Download Resource Lesson Nuclear Reaction When we speak of atoms, we

Resource Lesson
Nuclear Reaction
When we speak of atoms, we are speaking of collections of neutrons, protons, and electrons. Electrically,
atoms are naturally neutral - that is, there are as many electrons orbiting the nucleus of an atom as there are
protons within its nucleus.


The atomic number, Z, represents the number of protons in the atom and determines its chemical
properties.
The atomic mass number, A, represents the total number of protons and neutrons, collectively
known as nucleons.
The number of neutrons can be calculated by
subtracting the atomic number from the atomic mass
number.
Different isotopes of an element often have great
differences in nuclear stability. A nucleus is considered to
be stable if it remains together indefinitely. Stable isotopes
for light elements have the same number of protons as
neutrons; for heavy nuclei, stability arises when the
number of neutrons exceeds the number of protons. This
can be understood by noting that as the number of protons
increases, the strength of the Coulomb repulsive force
increases tending to break the nucleus apart. More
neutrons are needed to keep the nucleus stable since they
only experience attractive nuclear forces provided by the
strong nuclear force. When Z = 82, the repulsive force
between protons can no longer be buffered resulting in
elements that contain more than 82 protons having no
stable nuclei.
Refer to the following information for the next four questions.
Let's examine some isotopes of oxygen.
How many protons are in one atom of
17 O?
8
How many neutrons are in one atom of
17 O?
8
How many neutrons are in one atom of
18 O?
8
99.759% of commonly occurring oxygen atoms are 168O, while 0.037% are
17 O and 0.204% are 18 O. If the relative atomic masses for each isotope are
8
8
15.994915 amu, 16.999133 amu, and 17.999160 amu, then what is the
average atomic mass (in amu) of an oxygen atom?
Natural Radioactive Decay Modes
The three types of natural radioactive decay that can be emitted by a radioactive substance are: alpha radiation,
α, beta radiation, β, and gamma rays, γ.

α represents the nucleus of a helium atom. It carries a superscripted mass number of 4 and a subscripted
atomic number of 2. It has a charge of +2.
4 He = 4 α
2
2

β represents an electron, with a charge of -1, which has been released from the nucleus when a neutron
decays into a proton, an electron and an antineutrino by way of the weak force:
In formulas it can also appear as an e with a superscripted mass number of 0 and a subscripted atomic
number of -1
e-1 = 0-1e = 0-1β

γ represents radiation released when the nucleus settles to a lower energy level. It has short wavelengths
like x-rays [which are produced when excited electrons fall into vacancies in a ground state orbital of atoms
with high atomic numbers], is very energetic, and has no mass.
Balancing Nuclear Equations
Nuclear equations must be balanced by setting the superscripted mass numbers equal to each other on both sides
and by setting the subscripted atomic numbers equal to each other on both sides.
Single particles often found in these reactions are:



neutrons: 10n have a mass equal to 1.6750 x 10-27 kg
protons: 11H or 11p have a mass equal to 1.6726 x 10-27 kg
electrons: 0-1e have a mass equal to 9.11 x 10-31 kg
What is the value of X in the following equation?
What is the value of X in the following equation?
Refer to the following information for the next three questions.
Which reaction(s) represents a fusion reaction?
Which reaction(s) represents a fission-reaction?
Which reaction(s) represents a chain-reaction?
Refer to the following information for the next two questions.
There are three stable isotopes of silicon:
28
14Si
and 2914Si and 3014Si
Silicon-40, 4014Si, lasts for only a few seconds since it has too many neutrons for its 14 protons. One possible decay
product for Silicon-40, 4014Si, is calcium-40, 4020Ca.
Would this decay path most likely involve emission of alpha particles, beta particles, or gamma
particles?
What is the equation for this decay?
Detection Methods
These three particles can be distinguished by their levels of penetration and by passing them through magnetic
fields. In terms of penetration, alpha particles can be stopped by a sheet of paper, beta particles can make their
way through a few millimeters of aluminum, and gamma particles can penetrate several centimeters of lead
shielding.
The directions of curvature of charged particles passing through a magnetic field depends on the right-hand rule. In
this rule,



your fingers represent the external magnetic field,
your thumb represents a positive charge’s velocity, and
your palm represents the magnetic force exerted on the positively charged particle by the field.
Binding Energy
The formula used to calculate the amount of energy released during the complete transformation of mass into
energy is
ΔE = Δmc2
This formula can be used during radioactive decay to determine how much kinetic energy is present in either the
behavior of the reactants or the products. Customarily, atomic masses are stated in atomic mass units, or amu,
when given during nuclear reactions. Energies are also often given in electronvolts, eV, instead of Joules.
However, virtually all the formulas require standard SI units of kg and J instead. Subsequently, here are the
conversion factors to change amu to kg and eV to J:
1 amu = 1.6606 x 10-27 kg
1 eV = 1.6 x 10-19 J
Suppose you were asked to determine the energy released from the mass deficit in the reaction shown below.
where the atomic masses for the reactants and products are:
The first step in answering this question would be to determine the relative masses of the reactants and then the
products.
reactants
products
7
3Li
= 7.01601
4
2He
= 4.00260
1
1H
= 1.00718
4
2He
= 4.00260
8.02319 amu
8.00520 amu
The nuclear mass defect (deficit) is the difference in these two values, or 0.01799 amu. Since this additional mass
was present in the reactants, it will be released as energy with the two products. This energy can be calculated with
the equation E = Δmc2. Notice that the mass unit, amu, was first converted into kilograms before making this final
calculation.
E = Δmc2
E = 0.01799(1.66 x 10-27)(3 x 108)2
E = 2.59056 x 10-11 J, or
E = 16.8 MeV
The following graph shows a summary of when nuclear reactions "release energy." Note that both fusion and
fission reactions lower the mass per nucleon until the reaction reaches the element iron which has the greatest
binding energy per nucleon of 8.8 MeV.
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Name:_________________________________ Date:____________________ Period:_________
Balancing Nuclear Equations
1. Identify the missing particle in the following nuclear reaction:
214 Po
84
+ 242He + 20-1e → _____
2. Identify the missing particle in the following nuclear reaction:
1 n
0
+ 23592U → 2 10n + _____ + 13752Te
3. Identify the missing particle in the following nuclear reaction:
______ + 10n → 14256Ba +9136Kr + 310n
4. Identify the missing particle in the following nuclear reaction:
241 Am
95
+ 42He → 210n + _____
5. Identify the missing particle in the following nuclear reaction:
37 K
19
→ _____ + 0+1e
6. Identify the missing coefficient in the following nuclear reaction:
235 U
92
+ 10n → ____10n + 13956Ba + 9436Kr
7. Identify the missing particle in the following nuclear reaction:
239 U
92
+ 42He → _____ + 10n
8. Identify the missing particle in the following nuclear reaction:
239 Np
93
→ 23994Pu + _____
9. Identify the missing particle in the following nuclear reaction:
27 Al
13
+ _____ → 2411Na + 42He
10. Identify the missing particle in the following nuclear reaction:
99 Tc
43
→ _____ + 0-1e
11. Identify the missing particle in the following nuclear reaction:
9 Be
4
+ 11H → _____ + 42He
12. Identify the missing particle in the following nuclear reaction:
1 H
1
+ 31H → _____
13. Identify the missing coefficient in the following nuclear reaction:
+ 10n → ____10n + 13953I + 9539Y
235 U
92
14. Identify the missing particle in the following nuclear reaction:
6 Li
3
+ 10n → 0-1e + 42He + _____
15. Identify the missing particle in the following nuclear reaction:
226 Ra
88
→ 42He + _____
16. Identify the missing particle in the following nuclear reaction:
_____ → 42He + 20881Tl
17. Identify the missing particle in the following nuclear reaction:
27 Al
13
+ 42He → 3015P + _____
18. Identify the missing particle in the following nuclear reaction:
42 K
19
→ 0-1e + _____
19. Identify the missing particle in the following nuclear reaction:
3 He
2
+ 32He → 2 11H + _____
20. Identify the missing particle in the following nuclear reaction:
239 Pu
94
→ 42He + _____
n_ _5 half-lives
71.5 days
_14.3 days_/half-life
http://www.wwnorton.com/college/chemistry/gilbert/tutorials/interface.asp?chapter=chapter_02&folder=
modes_of_decay
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nuclear_reactions