Nuclear Chemistry

Learning Outcomes:
At the end of the lesson, the learner will be able to:
1. Determine the concept of nuclear reactions.
2. Explain the concept of nuclide of an atom.
3. Identify structures of isotopes.
4. Formulate, differentiate and balance nuclear equations.
5. Explain the nuclear stability.
6. Convert mass to energy from nuclear reactions
7. Explain the Radioactive Decay Law.
8. Solve problems on half-life reactions
9. Cite important uses of some radioactive nuclide.
10. Identify the industrial application of radioactive isotopes.
 NUCLEAR CHEMISTRY
▪ The subdiscipline of chemistry that is concerned with the changes in the nucleus
of elements. Hence, the changes in the nucleus of elements are the source of
radioactivity and nuclear power.
▪ Nuclear chemistry begins with the discovery of natural radioactivity by Antoine-
Henri Becquerel in 1896.
▪ In 1905 Albert Einstein proposed the theory of relativity, which states that mass
and energy are related by the equation E=mc2.And this grew as a result of
subsequent investigations by Pierre and Marie Curie and many others.
BASIC PARTS OF AN ATOM
NUCLEAR REACTIONS
▪ Nuclear reactions can be
radioactive decay. Nuclear
transmutation, nuclear fission and
nuclear fusion which are very
different from ordinary chemical
reactions.
▪ Involve changes in the composition
of nuclei
▪ Accompanied by the release of
tremendous amounts of energy
NUCLEAR RADIATION
COMPARISON OF CHEMICAL REACTIONS AND NUCLEAR REACTIONS
SUMMARY OF PROPERTIES OF NUCLEAR RADIATIONS
WRITING NUCLEAR REACTIONS
▪ Nuclear Reaction or transmutations involve a change in the
atomic mass and the mass number of the radioactive isotope.
Example:
𝟐𝟐𝟔
𝟖𝟖𝑹𝒂 → 𝟒𝟐𝑯𝒆 + 𝟐𝟐𝟐
𝟖𝟔𝑹𝒏

Mass number: 226 → 4 + 222

Atomic number: 88 → 2 + 86
BALANCING NUCLEAR REACTIONS
Nuclear reactions also follow conservation laws, and they are balanced in two ways:
1. Conserve mass number (A).
The total number of protons plus neutrons in the products and in the reactants
must be the same.
Example:
𝟐𝟑𝟓 𝟏 𝟏𝟑𝟖 𝟗𝟔 𝟏
𝟗𝟐 𝑼 + 𝟎 𝒏 → 𝟓𝟓 𝑪𝒔 + 𝟑𝟕 𝑹𝒃 + 𝟐 𝟎𝒏
235 + 1 = 138 + 96 + 2 x 1
2. Conserve atomic number (Z) or nuclear charge.
The total number of nuclear charges in the products and in the reactants must be
the same.
Example:
𝟐𝟑𝟓 𝟏 𝟏𝟑𝟖 𝟗𝟔 𝟏
𝟗𝟐 𝑼 + 𝟎 𝒏 → 𝟓𝟓 𝑪𝒔 + 𝟑𝟕 𝑹𝒃 + 𝟐 𝟎𝒏
92 + 0 = 55 + 37 + 2 x 0
SAMPLE PROBLEM:
Complete the given nuclear reaction and identify X.
𝟐𝟎𝟖 𝟒
𝟖𝟒 𝑷𝒐 → 𝑿 + 𝟐𝑯𝒆
Solving Process:
1. Find the mass number of the unknown product.
mass no of 4 + mass no of X = 208
mass no of X = 208 – mass no of 4
= 204
2. Find the atomic number of the unknown product. atomic no of 2 + atomic no of X =
84
atomic no of X = 84 – atomic no of 2
= 82
3. Determine the identity of the X (unknown product) and complete the nuclear
reaction. Turn to the periodic table and find which nuclide has an 82 atomic number.
This nuclide is lead. Hence, the completed nuclear reaction:
𝟐𝟎𝟖 𝟐𝟎𝟒 𝟒
𝟖𝟒𝑷𝒐 → 𝟖𝟐𝑷𝒃 + 𝟐𝑯𝒆
EXERCISE:
Identify the product X and balance the following nuclear equation.
1. 212
87𝐹𝑟 → 4
2𝐻𝑒 + 𝑿

2. 137
55𝐶𝑠 →
137
56𝐵𝑎 +𝑿
Nuclear reactions can be expressed in two ways: (a) long form or
complete and , (b) short form. Consider the following examples:
Long or complete form:
proton or p Alpha particle or α

26 1 23 4
12𝑀𝑔 + 1𝐻 → 11𝑁𝑎 + 2𝐻𝑒
parent nuclide projectile daughter nuclide ejected particle
or progeny

Short form:
Parent Nuclide ( projectile, ejected particle) daughter nuclide
26 23
12𝑀𝑔 (𝑝,α ) 11𝑁𝑎
NUCLEAR STABILITY
Radioisotope
▪ a naturally or artificially produced radioactive isotope.
▪ there are 264 stable nuclides found in nature.
▪ these nuclides fall within what is known as the band of stability.

https://www.qsstudy.com/physics/radioactive-decay
BAND OF STABILITY

▪ The figure shows the plot of neutrons vs

protons for various stable isotopes.
▪ The straight line represents the points at
which the neutron-to-proton ratio
equals 1.
▪ Stable nuclei are in the shaded are
represents the belt of stability.
▪ Most radioactive nuclei lie outside the
belt.
https://wps.prenhall.com/
FACTORS OF NUCLEAR STABILITY
1. Neutron-to-proton ratio (n/p)
▪ The stability of the nucleus is determined by the difference between
the repulsion and the short-range attraction.
▪ The nucleus disintegrates if repulsion outweighs attraction
▪ Larger excess of neutron is required to overcome the effect of forces
of repulsion
▪ n/p = 1 (stable)
▪n/p ≠ 1 (not stable)
❑ n/p > 1 undergo β-particle emission ( β- or −10𝑒 )
❑ n/p < 1 undergo positron emission ( β+ or +10𝑒 )
FACTORS OF NUCLEAR STABILITY
1. Neutron-to-proton ratio (n/p)
▪ Above the stability belt, the nuclei have higher neutron-to-proton
ratio. To lower this ratio, these nuclei undergo β-particle emission.
𝟏 𝟏 𝟎
𝟎 𝒏 → 𝟏 𝒑 + −𝟏𝜷

▪ Below the stability, the nuclei have lower neutron to proton ratio.
To increase this ratio, these nuclei emit a positron.
𝟏 𝟏 𝟎
𝟏 𝒑 → 𝟎 𝒏 + +𝟏𝜷
2. Pair Principle
▪ Nuclei with even of both protons and neutrons are generally stable
▪ A combination of an odd proton and even neutron or vice versa is less
stable
▪ There are more than 300 stable isotopes
3. Magic Numbers
▪ Nuclei that contains 2, 8, 20, 50, 82 and 126 protons or neutrons are
generally stable.
▪ The numbers 2, 8, 20, 50, 82 and 126 are called magic numbers.
▪ The significance of these numbers for nuclear stability is similar to
the number of electrons associated with the very stable noble gases.

4. Isotopes with atomic number greater than 83

▪ All isotopes of the elements with atomic number higher than 83 are
radioactive.
▪ All isotopes of technetium (Tc, Z = 43) and promethium (Pm, Z = 61)
are radioactive.
EXERCISE:
1. The following isotopes are unstable Use the Band Stability Graph to
predict whether they will undergo beta decay or positron emission.
(a.) 𝟏𝟑𝑩 (b.) 𝟏𝟖𝟖𝑨𝒖 .
Write a nuclear equation for each case.

2. Arrange the following in order of decreasing stability and explain.

35Cl, 4He, 22Na

3. Explain each of the following in terms of nuclear stability:

a. 212Po decays by alpha emission
b. 214Bi decays by beta emission
RADIOACTIVITY
▪ The emission of ionizing
radiation or particles caused
by the spontaneous
disintegration of atomic nuclei.
▪ Nuclides that are unstable are
known as radionuclides or said
to be radioactive.

https://www.env.go.jp/en/chemi/rhm/basic-info/
TYPES OF RADIOACTIVITY
▪ NATURAL
o radioactivity that exist in nature.
o Examples: uranium, actinium, thorium

▪ ARTIFICIAL
o radioactivity produced by man.
o Example: technetium, astatine, francium
 FOUR TYPES OF PROCESSES THAT INVOLVE
NUCLEAR TRANSFORMATION
1. RADIOACTIVE DECAY
▪ the spontaneous disintegration of a radionuclide accompanied by the
emission of ionizing radiation in the form of alpha or beta particles or
gamma rays.

www.chemwiki.ucdavis.edu
 TYPES OF RADIOACTIVE DECAY:
1. Alpha Decay (α)
▪ a type of radioactivity in which alpha particles are ejected. Alpha particle
contains 2 protons and 2 neutrons lightly bound together.
2. Beta Decay (β-)
▪ classified as beta decay if the atomic number of the nuclide changes but the
mass number does not change
▪ beta-minus decay, when the nucleus emits an electron and an antineutrino in a
process that changes a neutron to a proton.

https://upload.wikimedia.org/wikipedia/commons/8/82/Beta-decay.png
4. K-capture (electron capture)
▪ the nucleus may capture an orbiting electron, causing a proton to convert into
a neutron in a process called electron capture. All of these processes result in a
well-defined nuclear transmutation.
5. Gamma Decay (Ƴ)
▪ a radioactive nucleus first decays by the emission of an α or β particle.
The daughter nucleus that results is usually left in an excited state and
it can decay to a lower energy state by emitting a gamma ray photon.
SUMMARY:

https://opentextbc.ca/chemistry/chapter/21-3-radioactive-decay/
FOUR TYPES OF PROCESSES THAT INVOLVE NUCLEAR TRANSFORMATION
2. NUCLEAR TRANSMUTATION
▪ a process in which one nucleus is transformed into another through
bombardment by various sub-atomic particles or ions. Example:
14 4 17 1
7 𝑁 + 2 𝐻𝑒 → 8 𝑂 + 1𝐻
𝑂𝑟
14 4 17 1
7 𝑁 + 2 𝛼 → 8 𝑂 + 1𝑝

https://socratic.org/chemistry/nuclear-chemistry/nuclear-transmutation
3. NUCLEAR FISSION
▪ a process in which a heavy nucleus splits into lighter ones generating one or
more free neutrons. This process releases a huge amount of energy due to the
heavy nucleus which is less stable than its product

https://www.researchgate.net/figure/Nuclear-fission-of-Uranium-reproduced-from-reference-328-Creative-Commons-
Attribution_fig23_311807833
4. NUCLEAR FUSION
▪ a process in which two light nuclei combine/fuse together to form a heavier and
more stable nucleus. A significant amount of energy will be released on this
process.

www.chemwiki.ucdavis.edu
EXERCISE:
1. Complete the following nuclear equation:
a. Kyrpton-87 decays by beta emission

b. Uranium-232 decays by alpha emission

2. Write the nuclear equation for the reaction in which 221Fr decays
alpha emission
NUCLEAR BINDING ENERGY
▪ The quantitative measure of nuclear stability which is the energy required to
break up a nucleus into its component protons and neutrons.
▪ This quantity represents the conversion of mass energy to energy that occurs
during exothermic nuclear reaction.
▪ The energy required to separate a stable nucleus into its constituent protons
and neutrons.

http://thescienceweek.blogspot.com/2016/08/mass-defect-and-nuclear-binding-energy
MASS DEFECT
▪ The difference in mass of the sum of the individual masses of the
separated protons and neutrons to the mass of the stable nucleus.

Formulas:
E = 𝑚𝑐2 (Einstein’s mass-energy equivalence relationship)

∆E = (∆𝑚) 𝑐2 (Relationship between mass defect & energy released)

∆𝒎𝒂𝒔𝒔(𝒎𝒂𝒔𝒔 𝒅𝒆𝒇𝒆𝒄𝒕) = 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕 − 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒑𝒓𝒐𝒅𝒖𝒄𝒕

 Mass Defect and Energy in a Nuclear Fission
SAMPLE PROBLEM:
Calculate for the amount of energy produced when nitrogen undergoes nuclear fission
𝟏𝟒 𝟏 𝟏𝟒 𝟏
𝟕
N + 𝟎
n → C+ p
𝟔 𝟏
Mass reactants Mass products
14 14
7
N = 14.006700 amu 6
C = 12.010700 amu
1 1
0
n= 1.00865 amu 1
p = 1.007825 amu
15.015365 amu 13.018525 amu
∆𝑚𝑎𝑠𝑠(𝑚𝑎𝑠𝑠 𝑑𝑒𝑓𝑒𝑐𝑡) = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 − 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
∆𝒎𝒂𝒔𝒔(𝒎𝒂𝒔𝒔 𝒅𝒆𝒇𝒆𝒄𝒕) = 15.015365 amu – 13.018525 amu = 1.996840 amu
Conversion: 1 amu = 1.66054 x 10-27 Kg
∆𝑚𝑎𝑠𝑠 = 1.996840 amu x 1.66054 x 10-27 kg / 1 amu= 3.315832694 x 10-27 kg
∆𝑬 = (∆𝒎) 𝒄𝟐
= (3.315832694 x 10-27 kg) ((3.00 x 108 m/s)2) = 2.984249425 x 10-10 kg • m2/s2
Conversion: 1 Joule = 1 kg • m2/s2
∆𝑬 = 2.984249425 x 10-10 Joules
∆𝑬 / mole = 2.984249425 x 10-10 Joules x 6.022 x 1023 N-14 atom
N-14 atom 1 mol of N-14
∆𝑬 / mole = 1.797115003 x 1014 J/mol of N-14
This is a tremendously large quantity of energy produced since the enthalpies of ordinary chemical reactions are only 200
kJ/mole.
MASS DEFECT AND NUCLEAR BINDING ENERGY
SAMPLE PROBLEM:
Compute for the binding energy needed to break apart a nucleus of 237Np into its nucleons.
Mass Reactants
Protons: 93 x 1.007825 amu = 93.727725 amu
Neutrons: 144 x 1.008665 amu = 145.247760 amu
238.975485 amu
Mass Products
Mass 237Np = 237.0482 amu
∆𝑚𝑎𝑠𝑠(𝑚𝑎𝑠𝑠 𝑑𝑒𝑓𝑒𝑐𝑡) = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 − 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
∆𝒎𝒂𝒔𝒔 = 238.975485 𝑎𝑚𝑢 − 237.0482 𝑎𝑚𝑢 = 1.927285 amu
Conversion: 1 amu = 1.66054 x 10-27 kg
∆𝒎𝒂𝒔𝒔 = 1.927285 amu x 1.66054 x 10-27 kg / 1 amu = 3.2 x 10-27 kg
∆𝑬 = (∆𝒎) 𝒄𝟐
= (3.2 x 10-27 kg) ((3.00 x 108 m/s)2)
= 2.880 x 10-10 kg • m2/s2 = 2.880 x 10-10 J
Conversion: 1amu = 931.5 x 106 eV ; 1amu = 931.5 MeV
Δ E in MeV = 1.927285 amu x 931.5 MeV = 1795.27 MeV
1 amu
Δ E / nucleon = 1795.27 MeV / 237 nucleons = 7.575 MeV/ nucleon
This is the amount of energy needed to break the nucleus of Np per nucleon.
 Carbon- 14 dating was used to analyze the tissue samples
of Lyuba, preserved baby woolly mammoth found in
Siberia, the determined aged was over 40,000-year-old.

https://sites.google.com/site/internationalgcsephysic
s/section-7-radioactivity-and-particles/b-
radioactivity/4-half-lives-and-radio-dating https://www.sciencephoto.com/media/136309/view/lyuba-
preserved-woolly-mammoth
Exercises:
60
1. The radioactive nuclide 𝐶𝑜 has a half-life of 5.27 yrs. Calculate the mas of 60𝐶𝑜 that remains from a
27 27
0.0100 g sample of nuclide after 1.0 year has elapsed.
Given:
t1/2= 5.27 yrs
N0 = 0.0100 g of 6027Co
t = 1.0 year
Required to find: N at t = 1.0 yr
Solution:
i. t1/2 = 0.693 /k
k = 0.693 /5.27yrs = 0.132/yr
ii. log (N0/N) = k t /2.3
= [(0.132 / yr) (1 yr)]/ 2.3
log (N0/N) = 0.0574
N0/N = antilog 0.0574 = 1.14
N = N0 (1.14)
N = (0.0100 g 60
27Co) /(1.14)
N = 0.008771929825 g 𝟔𝟎 𝟐𝟕𝑪𝒐
Exercises:
103 103
2. The half-life of 𝑇𝑐 , a β emitter is 15.8 sec. How many atoms of 𝑇𝑐 are present in a sample with an
43 43
activity of 0.200 μCi?
Given: t1/2= 15.8 sec
Required: a) no. of atoms if A = 0.200 µCi
b) mass of sample in g
Solution:
a.) Solve for no. of atoms if A = 0.200 µCi
i. t1/2 =0.693/k
k = 0.693/15.8 sec
k = 0.0439/sec
ii. A is the activity of the sample in terms of the no.of atoms that disintegrate per second:
3.7 x 1010 atoms/sec 1 Ci
A = 0.200 µCi x x 106 µCi
1 Ci
A = 7.4 x 103 atoms/sec
iii. Since A = kN
N = A/k
= 7.4 x 103 atoms/sec / 0.0439/sec
𝟏𝟎𝟑
N = 1.686 x 105 atoms of 𝟒𝟑𝑻𝒄
Exercises:
103 103
2. The half-life of 𝑇𝑐 , a β emitter is 15.8 sec. How many atoms of 𝑇𝑐 are present in a sample with an
43 43
activity of 0.200 μCi?
Given: t1/2= 15.8 sec
Required: a) no. of atoms if A = 0.200 µCi
b) mass of sample in g
Solution:
b) Solve for mass of sample in grams
i. for 10343Tc: no. of protons = 43
no. of neutrons = 60
ii. mass of 1 mole 103 43Tc = 43(1.007825amu) + 60(1.008665amu)
103
mass of 1 mole 43Tc = 103.856 amu

103 1 mole 103

43Tc 103.856 g
iii. N = 1.686 x 105 atoms of 43Tc x x
6.022 x 1023 atoms 1 mole
𝟏𝟎𝟑
N = 2.908 x 10-17 g of 𝟒𝟑𝑻𝒄
Exercises:
3. A sample bone uncovered from the cave of Palawan has a specific activity of
6.33 disintegration/min. g of C. If the ratio of 12C : 14C in living organism
shows a specific activity of 15.38 d/min.g of C, find out the age of the bones
unearthed (t1/2 of C = 5730 years).
Given : ᾳo = 6.33 d/min.g t1/2 of C = 5730 years
ᾳt = 15.8 d/min.g
Required : age of the bone unearthed
Solutions:
ln 2 ln 2
i. k = 𝑡1/2
= 5730 years
k = 1.21 x 10-4/year
αt α0
ii. ln = −kt or ln = kt
α0 αt
α0
t = (1/k) x [ln ]
αt
d
15.8min.g
t = (1/1.21 x 10-4/yr) [ln d ]
6.33min.g

t = 7,559.58 years = 7,560 years

APPLICATION OF RADIOACTIVE NUCLIDES
▪ Food industry
▪ Manufacturing industries
▪ Medicine
▪ Chemistry
▪ Biology
▪ Geology
SOME RADIOACTIVE ISOTOPES USED IN MEDICINE
The biological effects of
radiation depends on:
▪ Energy of the radiation
▪ Penetrating ability of the
radiation
▪ Ionizing ability of the radiation
▪ Chemical properties of the
radiation source

https://emilms.fema.gov/IS3/FEMA_IS/is03/REM0203050.htm
RADIATION EXPOSURE CHART

https://www.env.go.jp/en/chemi/rhm/basic-info/1st/02-05-12.html
SOME RADIATION MONITORING INSTRUMENTS
Radiation Instrument
▪ Used to measure the
radiation exposure of
human and the
environment.

https://www.env.go.jp/en/chemi/rhm/basic-info/1st/02-04-
http://www- 01.html
naweb.iaea.org/nahu/DMRP/do
cuments/Chapter4.pdf
EXERCISE:
Direction: The copy of this activity will be given by your
chemistry instructor. Answer the activity sheet.
Background radiation: measuring your annual dose
https://www.iop.org/education/teacher/resources/radioactivity/file_41559.pdf
NUCLEAR WASTE CLASSIFICATION
▪ RADIOACTIVE WASTE CLASSIFICATION

EW
https://www-pub.iaea.org/MTCD/Publications/PDF/P1799_web.pdf
https://www.nrc.gov/waste/low-level-waste.html
 REFERENCES
▪ Chemistry for Engineering and Technologists Worktext 2019, Chemistry Department, COS, TUP-
Manila
▪ Chemistry 11th Edition by Chang & Goldsby 2013 @ Mc Graw-Hill Co., Inc.
▪ Chemistry & Chemical Reactivity by Kotz & Purcell 1987 @Saunders College Publishing
▪ Fundamentals of Chemistry by Redmore 1980 @ Prentice-Hall, Inc.
▪ Inorganic Chemistry 2nd Edition 200 by Housecroft and Sharpe @Pearson Education Limited
▪ Solving Problems in Chemistry by Smith & Himes 1984 @ Merrill Publishing Company
▪ The World of Chemistry by Joesten & Hogg 2012 @ Cengage Learning Asia Pte Ltd