CHEMISTRY 1000
Topic #4: Nuclear Chemistry
Fall 2010
Dr. Susan Lait
Radioactivity and Radiation

Nuclear reactions result in changing the nuclei of atoms. These
reactions are accompanied by emission of ionizing radiation
(which has enough energy to excite electrons out of molecules,
‘ionizing’ them). There are three main types of ionizing radiation:



alpha rays (): helium nuclei (2 protons + 2 neutrons)
beta rays (): electrons
gamma rays (): high energy photons (higher energy than x-rays)
which have no mass and no charge
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mass number (A)
Balancing Nuclear Reactions

atomic number (Z)
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C
In any nuclear reaction, two things are conserved:


The sum of the mass numbers of the products is equal to the sum
of the mass numbers of the reactants:
The sum of the atomic numbers of the products is equal to the sum
of the atomic numbers of the reactants:
The “atomic number” for an electron () is considered to be -1.

The exact masses of products and reactants are not the same.
The small mass difference between mass of products and
reactants results in release of energy (E = mc2) and the law
of conservation of energy still holds. (The law of conservation
of mass is a special case of the law of conservation of energy.)
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mass number (A)
Balancing Nuclear Reactions

atomic number (Z)
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C
Balance the following nuclear reactions.
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Be
4
0
-1 e
235
U
92
231
Th
90
218
At
85
4

2
206
Tl
81
236
92 U
0
-1 e
141
56 Ba
+ 3 10 n
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Classes of Nuclear Reactions

There are seven classes of nuclear reactions:

Alpha emission

Beta emission

Positron emission

Electron capture

Fusion

Fission

Bombardment (to make transuranium elements)
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Classes of Nuclear Reactions
reactants
products*
ΔZ
spontaneous?
alpha emission
1 nucleus
1 nucleus + 1 alpha particle
-2
yes
beta emission
1 nucleus
1 nucleus + 1 electron
+1
yes
positron emission
1 nucleus
1 nucleus + 1 positron**
-1
yes
electron capture
1 nucleus + 1 electron
1 nucleus
-1
yes
fission
1 nucleus
2 nuclei + neutron(s)
varies
no
fusion
2 light nuclei
1 nucleus + neutron(s)
varies
sometimes
2 heavy nuclei
1 nucleus + neutron(s)
varies
no
bombardment
Most nuclear reactions also emit electromagnetic radiation. Emitting an  or  particle leaves
the nucleus in an excited state so it emits a photon as it returns to the nuclear ground state.
The energy of the photon is specific to the nuclear reaction.
** As antimatter, positrons are not directly observable. A positron is annihilated as soon as it
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collides with an electron, releasing  radiation (a high energy photon).
*
“nuclide” = a specific type of nucleus (i.e. containing a specific #protons and #neutrons)
Classes of Nuclear Reactions

An unstable nuclide undergoes spontaneous nuclear reaction to
form a more stable nuclide. If this product is also unstable, it
undergoes another nuclear reaction (and another and another,
etc. until a stable nuclide is reached). Such a series of alphaand beta-emissions is called a radioactive decay series:
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Classes of Nuclear Reactions

Some classes of nuclear reaction, on the other hand, will never
occur spontaneously. Instead, they must be induced (often by
hitting the nucleus with a neutron to generate a highly unstable
nucleus which will then undergo the desired nuclear reaction).
This is true of fission and bombardment:
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Why Do Nuclear Reactions Occur?

What factors affect stability of a particular nuclide?




A nucleus consists of protons and neutrons held together by
nuclear binding energy.
At the same time, there is electrostatic repulsion between the
positively-charged protons. If this repulsion is too great, the
nucleus will be unstable.
Neutrons lessen this repulsion by increasing the distance between
protons; however, neutrons are inherently less stable than protons.
Excess neutrons will decompose into proton/electron pairs.
It is therefore possible to make a few generalizations:
 Nuclides containing more protons need more neutrons (to
keep the protons apart).
 Nuclides containing fewer protons need fewer neutrons (to
maximize stability).
 There is a maximum number of protons beyond which the
nuclear binding energy cannot hold the nuclide together stably
(because the electrostatic repulsion is too great).
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Why Do Nuclear Reactions Occur?

The number of stable nuclides is relatively small. Plotting
#protons (Z) vs. #neutrons (N) for all nuclides that have been
made/found gives a narrow band of stable nuclides (black dots)
surrounded by a wider band of unstable nuclides (red dots).
The stable nuclides form the band of stability.

Nuclides farthest from the band of stability are
least stable, decaying fastest.

Heavy nuclides decay faster than light ones.

N-to-Z ratio in stable nuclides is predictable:

If Z = 1-20 (H to Ca), N  Z is ideal
If Z = 20-83 (Sc to Bi), N > Z up to N  1.5 Z

If Z  84 (Po and larger), no stable nuclides exist


Even values for Z & N are conducive to stability.
Almost 60% of stable nuclides have both even.
Less than 2% of stable nuclides have both odd!
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Why Do Nuclear Reactions Occur?

The type of nuclear reaction which a nuclide is most likely to
undergo can be predicted from its N-to-Z ratio.



A nucleus which has “too many neutrons” (i.e. N/Z is too high) will
tend to undergo beta emission. How does this improve N/Z?
A small nucleus which has “too many protons” (i.e. N/Z is too low)
will tend to undergo either positron emission or electron capture.
How does this improve N/Z?
A large nucleus which has “too many protons” (i.e. N/Z is too low)
will tend to undergo alpha emission. How does this improve N/Z?
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Nuclear Binding Energy

When nucleons (protons and neutrons) come together to make
a nucleus, energy is released. This energy is referred to as
nuclear binding energy (E) and a nuclide’s nuclear binding
energy can be calculated using Einstein’s famous equation:
The nuclear binding energy for any nuclide can thus be
calculated by comparing its mass to the total mass of the
protons and neutrons it contains. You may have already
noticed that atomic masses are not exactly equal to the sum of
the masses of the protons, neutrons and electrons in the atom!

In order for a nuclide to be stable, its nuclear binding energy
must be greater than the electrostatic repulsion between its
protons.
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Nuclear Binding Energy


Since they release more energy upon formation, nuclides with
greater nuclear binding energies would at first seem to be more
stable; however, we must also factor in the number of
nucleons brought together. Otherwise, larger nuclides appear
to be excessively stable are simply because they contain more
nucleons.
So, a more useful quantity to calculate is the nuclear binding
energy per nucleon (Eb):
Eb 
E
A
where A = mass number = #nucleons = Z + N

Nuclides with larger Eb values are more stable.
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Nuclear Binding Energy

Mproton = 1.0072765 g/mol
Mneutron = 1.0086649 g/mol
Melectron = 0.0005486 g/mol
Calculate Eb for the helium isotopes: 3He (3.016029310 g/mol)
and 4He (4.002603250 g/mol). Which isotope is more stable?
For this type of calculation, ALWAYS use the mass of the specific isotope or nuclide.
NEVER use the average atomic mass listed on the periodic table
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Nuclear Binding Energy


4He
is one of the most stable nuclide. When Eb is plotted as a
function of A, the most stable nuclide is found to be 56Fe:
This plot also shows which nuclides can undergo fusion
(increasing Eb by increasing A) and which nuclides can undergo
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fission (increasing Eb by decreasing A).
Nuclear Binding Energy


Since a nuclear reaction involves conversion of one (or more)
nucleus to another, it can be modeled as destruction of the
original nucleus followed by creation of a new nucleus:
So, we can calculate the energy released by any nuclear reaction
as long as we know the masses of the nuclides involved.
e.g. A neutron (1n, 1.0087 g/mol) strikes 235U (235.0439 g/mol)
to give 138Xe (137.908 g/mol), 95Sr (94.913 g/mol) and 3 1n.
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