Read Sections 11.1, 11.2, and 11.3 before viewing the slide show.
Unit 31
Nuclear Chemistry (Chapter 11)
•Radiation sources (11.1)
•Identification of nuclear particles (11.2)
•Completing nuclear equations (11.2)
•Half-lives (11.3)
Radiation Sources (11.1)
Above is a breakdown of the sources of radiation in our lives. Notice 82% is
natural. The radiation we are most concerned with is called ionizing radiation radiation that knocks electrons out of atoms and forms ions.
Radiation Sources (11.1)
•If you are interested in knowing your personal radiation exposure, you
can go to http://www.new.ans.org/pi/resources/dosechart/ to figure it out.
The basic unit for measurement is the millirem – the average dosage in
the USA is about 620 mrems.
•Several of you indicated you were interested in teaching and, I am not
sure it is age-appropriate, but the EPA sponsors an interactive activity
called USA Radtown. Click the link and you can see the town and, clicking
on any of the activities going on, gives you some information about
radiation exposure due to that activity.
Identification of Nuclear Particles (11.2)
To this point, all of the chemical reactions we have looked at dealt with the transfer
or sharing of electrons. Nuclear reactions are focused on the reactions of the
nucleus, typically disintegrations and sometimes the reaction between nuclei
colliding at high energies.
There are basically five types of nuclear processes we will look at, each involving
the loss or gain of particles or energy from the nucleus. The following table
contains a list of the types of decay and the particles or energy involved.
Type of Decay
Decay
Particle
Particle Mass
(u)
Particle
Charge
alpha
α
4
2+
beta
β
0
1-
gamma
γ
0
0
positron emission
β+
0
1+
electron capture
e-
0
1-
More Description of Nuclear Particles
(11.2)
•The alpha particle is comprised of two protons and two neutrons which is a helium nucleus.
•The beta particle is an electron that has been emitted from the nucleus (We didn’t think about
electrons being in the nucleus. Think of the neutron as a proton and an electron “stuck” together. If an
electron leaves, it will leave a proton behind.)
•The gamma ray is a burst of energy – not a particle. Thus, it has no mass and does not affect the
charge.
•Think of the positron as a positive electron – same mass as an electron, opposite charge.
•Electron capture occurs when an electron (one of those you are used to in an electron shell) is
captured by the nucleus.
Type of Decay
Decay
Particle
Particle Mass
(u)
Particle
Charge
alpha
α
4
2+
beta
β
0
1-
gamma
γ
0
0
positron emission
β+
0
1+
electron capture
e-
0
1-
Symbols for Isotopes
(Review from Unit 11)
• The same system you have seen previously is used to
represent the various isotopes of elements. The base of
the system is:
mass number
atomic number
ChemicalSymbol
or
A
Z
X
where A represents the mass number and Z the atomic
number
• Consider12the examples below:
6
C
6 protons, 12 - 6 = 6 neutrons
35
17
Cl
17 protons, 35 - 17 = 18 neutrons
1
1
H
1 protons, 1 - 1 = 0 neutrons
Symbols for Nuclear Change (11.2)
•We can use the nuclide symbols from the previous page to write symbols for
the decay particles in the table below.
•Notice the alpha particle is identical to a helium nucleus – two protons, two
neutrons.
Type of Decay
Decay
Particle
Particle
Mass (u)
Particle
Charge
alpha
α
4
2+
beta
β
0
1-
gamma
γ
0
0
positron emission
β+
0
1+
electron capture
e-
0
1-
Symbol
4
2
He
0
1 
0
0
0
1 
0
1 e
Writing Nuclear Equations (11.2)
•Writing balanced nuclear equations is a matter of making sure the sum of the
superscripts on the left side of equation matches the sum of superscripts on
the right and that the subscripts also match.
•For example:
238
92
U

Th 
234
90
4
2
He
•Notice how the superscripts add up: 238 on the left = 234 + 4 on the right
and the subscripts add up: 92 on the left = 90 + 2 on the right
•The “trick” to completing nuclear equations
is to ensure that, whatever the species
missing is, its insertion will make the
superscripts and the subscripts add up
on each side.
Type of Decay
alpha
beta
gamma
positron emission
electron capture
Symbol
4
2
He
0
1 
0
0
0
1 
0
1 e
Completing Nuclear Equations (11.2)
•Suppose we are faced with the task of completing the following nuclear
equation:
226
88
Ra 
A
Z
X

4
2
He
•We must find out what nuclide corresponds to the X and its A and Z values.
•On the left the superscripts add to 226, on the right to A+4. So, A has to be
222 to make those two sides equal.
•On the left the subscripts add to 88, on the right to Z+4. So, Z has to be 86
to make those two sides equal.
•The subscript identifies the element’s atomic
Type of Decay
Symbol
number. The element is Rn since its atomic
4
He
alpha
2
number is 86.
0
beta
•The completed equation is:
1 
226
88
Ra 
222
86
Rn 
4
2
He
gamma
positron emission
electron capture


e
0
0
0
1
0
1
Completing Nuclear Equations cont.
(11.2)
•Sometimes (actually most often in life) these sorts of situations present
themselves as word problems.
•As an example: What particle is emitted when thorium-231 decays to
protactinium-231?
•First write out an equation representing what you know (the subscripts are
the atomic numbers of the respective elements):
Th 
231
90
A
Z
? 
231
91
Pa
•Superscripts: 231 on the left, A + 231 on the right, so A = 0
•Subscripts: 90 on the left, Z + 91 on the right
Type of Decay
so Z = -1.
•The equation looks like:
alpha
Th 
231
90
0
1
? 
231
91
Pa
beta
gamma
•Based on the numbers in front of the
question mark, the missing particle must be
positron emission
a beta particle, β. Notice that even though
electron capture
the numbers match the electron capture
process, that process can only occur if the
electron is on the left-hand side of the equation.
Symbol
4
2
He
0
1 
0
0
0
1 
0
1 e
Half-lives (11.3)
•The rate of the decay process for a radioactive nuclide is often represented by
its half-life.
•The half-life is the time required for one-half of the original sample to decay
and does not vary with the size of the original sample.
•Half-lives for different radioactive isotopes can range from mere tiny fractions
of a second to millions of years.
•Since the sample size of the radioactive material gets cut in half for each halflife, finding the fraction of remaining sample after “n” half-lives is just a matter of
multiplying ½ by itself “n” times.
n
1
Fraction remaining =   for n half-lives
2
Applications of Half-lives (11.4)
•One of the most well-known uses of the half-life concept is radiocarbon dating.
In the upper atmosphere, carbon-14 is formed from the bombardment of
nitrogen-14 with a neutron according to the equation:
14
7
N  01n 
14
6
C

1
1
H
•Carbon-14 has a half-life of 5730 years and is incorporated in all living
organisms. Once the living organism has died, there is no mechanism to
replenish the carbon-14 so its level continuously drops.
•The carbon-14 level can be compared against that found in living organisms to
determine how long something has been dead.
•Carbon-14 dating is reliable to about 50000 years but longer than that and its
level becomes too low to detect reliably.
•Other isotopes can be used to determine the age of objects much older or
younger than 50000 years. For example, hydrogen-3 (tritium) is reliable for 1 to
100 year measurements. Uranium-238 is good from 107 years back to the
oldest Earth samples.
Radioisotopic Dating Example (11.4)
•A piece of wood from an Egyptian tomb has carbon-14 activity of 980 counts
per hour. A piece of new wood of the same size gave 3920 counts per hour.
What is the age of the wood from the tomb?
•First figure out how many half-lives have passed. Notice that 980 is ¼ of
3920. That means two half-lives have passed. (980 is one-fourth of 3920:
½ x ½ = ¼)
•Since each half-life for carbon-14 is 5730 years, the time elapsed is 2 x 5730
years = 11460 years.
End of Unit 31 (and all of the) Slide Shows