Lesson 2. Nuclear Chemistry & Energy
Outline
2.1 Nuclear Reactions and Radioactivity
2.2 Nuclear Stability
2.3 Kinetics of Radioactive Decay
2.4 Energetics of Nuclear Reactions
2.5 Fission, Transmutation, Fusion
2.6 Interaction of Radiation and Matter
Nuclear Chemistry
Nuclear chemistry is a study of nuclear reactions. Nuclear reactions are changes in matter originating in the
nucleus of an atom. When a nuclei change spontaneously, it emits radiation and thus said to be radioactive.
Radioactive elements are widely used in medicine as diagnostic tools and as a means of treatment especially for
cancer. They are also used to help determine the mechanisms of chemical reactions, to trace the movement of
atoms in biological systems, and to date important historical artifacts. Nuclear reactions are also used both to
generate electricity and to create weapons of massive destruction.
This lesson discusses nuclear reactions and its applications.
2.1 Nuclear Reactions and Radioactivity
Nuclear Reactions
One of the nuclear reactions in the universe is the formation of carbon-14 which is a result of collision of gas
molecules in the atmosphere with cosmic rays. Cosmic rays are particles traveling at high speeds. Majority of
cosmic rays are atomic nuclei. Roughly, 87% of cosmic rays are hydrogen nuclei and about 12% are helium
nuclei. Some cosmic rays emanate from the sun and others originate outside the solar system. The energies of
cosmic rays are far higher than chemical energies. Chemical energies are usually expressed in kilojoules/mol
(kJ/mol) but that of cosmic rays are expressed in electron volts (eV). One (1) eV is equal to 96.5853 kJ/mol.
As cosmic rays enter the atmosphere, it collides with gas molecules. These collisions induce nuclear reactions in
the atmosphere. One of this nuclear reactions leads to the formation of the radioactive carbon-14. Carbon-14 is
unstable isotope of carbon, it undergoes spontaneous radioactive decay or disintegration, ejecting particles
from the nucleus. Carbon-14 is formed when a free neutron is absorbed by nitrogen nucleus resulting to carbon
nucleus and proton. Nitrogen nucleus and neutron are the reactants and carbon nucleus and proton are the
products. The nuclear reaction is written as:
14
7N
+ 10n
14
6C
+
1
1p
Take note that reactants and products of a nuclear reaction are written in a form of atomic notation, AZE, where
E is the symbol of the atom, A is the mass number and Z is the atomic number or nuclear charge. Similar
notation is used for subatomic particles, neutrons (10n), protons (11p), and electrons (0-1e).
Nuclear reactions also obey conservation rules, hence, nuclear reactions are balanced. Balancing nuclear
equation is with respect to both mass number and nuclear charge. For the nuclear equation to form carbon-14,
the sum of the mass numbers for each side of the equation is 15, and the sum of the nuclear charges on each
side of the equation is 7, therefore the nuclear equation is balanced.
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Radioactivity
Radioactive decay or nuclear decay is any process by which an unstable atom or nucleus spontaneously emits
subatomic particles. Radiation refers to particles or photons emitted in radioactive decay or nuclear decay.
Soon after radioactivity was discovered, physicist Ernest Rutherford demonstrated three types of radiation,
namely: alpha rays, beta rays and gamma rays. The different types of radioactive decay are discussed below.
○ Alpha Decay
When a nucleus undergoes an alpha decay it ejects an alpha particle (α), which is actually a helium nucleus (42He).
During alpha decay, the mass number decreases by 4 and its atomic number decreases by 2.
Example 1.
a.
b.
238
92U
230
90Th
234
4
90Th + 2He
226
4
88Ra + 2He
○ Beta Decay
When a nucleus undergoes beta decay it emits a beta particle (0-1β), which is an electron ejected from the
nucleus. But how can an electron be ejected from the nucleus where the nucleus does not have electrons? The
answer is, a neutron in the nucleus decays into a proton and an electron. The electron is ejected and the proton
remains in the nucleus. Detailed studies of the energetics of beta decay also shows that an additional particle
with no charge and no mass, is emitted. This particle is called antineutrino (υ).
1
0n
1
0
1p + -1β
+υ
Since the proton remains in the nucleus, beta decay results to an increase in atomic number by 1.
Example 2.
a. 146C
b. 23491Pa
14
0
7N + -1β + υ
234
0
92U + -1β +
υ
○ Gamma Decay
Gamma decay is the emission of high energy photons and tends to accompany other types of decay. When an
alpha and beta particles leave the nucleus, some energy levels in the nucleus are no longer occupied, making
the nucleus in an excited state. To return the nucleus to its ground state, it emits a photon. This photon takes
the form of a very high energy gamma radiation (00γ). A photon is a quantum of electromagnetic radiation.
Gamma decay accompanies the beta decay of most nuclei.
Example 3.
a.
14
6C
14
7N
+ 0-1β + υ + 00γ
○ Electron Capture
In electron capture, the nucleus captures an electron from the first (n=1) shell or energy level in the atom.
Because the first energy level is called the “K shell”, the electron capture is referred to as K capture. The result
of electron capture is that a proton in the nucleus is converted to a neutron and an additional particle is emitted.
This particle is called neutrino (ν).
1
1p
+ 0-1e
1
0n
+ ν
The electron capture causes the nuclear charge to decrease by one.
Example 4.
a.
b.
26
13Al
81
37Rb
+ 0-1e
+ 0-1e
26
12Mg + ν
81
36Kr + ν
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○ Positron Emission
A positron is a positively charged electron (01β). Positron and electron are identical in mass and spin but opposite
in charge. In positron decay, a proton decays into a neutron and a positron.
1
1p
1
0n
+ 01β + ν
Positron decay results to a decrease in nuclear charge by 1.
Example 5.
a. 158O
b. 116C
15
7N
11
5B
+ 01β + ν
+ 01 β + ν
Clicker Exercise 1
1. Complete the following equations and identify the mode of decay:
a. 4019K
___ + 0-1β + υ
40
b. 4019K + ___
18Ar + ν
232
4
c. ____
92U + 2He
d. 4019K
___ + 01β + ν
2. Complete the following nuclear equations:
9
6
4
a.
4Be + ____
3Li + 2He
24
4
b.
____ + 10n
11Na + 2He
40
1
c. 4020Ca + ____
19K + 1H
243
d. 24195Am + 42He
97 Bk + ____
2.2 Nuclear Stability
The stability of a particular nucleus depends on a variety of factors, and no single rule can predict whether a
particular nucleus is radioactive and how it will decay. Nuclear stability can be visualized through the patterns
in the chart of the nuclides (Figure 2.1). The chart of nuclides is a plot of the number of protons versus the
number of neutrons of all known nuclei.
The blue colored dots in the chart of nuclides are stable nuclides. All stable nuclides fall in a central region and
this area with blue color is often called the band of stability. The red colored dots are unstable nuclides and
this area with red color is referred to as the sea of instability. At low atomic numbers, the stable nuclides lie
along the line with N and Z approximately equal. But, beginning with Z = 20, the band of stability begins to
deviate from the line. Eventually, when Z>83 all nuclides are unstable.
Nuclides below or to the right of the band of stability tend to emit beta particles to gain stability. While, nuclides
above or to the left of the band of stability undergo positron emission or electron capture to increase stability.
Heavier nuclei tend to emit alpha particles, decaying successively, and often with beta particle emissions as well,
until a stable nucleus is formed.
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Figure 2.1 Chart of nuclides
(Source: Brown,WH & Holme, TA. (2011). Chemistry for Engineering Students. 2nd edition)
2.3 Kinetics of Radioactive Decay
Each radioactive decay produces a high energy particle or photon, which allows us to count the decays in a given
time period. The rate at which a sample decays or disintegrate is called the activity of the sample. For a sample
of N nuclei, the rate of disintegration is ΔN/Δt, where N is the number of nuclei and t is time. The SI unit of
nuclear activity is Becquerel (Bq), defined as one disintegration per second. Another unit of nuclear activity is
Curie (Ci) defined as the number of disintegrations per second in 1 gram of radium-226. One Ci is equal to 3.7 x
1010 Bq.
Kinetic Equation of Radioactive Decay
The rate at which a sample decay or disintegrate is a first order reaction. The activity decreases exponentially
with time (t) as shown in Figure 2.2. Since the activity is proportional to the number of nuclei present, N also
decreases exponentially.
The equation for the curve in Figure 2.2 is: Nt = N0 e-kt
where: No is the initial number of nuclei or initial activity
Nt is the number of nuclei at time t or activity at time t
k is the decay constant
t is time
Taking the natural logarithm of each side of the equation and rearranging the equation
Nt = N0 e-kt
ln Nt = ln N0 – kt
ln Nt - ln N0 = - kt
(ln Nt - ln N0 = - kt) * -1
ln N0 – ln Nt = kt
ln N0/ Nt = kt
(kinetic equation for radioactive decay)
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Figure 2.2 Radioactive decay following first order kinetics
(Source: Brown,WH & Holme, TA. (2011). Chemistry for Engineering Students. 2nd edition)
Half-life (t1/2)
Half- life (t1/2) is the time it takes for N0 to decay or disintegrate to one half its original value (Nt = ½ * N0).
From the kinetic equation:
ln N0/ Nt = kt
Substitute Nt with ½* N0 @ t1/2 in the kinetic equation
ln N0 / (1/2 *N0) = k t1/2
ln N0 / (N0/2) = k t1/2
ln 2 = k t1/2
t1/2 = ln 2 / k
t1/2 = 0.693/k
Example 6. The half-life of carbon-14 used in radiocarbon dating is 5730 years. What is the decay constant for
carbon- 14?
Given:
Required:
t1/2 = 5730 years
k
Solution:
Solve for the decay constant (k) using the formula of t1/2: t1/2 = 0.693/k
t1/2 = 0.693/k
k = 0.693/ t1/2
k = 0.693/ 5730 yr
k = 1.21 x 10-4 yr-1
Radiocarbon Dating
The kinetic equation of radioactive decay is useful in determining the age of artifacts. Determining the age of
the artifacts is known as radiocarbon dating. Why carbon is used in dating the artifacts? As mentioned in section
2.1, carbon-14 is constantly formed through the interaction of cosmic rays and the atmosphere. This carbon-14
is incorporated into plants and animals, although carbon-14 decays, it is constantly replenished in living
organisms, so that the 14C/12C ratio in living organisms remains relatively constant over time. Once a plant or
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animal dies, the decay of 14C continues and the ratio of 14C/12C in the remains of plants and animals decreases
as time passes. So, when a plant or animal dies, the 14C/12C ratio becomes the initial point for decay because that
is when the 14C/12C ratio starts to decrease due to decay of 14C. If the current value of 14C/12C ratio is known, then,
by using the kinetic equation for radioactive decay, the age of the artifacts can be determined.
Example 7. A piece of cloth is discovered in a burial pit in the southwestern United States. A tiny sample of the
cloth is burned to CO2, and the 14C/12C ratio is 0.250 times the ratio in today’s atmosphere. How old is the cloth?
Given:
N0/N = 1/0.250
k = 1.21 x 10-4 yr-1 (from example 6)
t
Required:
Solution:
Solve for the age of cloth or time (t) using the kinetic equation: ln N0/Nt = kt
ln N0/Nt = kt
t = (ln N0/Nt) / k
t = ln (1/0.250) / 1.21 x 10-4 yr-1
t = 1.386 / 1.21 x 10-4 yr-1
t = 11,454 yr
Example 8.
137
Cs has a half- life of 30.2 years. How many years will it take for 100 g sample to decay to 0.01 g?
Given: Nt = 0.01 g
N0 = 100 g
t1/2 = 30.2 yr
Required:
t
Solution:
a. Solve for the decay constant (k) using t1/2: t1/2 = 0.693/k
t1/2 = 0.693/k
k = 0.693 / t1/2
k = 0.693 / 30.2 yr
k = 0.023 yr-1
b. Solve for time (t) using the kinetic equation: ln N0/Nt = kt
ln N0/Nt = kt
t = ln (N0/Nt) / k
t = ln (100 g/0.01 g) / 0.023 yr-1
t = 9.21 / 0.023 yr-1
t = 400 yr
Example 9. Suppose an archeologist found a carbonized wheat grain in a fire pit at a dig site. Measurement of
the decay rate showed 0.070 Bq/g of carbon. How long ago was the wheat harvested?
Given:
Required:
Nt = 0.070 Bq/g of carbon
k = 1.21 x 10-4 yr-1 (from example 6)
t
18
Solution:
Solve for time (t) using the kinetic equation: ln N0/Nt = kt
(Take note that the activity of 14C in the atmosphere (and in all living things) today is 0.255 Bq/g of total carbon,
so No = 0.255 Bq/g of carbon)
ln N0/Nt = kt
t = ln (N0/Nt) / k
t = ln (0.255 /0.070) / 1.21 x 10-4 yr-1
t = 1.29 / 1.21 x 10-4 yr-1
t = 10,684 yr
Clicker Exercise 2
1. What is the half- life of a radioisotope if it decays to 12.5% of its radioactivity in 12 years?
2. A wooden artifact is burned and found to contain 21g of carbon. The
disintegrations/min. What is the age of the artifact?
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C activity of the sample is 105
2.4 Energetics of Nuclear Reactions
Nuclear technology is used in nuclear weapons and power plants because it can release tremendous amount of
energy. But, where this energy comes from?
Binding Energy
Binding energy is the energy released by the nucleus. The binding energy of a nucleus is the energy that would
be released if the nucleus were formed from a collection of free nucleons. The greater the binding energy, the
more stable the nucleus. The calculation of the binding energy is based on the Einstein’s famous equation, which
describes the interconversion of mass and energy: E = mc2
where: E is energy
m is mass
c is the speed of light equal to 2.998 x 108 m/s
The mass that is converted to energy is the mass defect (Δm). Mass defect is the mass difference between
calculated and observed masses of the atom.
Example 10. Calculate the binding energy of helium atom with an observed experimental mass of 4.002603
amu.
Given:
Required:
Experimental mass of He = 4.002603 amu
Binding energy (Eb)
Solution:
Helium atom has 2 protons and 2 neutrons
Masses of subatomic particles: p = 1.007825 amu
n = 1.008665 amu
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a. Solve for the calculated mass of helium atom (42He)
Mass of helium atom = mass of protons + mass of neutrons
Mass of helium atom = 2(1.007825) + 2(1.008665)
Mass of helium atom = 4.032980 amu
b. Solve for the mass defect (Δm)
Δm = calculated mass – experimental mass
Δm = 4.032980 amu – 4.002603 amu
Δm = 0.030377 amu
c. Convert the mass defect to binding energy using Einstein’s equation: E = mc2
Eb = Δm * c2
Eb = 0.030377 amu (2.998 x 108 m/s)2
Eb = 2.7303 x 1015 amu m2/s2
Convert amu to kg: 1 amu = 1.66054 x 10-27 kg
Eb = 2.7303 x 1015 amu m2/s2 * 1.66054 x 10-27 kg/amu
Eb = 4.5335 x 10-12 kg m2 s-2
Convert kg m2 s-2 to Joules:
1 J = 1 kg m2 s-2
Eb = 4.5335 x 10-12 kg m2 s-2 * 1 J/ 1 kg m2 s-2
Eb = 4.5335 x 10-12 J
The binding energy for 1 mole of He:
Eb = 4.5335 x 10-12 J * 6.02214 x 1023 mol-1
Eb = 2.7301 x 1012 J/mol
(6.02214 x 1023 is Avogadro’s No.)
This binding energy of one mole of Helium is equivalent to the energy required to drive an average automobile
about 30 times around the Earth at the equator. This is much more energy that can be derived chemically than
by just burning a comparable amount of any conceivable fuel.
Energy from Nuclear Reactions
In power generation and nuclear weapons, the fission reactions of nuclides release tremendous amount of
energy. Fission reaction is splitting of heavy nucleus into two smaller ones. Fission reaction may be spontaneous
which means the large nucleus simply breaks into smaller nucleus or induced by neutron bombardment. The
fission of 235U, for example, is the heart of nuclear power industry. How much energy is released from the fission
of 235U? The calculation of energy in this case also uses the Einstein’s equation (E = mc2). The mass that is
converted to energy is the mass difference between the fissile nuclide and the resulting products.
Example 11. Calculate the energy released by a nucleus of uranium-235 if it splits into a barium-141 nucleus and
a krypton-92 nucleus according to the following equation:
235
92U
Given:
Required:
+ 10n
236
92U*
141
56Ba
+ 9236Kr + 3 10n
Particles in the nuclear equation are: Uranium-235, Barium-141, Krypton-91, and neutron
amount of energy released by fission of Uranium-235
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Solution:
The masses of the particles involved in the nuclear reaction are:
Particle
Mass (amu)
Uranium-235
235.0439231
Barium-141
141.9144064
Krypton-92
91.9261528
Neutron
1.0086649
a. Calculate the mass of reactants
Mass of reactants = mass of uranium-235 + mass of neutron
Mass of reactants = 235.0439231 amu + 1.0086649 amu
Mass of reactants = 236.0525880 amu
b. Calculate the mass of products
Mass of products = mass of barium-141 + mass of krypton-92 + mass of neutron
Mass of products = 141.9144064 amu + 91.9261528 amu + 3 * 1.0086649 amu
Mass of products = 235.8665539 amu
c. Calculate the mass defect
Δm = mass products – mass reactants
Δm = 235.8665539 amu – 236.0525880 amu
Δm = - 0.1860341 amu
d. Convert mass defect to energy
E = Δm c2
E = - 0.1860341 amu * * (2.998 x 108 m/s)2
E = - 1.672075 x 1016 amu m2 s-2 * 1.66053886 x 10-27 kg/amu
E = - 2.7765 x 10-11 kgm2 s-2 * 1 J/ 1 kg m2 s-2
E = - 2.7765 x 10-11 J
Converting to kJ/mol
E = - 2.7765 x 10-11 J * 1kJ/1000J * 6.023 x 1023 mol-1
E = -1.67 x 1010 kJ/mol
Notice that the nuclear equation for the fission of uranium-235 shows that bombardment the uranium-235 with
one neutron produces three neutrons. Each neutron product is capable of inducing another fission reaction. So,
the three neutron product can induce three fission reactions that will release nine neutrons to induce nine more
fissions from which 27 neutrons are obtained, and so on. The neutron induced fission of uranium-235 is
extremely rapid, this sequence of reactants can lead to an explosive chain reaction as illustrated in Figure 2.3.
.
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Figure 2.3 A self- propagating nuclear chain reaction initiated by capture of neutron
(Source: Brown,WH & Holme, TA. (2011). Chemistry for Engineering Students. 2nd edition)
In nuclear power plants, nuclear engineers control the fission reaction to run away explosive chain reaction. This
is done by limiting the number and energy of the neutrons available so that energy can be derived safely and
used as a heat source. This is done by inserting cadmium rods or other neutron absorbers to the nuclear reactor.
The rods absorb neutrons that would propagate fission reactions.
2.5 Fission, Transmutation and Fusion
Nuclear Fission
Fission is splitting of heavy nucleus into two smaller ones. Not all nuclei can undergo fission, but those that can
do are said to be fissionable or fissile. Some fission reactions are spontaneous which means the large nucleus
simply breaks into smaller nucleus. Other fission reactions are induced by neutron bombardment. In induced
fission, a neutron is absorbed by a fissile nucleus producing a highly unstable intermediate compound nucleus.
This compound nucleus breaks into two smaller nuclei, emitting neutrons in the process.
Example 12.
235
92U
+ 10n
236
92U*
141
56Ba
+ 9236Kr + 3 10n
compound nucleus
Nuclear Transmutation
Transmutation is changing one nucleus to another either by natural decay or in response to some outside
intervention like neutron bombardment. Transmutation reactions are used to produce a number of medically
useful radioisotopes. Like for example the radioactive 131I is often used to diagnose thyroid problems. The
patient is given an injection of a small amount of the radioisotope, and through natural biochemical pathways
carry the 131I to the thyroid. 131I undergoes beta decay, so detection of the radiation emanating from the thyroid
is used to produce an image of the gland.
Example 13.
a. 105B + 10n
b. 24094Pu + 10n
11
5B*
7
3Li
+
compound nucleus
241
94Pu*
compound nucleus
4
2He
241
95Am
+ 0-1e
22
Nuclear Fusion
Fusion is combining small nuclei to form larger and more stable nuclei. The energy of the sun originates in a
fusion reaction where four hydrogen nuclei combined to form helium nucleus:
4 11H
Example 14.
a. 21H + 21H
b. 11H + 32He
c. 21H + 31H
4
2He
+ 2 01β + 2ν + energy
3
2He
4
2He
4
2He
+ 10n
+ 0-1e
+ 10n
All the fusion reactions in example 14 generates tremendous amount of energy. Development of nuclear fusion
as commercial source of energy is appealing because hydrogen isotopes are available. Deuterium ( 21H) for
example is naturally occurring, its supply is practically unlimited. While, tritium (31H) can be produced from 6Li
4
3
through this reaction: 63Li + 10n
2He + 1H
Aside from the availability of hydrogen isotopes, fusion does not produce the high level radioactive wastes that
fission generates. However, the use of fusion to generate electricity is complicated by a number of factors and
these factors present enormous engineering challenges.
2.6 Interaction of Radiation and Matter
The effects of radiation on matter are governed by three factors. The first factor is the amount of radiation to
which matter is exposed. Higher doses of radiation have more serious effects. The second factor is ionizing
power of radiation and the third factor is penetrating power of radiation. The impact of radiation exposure
depends on the characteristics of the radiation which are ionizing power and penetrating power.
Ionizing and Penetrating Power of Radiation
Radiation can be classified as ionizing and non-ionizing; the distinction between ionizing and non-ionizing
radiation is based on the energy carried by the photon or particle. If the energy carried by the photon or particle
is greater than the ionization energy of typical atoms or molecules, then the radiation will induce ionization of
whatever material it encounters. If the photon or particle energies are smaller than the typical ionization energy,
then the radiation is non-ionizing. Non-ionizing radiation includes visible light, radio waves, and microwaves.
Ionizing radiation includes X-rays, gamma rays, and alpha and beta particles. Ionizing radiation is much more
likely to cause damage to any materials that it encounters including living tissues.
Looking into ionizing power alone cannot predict the impact of radiation on matter, the penetrating power must
also be considered. Penetrating power refers to how far a photon or particle penetrates into a material before
its energy is absorbed or dissipated. Alpha particles, for example, have high energies and therefore have high
ionizing power. But because alpha particles are relatively large, it does not tend to penetrate deeply into matter.
The energy of the alpha particles is dissipated in the skin. Alpha particles originating outside the body usually
do not cause serious harm because they do not penetrate sufficiently to reach internal organs. But alpha
particles produced inside the body is very dangerous because their energy will be deposited in internal organs.
Just like for example inhaling radon gas, radon is a radioactive material that once inhaled, the radon atom
undergoes decay in the lungs. The dissipation of energy of the alpha particles produced during radon decay can
cause serious tissue damage.
Beta particles have lower energy than alpha particles, but because beta particles are smaller than alpha
particles, they can penetrate several centimeters into the body. Beta radiation is often more dangerous than
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alpha radiation because of its penetrating power. Gamma rays can pass entirely through the body, and their
interactions with atoms and molecules in vital organs can do great damage.
Both ionizing and penetrating power are important in assessing possible health effects of radiation exposure.
Measuring Radiation Dose
There are different ways to express radiation because of the interplay between ionizing and penetrating power.
One is exposure; this measures the number of ions produced in air. Second is absorbed dose; this measures the
amount of radiation actually absorbed by a particular material. Third is equivalent dose; this quantify the
resulting damage to human tissue. The quality factor (Q) is used in calculating equivalent dose. It varies from a
value of 1 for high energy photons (gamma and X-rays) to about 20 for alpha particles. The units used in
measuring dose are:
Expression of dose
a. Exposure
US
Roentgen (R)
1 R = 2.58 x 104 C/kg of dry air
SI
b. Absorbed dose
Radiation absorbed dose (rad)
1 rad = 1 erg/g
(1 erg = 10-2 J)
Gray (Gy)
1 Gy = 100 rads
c. Equivalent dose
Roentgen equivalent man (rem)
1 rem= Q*absorbed dose in rad
Sievert (Sv)
1 Sv = 100 rem
1 Sv = Q*absorbed dose in Gy
Clicker Exercise 3
1. Compute the binding energy of the 7Li nucleus; the experimentally determined mass of 7Li is 7.016004 amu.
2. How much energy is released in the fission of 1 kg of 235U according to the equation below? The experimentally
determine masses of the products are 136.92532 amu (Te-137) and 96.91095 amu (Zr-97).
235
92U
+ 10n
137
52Te
+ 9740Zr + 2 10n
3. An average person is exposed to about 360 millirem of background radiation a year from a variety of sources.
Two-thirds of that comes from inhalation of 222Ra produced in soil. Given that 222Ra decays by alpha emission,
estimate the absorbed dose in a) Joules and b) Grays for a 60 kg person from one-year inhalation of radon.
References
1. Brown, WH & Holme, TA. (2011). Chemistry for Engineering Students. 2nd edition, Brooks/Cole, Centage
Learning, USA.
2. Gaffney J & Marley N. (2017). General Chemistry for Engineers. 1st edition, Elsevier Publishing.
3. Moore, JW, Stanitski, CL & Jurs, PC. (2005). Chemistry: The Molecular Science. 2nd edition, Brooks/Cole,
Thomson learning, Canada.
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