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Natural Sciences I
Lecture 21: Radioactivity and Nuclear Reactions
Introduction
A basic understanding of radioactivity and related topics is important
not only to scientists but also to citizens of a modern society. Knowledge
about radioactivity should enter into our thinking about nuclear energy,
nuclear weapons, and environmental health concerns such as radon in
homes.
Discovery of Natural Radioactivity
Like many other phenomena, radioactivity was in a sense
discovered by accident. In 1896 Henri Becquerel, intrigued by the recent
discovery of X rays, was trying to ascertain whether natural phosphorescent minerals produced X rays. (Phosphorescence is actually a type of
luminescence – i.e., light produced by means other than heat. It actually
has nothing to do with radioactivity; rather, it is a chemical effect following
exposure to light.) Becquerel had planned to expose a photographic
plate to a lump of natural phosphorescent salt (potassium uranyl
disulfate) that had been sitting in the sun for a few hours. Waiting for a
sunny day, he put his experimental materials in a desk drawer, wrapped
in a opaque cloth. His phosphorescent salt just happened to contain
uranium, and his photographic plate recorded something really
interesting that clearly had nothing to do with sunlight...
Becquerel's accidental
discovery of radioactivity
radioactive mineral
copper cross
opaque cover
photographic plate
2
Becquerel referred to his newly-discovered invisible radiation as
radioactivity, a term still used today. Materials that exhibit this property
are described as radioactive. Becquerel's finding precipitated a flurry of
research activity, perhaps most notably by Ernest Rutherford (remember
the "size of the nucleus" guy?), who identified three types of radioactivity. We can learn something about these types from the ways in which
they respond to a magnetic field...
magnet
lead container
observations...
g rays not deflected
b particles strongly
deflected
a particles slightly
deflected the other
way
radioactive
source
b
S
g
a
N
Remember that the path of a charged particle is deflected to an
extent proportional to its charge and inversely proportional to its mass.
Electromagnetic radiation is not deflected at all. The observations above
indicate that:
a and b "rays" are really particles; they have opposite charge
and as are much more massive.
g rays are electromagnetic radiation
Today we know a lot more about these types of radioactivity...
An a particle consists of
2 neutrons and 2 protons;
the charge is therefore +2
A b particle is an electron;
the charge is therefore -1
g rays are here in the EM spectrum
10
-14
g rays
10
-12
l in meters
X-rays
10
-10
10
UV
-8
10
-6
IR
3
Let's be clear about our symbols before we begin to work with nuclear
equations.
For isotopes...
For particles...
chemical symbol
particle symbol
mass number
(no. of nucleons)
238
92 U
atomic number
(no. of protons)
Summary table of
common particles
and photons
0
e
-1
mass number
charge
symbol mass
no. charge
name
proton
electron
neutron
g photon
1
1H
0
e
-1
1
n
0
0
0
or
or
g
1
p
1
0
b
-1
1
+1
0
-1
1
0
0
0
Nuclear reactions involve reactants and products just as chemical
reactions do; e.g.
238
U
92
234
Th
90
4
+ 2 He
This reaction describes the initial a decay of the naturally radioactive
isotope uranium-238.
Like chemical reactions, nuclear reactions must be balanced. For
our purposes, this means simply that the number of nucleons (protons +
neutrons) must be conserved. Knowing this is useful in answering
questions such as What is the identity of the nucleus produced by a
decay of plutonium-242?
242
4
He + ?
Pu
94
2
242 - 4 = 238 nucleons (= the mass number of the product nucleus)
94 - 2 = 92 protons (= the number of protons in the product nucleus)
So...
242
Pu
94
4
He
2
+
238
U
92
4
Nature and Stability of the Nucleus
The nucleus is now understood to have a shell structure analogous
to that of the electrons in the outermost part of the atom. Protons and
neutrons are envisioned as filling energy levels in a manner that
maximizes the overall stability of the nucleus given the available
nucleons. Some nuclei are more stable than others, and some are not
stable at all.
Consideration of the nucleus requires recognition of forces that we
have not discussed previously in this course. A logical question to ask is
Why are atomic nuclei stable at all, given that they are crammed with
positive charges that should all repel one another? The answer lies in
the existence of another force – the nuclear force – that operates only
-15
at very short distances (<10 meters). At these distances , the nuclear
force is strong enough to overwhelm the mutual electrostatic repulsion of
like charges, and protons strongly attract one another when they get
very close.
As noted above, certain nuclear shell configurations tend to
stabilize the nucleus: Nuclei having 2, 8, 20, 28 50, 82 or 126 protons or
neutrons are particularly stable (much like noble-gas electron configurations). The helium nucleus, with 2 protons and 2 neutrons, is extremely
stable – suggesting that pairing of protons and neutrons stabilizes
the nucleus just as
no. of
pairing of electrons
protons neutrons nucleons stable nuclei
stabilizes a covalentlybonded molecule. Nu50
odd
odd
even
clei with an even number of both protons
55
odd
even
odd
and neutrons are
4
even
odd
odd
extremely stable.
165
even
even
even
An additional factor in nuclear stability
is that more and more "excess" neutrons (i.e., over and above the
number of protons) are needed to stabilize the nuclei as it becomes
progressively larger. As you know, the largest stable nucleus is atomic
number 83. The net result of these various factors affecting nuclear
stability is the band of stability on a plot of the number of protons vs.
the number of neutrons...
(see next page)
5
140
band of
stability
number of neutrons (N)
120
neutron : protons
= 1.5 : 1
100
80
1.25 : 1
1
1:
60
(n
tro
u
e
:
s
n
s
on
t
o
pr
)
40 1.1 : 1
20
20
40
60
80
number of protons (Z)
Radioactive Decay Modes
There are three common processes through which unstable nuclei
change into stable ones. These are..
Alpha emission
An a particle is ejected from the nucleus to reduce both Z and N by
2 (the product nucleus is also unstable and usually continues to
disintegrate by b decay). We've already seen a couple of examples of a
emission, but here's one more
222
86 Rn
radon
218
84 Po
+
4
2 He
the expelled a particle
will acquire two electrons
to become a helium atom
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a DANGER
Alpha particles can do damage to the cells of
living organisms, but they are relatively easy to protect ourselves against, and can't do much harm unless they are emitted by an ingested radioisotope. They travel 2 to 12 cm in air,
depending upon the energy with which they are ejected from
the disintegrating nucleus. However, they are easily stopped
by dense matter – including an ordinary piece of paper.
Beta emission
As its name suggests, beta (b ) emission involves the ejection of
an electron at high speed from a disintegrating nucleus – in effect
converting a neutron into a proton:
1
n
0
0
e
-1
+
1
p
1
The number of protons thus increases by 1 and the number of
neutrons decreases by 1 (total nucleons remain constant). The classic
example is the decay of carbon-14:
14
C
6
14
N
7
0
+ -1
e
b DANGER
Depending upon its energy, a b particle can
travel hundreds of centimeters through the air. Dense matter
will stop a b particle in a fraction of a millimeter to a few
millimeters. A source of b particles in the body is dangerous
because of the potential for ionization of local atoms.
Gamma emission
The ejection of an a or b particle often leaves the immediate
product nucleus in an excited state. By analogy with the energetics of
electron orbitals, this condition would be like an electron occupying a
high energy level when a lower one is available. The result is that a
spontaneous rearrangement of nucleons occurs and a photon of very
energetic (short l) radiation is emitted by the nucleus...
(next page)
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Gamma emission (cont'd)
Example: The decay of radium-226 (and most other isotopes that
decay by a emission) is actually a two step process that involves
ejection of an a particle (page 5) to produce radon-222 in an excited
nuclear state, followed immediately by g emission as the nucleus
reorganizes to a lower-energy state:
226
88 Ra
radium
222
*
86 Rn
+
4
2 He
222
*
86 Rn
then...
222
86 Rn
+
g
denotes excited state of radon
g DANGER
Like the more familiar X rays, g rays are capable
of penetrating right through the human body. Of the three
common types of radioactive emission (a, b, g), they are the
most difficult to protect ourselves against, since they can
penetrate several centimeters of lead.
Summary Table
unstable
condition
resulting
nucleus
type of
decay
emitted
> 83 protons
a emission
4
2 He
minus 2p and 2n
N:Z too high
b emission
0
e
-1
plus 1p; no mass change
excited nucleus
g emission
0
0
no change
N:Z too low
e+ emission
g
0
e
1
minus 1p; no mass change
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Systematics of Decay
The decay of a single unstable nucleus is a matter of probability
– it cannot be predicted when an individual nucleus of say, uranium238 will eject an alpha particle. When large numbers of unstable
nuclei are considered, however, the overall behavior is simple and
predictable. Each radioactive nucleus has a characteristic
radioactive decay constant, k:
decay rate
rate
or
k = number of nuclei present
k =
n
(where rate = no. of nuclei that disintegrate each second)
We usually express the decay process in terms of the half-life, or
time required for the number (n) of radioactive nuclei present to drop
to n/2. The half-life has a simple relationship to the decay constant:
t½ =
ln2
k
=
0.693
k
mass of strontium-90 (grams)
Let's see what this means with a simple example: strontium-90 (this
is a fission product resulting from nuclear weapons tests; also
produced in nuclear reactors). The half-life of strontium-90 is 28
years; it decays by beta decay. Let's start with a kilogram of
strontium-90...
no. of half-lives
0
1000
1
2
28
56
3
4
5
6
7
8
800
600
400
200
0
84 112 140 168 196 224
years
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Table of selected isotopes, half-lives
and decay modes
isotope
half-life
decay
mode
3
1H
12.26 yr
beta
14
6C
5,730 yr
beta
40
K
19
1.3E9 yr
multiple
90
Sr
38
28 yr
beta
131
53 I
8 days
beta
133
54 Xe
5.27 days
beta
232
90 Th
1.39E10 yr
alpha
235
92 U
7.13E8 yr
alpha
238
92 U
4.51E9 yr
alpha
239
94 Pu
2.44E4 yr
alpha
240
94 Pu
6,760 yr
alpha
242
94 Pu
3.79E5 yr
alpha
naturally-occurring
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Radiation Units
The curie is the most basic of radiation units – it is simply a
number of disintegrations per unit time:
10
1 curie = 3.7 x 10 disintegrations/second
The rad (short for radiation absorbed dose) takes into account
the energy absorbed by a material:
-2
1 rad = 1 x 10 J/kg
The rem (or roentgen equivalent man) takes into account the
damage inflicted on living tissue, so the nature of the radiation or
particles enters in (see Table 15.4 of your text).
Nuclear Energy
Energy from nuclear reactions can come from either fission (the
breaking apart of massive nuclei) or fusion (the combining of small
nuclei – as in stars). In both cases, the energy comes from the mass
difference between the reactants and products of the relevant nuclear
equations.
Here's an example of a mass-to-energy conversion...
238
92 U
234
90 Th
+
4
2 He
Summing the masses of reactants and products...
238.0003 u
233.9942 u + 4.0015 u
233.9942 + 4.0015 - 238.0003 = -0.0046 u
...so we've lost some mass in this nuclear reaction. This mass was
converted to energy.
E = mc
2
DE = Dm c
In the above example, Dm = -0.0046 u.
2
11
The nuclear reactors in use today use sustained fission reactions
as the source of energy. Fission involves the disintegration of a
heavy nucleus following the absorption of a neutron:
92
36
Kr
neutron
more neutrons
235
92
U
141
56
92
36
Kr and
141
56
Ba
Ba are example fission products – two resulting
nuclei of sub-equal mass, usually themselves radioactive.
Uranium-238 is abundant in the nuclear "fuel"; this leads to the
production of plutonium.
Nuclear fusion has the potential to yield enormous amounts of
energy if it can be done in a controlled, sustained way. Consider
the mass-to-energy numbers...
2 protons = 2 x 1.00728 u = 2.01456 u
2 neutrons = 2 x 1.00867 u = 2.01734 u
sum = 4.03190 u
The mass of a helium nucleus is 4.00150 u – significantly
less than the sum of the nucleons – so a large amount of energy
results from hydrogen fusion (this would actually take place in a
3-step process like H-"burning" in stars.