculate the BE of 54Fe which has an
average mass of 53.9396 u.
• Mass p+ = 1.00782 u
• Mass no = 1.00866 u
Calculate mass defect & BE/nucleon
•
•
•
•
26 p+ x 1.00782 u =
28 no x 1.00866 u =
26 e- x 0.000549
Mass constituents
mass Fe nucl
subtract
Mass defect
energy
26.20332
28.24248
0.014274
54.4458 u
53.9396 u
0.5062 u
x 931.5 MeV/u
• BE = 471.5 MeV ÷ 54 nucl = 8.7 MeV/nucl
23
2. Find the binding energy of He-4 which has a known
mass of 4.002602 u. (I’m including the e- this time).
Neutral He-4 consists of 2e-, 2p+, and 2no.
Find the sum of the parts in u.
Look up in table:
e- rest mass
p+ rest mass
no rest mass
Mass Defect
2 x (0.000549u)
+2 x (1.007277u)
+2 x (1.008665u)
0.001098
2.014554
2.017300
constituents
4.032982 u
nucleus
4.002602 u
defect
0.3038 u x 931.5 MeV/nucl
28.29897 MeV ÷ 4 nucl = 7.0747 MeV/nucl
25
25
The nucleus of a deuterium atom consists
of a proton and a neutron. If the mass of
deuterium is 2.014102 u, calculate the BE
in MeV.
26
Binding Energy
• The amount of work needed to pull apart
nucleus.
• The nucleus converts the missing mass to BE,
it’s like a glue to hold nucleus together.
• When nucleus transitions to lower energy
state, mass defect E is released!
27
27
BE and atomic size 3 min
• http://www.youtube.com/watch?v=UkLkiXi
OCWU
To calculate the BE/nucl:
1. Find mass defect - the difference between the mass
of the separate nucleons (unbound) and the mass of
the bound nucleus.
2.Calculate the mass in atomic mass units.
3. Each unit has an energy equal to 931.5 MeV.
4. Multiply the mass defect by 931.5 MeV to convert
the mass to energy.
5. Divide by # nucleons.
29
Do Hamper pg 154 #9. and addl
prob’s Holt
30
Radioactive Decay
Radioactivity – emission of energy.
• 1896 Antoine Becquerel discovered that
certain U salts emitted rays that could
penetrate dark paper to expose a
photographic plate. Called penetrating rays
radiation.
• Marie Curie followed up & discovered other
elements with the same properties (Th).
31
Two types of Transmutation
• Spontaneous
• Artificial/Induced
• Unstable Nucleus
• Bombard Nucleus with
a particle.
32
Emitted From Nucleus
•
•
•
•
•
•
Alpha
Beta
Positron
Gamma
Neutrino
Antineutrino
33
Spontaneous Radioactive Decay is
random process
• The type of decay & rate of decays do not
depend on pressure, temperature, chemical
bonds.
34
1899 Rutherford found U emits two
dif types of “radiation” a and b.
b rays were more penetrating.
Later, Villard found another type of
radiation with even more penetrating
power - gamma g.
35
Alpha Rays
a rays are helium nuclei, (2p+ and 2no), that are
emitted from nucleus. They are positively charged
since e- missing.
When nucleus is
too large, it may emit
alpha particle.
36
Alpha Decay of Americium-241 to Neptunium-237
Parent
Daughter
He nucleus
37
a particles are easily stopped by skin or thin
sheet of paper.
Likely to knock e- from orbits if they hit
them. a loses all its KE at once when it’s
stopped.
Charge = +2e.
Mass = 4 units.
Energy is KE = ½
mv2.
38
Alpha a Decay
Alpha a have KE ~5 MeV
39
Beta b Particles
• Fast moving electrons emitted from nucleus.
• More penetrating than alpha because they are
smaller.
• Think of B as a no that emits e-, becomes a p+.
40
Less capable of ionizing (knocking out e-) Charge = -1 or -e.
mass = e, but considered massless in mass #
calcs.
KE = ½ mv2. v can be sig portion of c.
Need a few mm of Al to stop them.
41
42
Positron Decay like beta decay but
p+ loses b+ (positive electron )becomes a
neutron:
Gamma g Radiation
A highly penetrating type of nuclear radiation, similar
to x-rays and light, except that it comes from within
the nucleus of an atom, and, has a shorter l. Gamma
ray emission is a decay mode by which excited state of
a nucleus de-excite to lower (more stable) state in the
same nucleus. In diagrams, a gamma ray is
represented by this:
In equations = g.
Can pass thru human body, concrete, and lead.
44
Gamma Radiation
45
Gamma rays are EM waves. They have:
Lowest ionizing power.
No charge. No mass.
Energy described by E = hf.
Travel with vel of light in vacuum.
No maximum stopping range.
46
Neutrinos are particles that are emitted with
beta and positron decay.
No charge.
Almost no mass.
They do not interact with matter.
47
Which emitter would be safer to
swallow?
48
How could we distinguish the
different types of radiation? What
could we observe?
49
50
51
52
7.2.4 Biological Effects of
Ionizing Radiation
Outline them.
Becquerel Rays 9 min.
• https://www.youtube.com/watch?v=INF9y1
54EZA
54
Products of Decay
When a parent nucleus decays, a daughter
product, is produced, a new element could be
produced.
We can identify products by balancing mass
and atomic numbers.
55
55
1. U – 232 decays by alpha
emission. Write the nuclear
equation and determine the
daughter of the decay.
56
alpha
232
4
U
92
A
He
2
+
X
Z
X will have 4 less nucleons than U-232.
X will have 2 less protons than U-232.
57
Daughter will be:
228
X
90
Look up on periodic table element with 90 p+ (Thorium).
Answer is:
228
Th
90
58
58
Positron/Beta Decay
• During beta decay a beta, b- (e-) or a positron b+
(+e) is emitted from the nucleus.
• When an e- (b-) emitted a no changes to a p+, the
atomic number increases by 1.
• When an e+ (b+) emitted a p+ changes to a no,
the atomic number decreases by 1.
59
A neutrino v or antineutrino v is also emitted
in positron or beta decay.
b+ decay emits neutrino v.
b- decay emits an antineutrino v.
60
Positron Decay
like beta decay but:
• Proton
no + b+ + n
neutrino
A positron particle b+ is the antimatter e-.
•
•
90
•
•
90
44
44
b+ + ___ + n.
Ru
+1
b+ + 90 Tc + n.
Ru
+1
43
61
2. C-14 undergoes beta decay. What will be
the daughter. Remember a beta particle is
an e- ejected from the nucleus so that a
neutron becomes a proton.
62
14
C
6
0
A
e
-1
+
X + v.
Z
A beta particle/e- has atomic number –1
and is considered massless.
63
Balance – make right side add up
to left side.
14
C
6
A will be 14.
0
e +
-1
A
X + v.
Z
Z will be ??
64
Z will be 7 since 7 and -1 = 6. So:
14
0
C
14
e +
6
-1
X+v
7
Z=7 is nitrogen. So daughter is :
14
N
7
65
3. Radium-226 decays by
alpha emission. What is the
resulting daughter element?
66
226
88
Ra
X + 4He
2
Mass X = 226 – 4 = 222.
Atomic number X
88 – 2 = 86.
222 Rn
86
67
4. Sulfur 35 emits b- particles
when it decays. Write the
equation. What will be the
daughter product?
68
35
16
S
0
-1
e +
A
X + v.
Z
Atomic number X is 17 since neutron went to
proton, so daughter is 17Cl – 35.
69
5. Complete the equation:
•
•
23
12
•
•
23
12
b+ + __ __ + __.
Mg
+1
__
b+ + 23Na+ n.
Mg
+1
11
70
Nuclear Stability
The more stable the nucleus, the less likely that
it will decay.
Electrostatic repulsion in the
positively charged nucleus
makes it want to decay.
The strong nuclear force holds
the nucleus together.
72
The extra neutrons increase the strong force
& help shield against electric repulsion.
73
As more p+ added to nucleus more no needed
for stability (for nucleus not to decay).
There seems to be a ratio between p+ and no
for each element to maintain enough strong
force to keep nucleus from flying apart.
In general heavier elements require more no.
p+– no ratio determines stability.
74
Nuclides above the
band are too large decay by a.
To the left b- decay
occurs.
Nuclides below the
band have too few no,
positron decay occurs.
A p+ becomes a no.
75
Summery: Natural Decay occurs spontaneously.
a decay reduces the mass # by 4 and the atomic # by
2.
b- decay does not affect the mass #,
but increases daughter's the atomic # by 1.
b+ decay does not affect the mass #,
but decreases the daughter's atomic # by 1.
g decay affects neither the atomic nor the mass # but
returns excited nucleus to ground state.
76
Properties of Becquerel Rays 9 min
http://www.youtube.com/watch?v=INF9y154EZ
Ahttp://www.youtube.com/watch?v=I7WTQD2x
YtQ
77
Hwk Read Hamper 7.3 do #11-13 But Write out the
balanced nuclear reaction for each.
78
Nuclear Reactions:
Natural & Artificial Transmutation
A process changing the nucleus =nuclear reaction.
Typical reactions are nuclear fission & fusion.
Reactions can occur spontaneously or be artificially
induced.
Transmutation – can be artificially induced nuclear
reaction.
79
Fission
We looked at fission by spontaneous
radioactive decay.
nucleus splits
to 2 or more parts.
Parent mass # decreases.
Fusion
material is added
and “fused” to the nucleus.
80
Nuclear Fusion
1919 Rutherford bombarded gas with a
particles. New particles were produced in the
gas which were not a particles. From the
deflection he concluded they were p+.
He later discovered that the nuclei of the gas
absorbed the a particle & emitted a p+ .
The a fused to form a larger nucleus.
Fusion occurs naturally in stars.
81
Equation for Fusion Rx.
Mass # of parent increases
4
2
He +
a
14
7
N
17
8
O+
1H
1
Proton
82
Other types of particles can be used to
bombard the nucleus.
Neutrons, protons, and H-2 are common.
Which rx is this?
83
16O
8
+
1n
AX
0
z
+
2H
1
• Identify X if a 2H is emitted in the
reaction.
• Balance mass numbers & proton
numbers on the left must and the right
first.
84
16O
+
8
1n
AX
0
z
+
2H
1
Solve for the mass & proton number for X by balancing.
•
16O
8
+
1n
15X
0
7
+
2H
1
Element must have 7 p+. It will be
15N.
7
85
Energy Considerations in Fission &
Fusion
86
Nuclear decay seeks to stabilize nucleus. When nucleus
goes from lower to higher binding energy ratio/nucleon,
energy is released in process.
87
Release of E occurs when element
goes to more stable state.
• Use BE table to predict the total energy release.
•
•
235
92
U
2
117
Pd fragments
46
• Find the energy released. Use the BE/nucleon
table.
88
•
235
U ~ 7.6 MeV per nucleon
•
117
Pd ~ 8.4 MeV per nucleon.
• Take difference. This E is released per nucleon.
• 8.4 – 7.6 = 0.8 MeV/nucleon.
• But 235 U has 235 nucleons so:
• 235 x 0.8 MeV = 188 MeV released!
89
BE Released in Rx can be converted to
KE and/or heat.
90
3: Lithium can be bombarded with no to induce the
following rx:
6 Li
3
+
1n
3H
0
1
+
• Mass Li-6
• Mass no =
6.015126
1.008665
• Mass H-3
• Mass He-4
3.016030
4.002604
How much E is released?
4He
2
Calculate mass for each side.
6 Li
3
+
1n
3H
0
1
(7.023791u)
+
4He
2
(7.018634)
Dm = (7.023791 – 7.018634) u
•
•
•
•
Dm = 0.005175 u x 931.5 MeV/u =
4.804 MeV
Available as KE
some may be released as heat.
92
8.1 Natural transmutations and half lives 10
min
http://www.youtube.com/watch?v=I7WTQD2xYtQ
9. Energy From Reactions 10 min
• http://www.youtube.com/watch?v=YMgacsJyD0&playnext=1&list=PL0191606751B22A12&
feature=results_main
93
Read Hamper 7.4 and 7.5
• ForHwk. Hamper pg 162 15-17, pg 163
#18 – 19, 20 and pg 166 #21,22, pg
Decay Series
Heavier elements – even a decay often does
not decrease the mass of heavy elements
enough for stability. These elements go
through a series of decays before reaching
a stable state.
95
Radioactive Decay Series
All elements above Z=83 are radioactive.
Most of these decay through a series of
transitions. The daughters are radioactive
and decay a number of times before they
become a stable element.
Many elements decay through many a and
b steps.
U-238 decay series
97
Spontaneous Random Decay & Half life
Radioactive decay is random & not affected by
outside factors. There is no way of knowing
whether a particular nucleus will decay.
Changing temperature or pressure does not
change chances of a decay, a chemical rx does not
alter its activity.
99
We can only know the chances of a
decay happening.
100
Rate of Decay is proportional to number (N) of
atoms in the sample.
If large number of atoms, the number decaying
would be larger.
Radioactive decay is an exponential process.
The number, N, of parent atoms decreases
exponentially over time so do the number of
decays/ unit time.
101
As number of
parent atoms
decreases, the
rate in units per
second
decreases
exponentially.
The daughter
product grows
as a mirror
image.
102
Half Life t1/2 defined as:
time it takes for half of original parent in a
sample to decay, or time for original mass to be
halved, or time for activity to be halved.
103
Activity in counts/sec (Bq) easier to measure.
It’s the decay rate of the sample.
104
What is the half life of this sample?
105
Ex 1: The half life of Quinnium is 4
days. If we start with 160 g of pure
Quinnium , how much will be left after
12 days??
106
Exact multiple of half lives:
160 g
80 g
4 days
40 g
8 days
20 g
12 days
3 half lives have passed
107
Rd Hamper 7.4
• Do 15-17 pg 162 and finish #2 pg 167
108
Radioactive Decay Equations
The activity defined av # of disint/ sec
for a particular sample
unit = becquerel (Bq).
Activity of sample:
DN
ADt
A = activity (Bq) counts or decays/s
DN =# disintegrations
DT = time (s)
Negative sign b/c disintegrations are decreasing as
N decreases.
Ex 1: If a sample starts with 1000 parent
atoms and decreases to 100 parent atoms in
3600 seconds, what is the activity of the
sample?
DN
ADt
• 900 atoms/ 3600 s = 0.25 Bq.
110
Activity is proportional to N (parent atoms) at any
time so:
A  lN
l is radioactive decay constant s-1.
We can’t predict which nucleus will decay but
we can predict the number of nuclei decaying
in a specific period of time.
111
Ex 2: A sample has an activity of 2 x 103 Bq
and a l of 5 x 10-5 s-1. What is the number
of remaining parent atoms at this time?
• A/l = N
• 2 x 103 Bq/ 5 x 10-5 s-1 =
• 4 x 107 parent atoms remain.
112
Derive the half life equation.
113
The graph below shows the fraction of original
atoms remaining with the passage of ½ lives.
N
The number of parent atoms decays exponentially
over time. The Y axis is the activity of the sample.
Radioactive Decay Law
The equation of the graph is:
N = Noe-lt.
No original # atoms (parent)
l = decay constant.
N # (parent)atoms remaining
e is natural log
115
Relating ½ life to decay constant
e-lt.
N = No
N
N
e
- lt
o
When t is t1/2, then
N = No.
2
Rearrange:
2 = No.
N
116
N
N
e
- lt
flip
N
N
o
o
e
lt
But when t = t1/2
2 = No.
N
so
lT1 / 2
2e
Take ln both sides:
ln 2 = lT1/2.
T1/2 = ln2
l
N = Noe-lt. can be written in terms of
original sample activity rather than
number of atoms.
A = Aoe-lt
118
Since:
A  lN
&
• N = Noe-lt.
• Multiply both sides by l.
l N = l Noe-lt.
A = l Noe-lt
A = Aoe-lt
119
Ex 3: A sample of radioactive isotope originally contains
1.0 x 1024 atoms and has a half-life of 6.0 hours. Find
the:
a. decay constant
b. initial activity
c. number of atoms remaining after 12 h.
d. number of atoms remaining after 30 min.
120
How to determine the half life.
1. Measure the activity using ionizing
properties (Geiger Counter)
121
121
Short Lived Nuclei – read it off
122
Longer Lived Isotopes (long half life)
• Plot N vs t on semi log graph
• Find gradient
123
Detection Devices
Electroscope - charged electroscope
can be discharged by oppositely
charged ions. Rate of discharge
related to greater radioactivity.
124
Other Detection Methods
• Ionization Chamber
Geiger-Muller Detector
• Gas filled tubes. Use the fact that radiation
can ionize other atoms thereby giving
them a charge. Current is generated and
detected.
125