Unformatted text preview: 12/19/21, 1:03 PM Nuclear Reactions | Boundless Chemistry Boundless Chemistry
Nuclear Chemistry Nuclear Reactions 1/10 12/19/21, 1:03 PM Nuclear Reactions | Boundless Chemistry Balancing Nuclear Equations
To balance a nuclear equation, the mass number and atomic numbers of
all particles on either side of the arrow must be equal. LEARNING OBJECTIVES Produce a balanced nuclear equation KEY TAKEAWAYS Key Points A balanced nuclear equation is one where the sum of
the mass numbers (the top number in notation) and the
sum of the atomic numbers balance on either side of an
equation.
Nuclear equation problems will often be given such that
one particle is missing.
Instead of using the full equations, in many situations a
compact notation is used to describe nuclear reactions.
Key Terms baryon: A heavy subatomic particle created by the binding of quarks by gluons; a hadron containing three
quarks. They have half-odd integral spin and are thus
fermions. Nuclear reactions may be shown in a form similar to chemical equations,
for which invariant mass, which is the mass not considering the mass de- 2/10 12/19/21, 1:03 PM Nuclear Reactions | Boundless Chemistry fect, must balance for each side of the equation. The transformations of
particles must follow certain conservation laws, such as conservation of
charge and baryon number, which is the total atomic mass number. An
example of this notation follows:
6
3 2 4 1 2 Li + H → He + ? To balance the equation above for mass, charge, and mass number, the
second nucleus on the right side must have atomic number 2 and mass
number 4; it is therefore also helium-4. The complete equation therefore
reads:
6
3 2 4 4 1 2 2 Li + H → He + He Or, more simply:
6
3 2 4 1 2 Li + H → 2 He Compact Notation of
Radioactive Decay
Instead of using the full equations
in the style above, in many situations a compact notation is used
to describe nuclear reactions.
This style is of the form A(b,c)D,
which is equivalent to A + b gives
c + D. Common light particles are
often abbreviated in this shorthand, typically p for proton, n for
neutron, d for deuteron, α representing an alpha particle or helium-4, β for beta particle or electron, γ for gamma photon, etc.
The reaction in our example Lithium-6 plus deuterium gives two
helium-4s.:
The
visual
representation of the equation we
used as an example. above would be written as Li6(d,α)α. 3/10 12/19/21, 1:03 PM Nuclear Reactions | Boundless Chemistry Balancing a Radioactive Decay Equation
In balancing a nuclear equation, it is important to remember that the sum
of all the mass numbers and atomic numbers, given on the upper left
and lower left side of the element symbol, respectively, must be equal
for both sides of the equation. In addition, problems will also often be
given as word problems, so it is useful to know the various names of radioactively emitted particles. EXAMPLE 235
92 U → 231
90 Th + ? This could be written out as uranium-235 gives thorium-231
plus what? In order to solve, we find the difference between
the atomic masses and atomic numbers in the reactant and
product. The result is an atomic mass difference of 4 and an
atomic number difference of 2. This fits the description of an
alpha particle. Thus, we arrive at our answer:
235
92 U → 231
90 4 Th + He
2 EXAMPLE 214
84 4 0 2 −1 Po + 2 He + 2 e → ? This could also be written out as polonium-214, plus two alpha
particles, plus two electrons, give what? In order to solve this
equation, we simply add the mass numbers, 214 for polonium,
plus 8 (two times four) for helium (two alpha particles), plus
zero for the electrons, to give a mass number of 222. For the
atomic number, we take 84 for polonium, add 4 (two times two)
for helium, then subtract two (two times -1) for two electrons
lost through beta emission, to give 86; this is the atomic number for radon (Rn). Therefore, the equation should read: 4/10 12/19/21, 1:03 PM 214
84 Nuclear Reactions | Boundless Chemistry 4 Po + 22 He + 2 0
−1 e → 222
86 Rn Writing Nucl…
Nucl… Writing nuclear equations: Describes how to write the nuclear equa- tions for alpha and beta decay. Nuclear Binding Energy and Mass Defect
A nucleus weighs less than its sum of nucleons, a quantity known as the
mass defect, caused by release of energy when the nucleus formed. LEARNING OBJECTIVES Calculate the mass defect and nuclear binding energy of an
atom KEY TAKEAWAYS 5/10 12/19/21, 1:03 PM Nuclear Reactions | Boundless Chemistry Key Points Nuclear binding energy is the energy required to split a
nucleus of an atom into its components.
Nuclear binding energy is used to determine whether
fission or fusion will be a favorable process.
The mass defect of a nucleus represents the mass of
the energy binding the nucleus, and is the difference
between the mass of a nucleus and the sum of the
masses of the nucleons of which it is composed.
Key Terms nucleon: One of the subatomic particles of the atomic nucleus, i.e. a proton or a neutron.
strong force: The nuclear force, a residual force re- sponsible for the interactions between nucleons, deriving from the color force.
mass defect: The difference between the calculated mass of the unbound system and the experimentally
measured mass of the nucleus. Binding Energy
Nuclear binding energy is the energy required to split a nucleus of an
atom into its component parts: protons and neutrons, or, collectively, the
nucleons. The binding energy of nuclei is always a positive number,
since all nuclei require net energy to separate them into individual protons and neutrons. Mass Defect
Nuclear binding energy accounts for a noticeable difference between
the actual mass of an atom’s nucleus and its expected mass based on
the sum of the masses of its non-bound components. 6/10 12/19/21, 1:03 PM Nuclear Reactions | Boundless Chemistry Recall that energy (E) and mass (m) are related by the equation: E = mc 2 Here, c is the speed of light. In the case of nuclei, the binding energy is
so great that it accounts for a significant amount of mass.
The actual mass is always less than the sum of the individual masses of
the constituent protons and neutrons because energy is removed when
when the nucleus is formed. This energy has mass, which is removed
from the total mass of the original particles. This mass, known as the
mass defect, is missing in the resulting nucleus and represents the energy released when the nucleus is formed.
Mass defect (Md) can be calculated as the difference between observed
atomic mass (mo) and that expected from the combined masses of its
protons (mp, each proton having a mass of 1.00728 amu) and neutrons
(mn, 1.00867 amu):
Md = (mn + mp ) − mo Nuclear Binding Energy
Once mass defect is known, nuclear binding energy can be calculated
by converting that mass to energy by using E=mc2. Mass must be in
units of kg.
Once this energy, which is a quantity of joules for one nucleus, is known,
it can be scaled into per-nucleon and per- mole quantities. To convert to
joules/mole, simply multiply by Avogadro’s number. To convert to joules
per nucleon, simply divide by the number of nucleons.
Nuclear binding energy can also apply to situations when the nucleus
splits into fragments composed of more than one nucleon; in these
cases, the binding energies for the fragments, as compared to the
whole, may be either positive or negative, depending on where the parent nucleus and the daughter fragments fall on the nuclear binding en- 7/10 12/19/21, 1:03 PM Nuclear Reactions | Boundless Chemistry ergy curve. If new binding energy is available when light nuclei fuse, or
when heavy nuclei split, either of these processes result in the release of
the binding energy. This energy—available as nuclear energy—can be
used to produce nuclear power or build nuclear weapons. When a large
nucleus splits into pieces, excess energy is emitted as photons, or
gamma rays, and as kinetic energy, as a number of different particles are
ejected.
Nuclear binding energy is also used to determine whether fission or fusion will be a favorable process. For elements lighter than iron-56, fusion
will release energy because the nuclear binding energy increases with
increasing mass. Elements heavier than iron-56 will generally release energy upon fission, as the lighter elements produced contain greater nuclear binding energy. As such, there is a peak at iron-56 on the nuclear
binding energy curve.
9 Average binding energy per nucleon (MeV) O 16
8 C 12
7
6
5 U 235
U 238 Fe 56 He 4 Li 7
Li 6 4
3 H3
He 3 2
2
1 H
1
0 H
0 30 60 90
120
150
180
Number of nucleons in nucleus 210 240 270 Nuclear binding energy curve: This graph shows the nuclear binding
energy (in MeV) per nucleon as a function of the number of nucleons in the
nucleus. Notice that iron-56 has the most binding energy per nucleon,
making it the most stable nucleus. The rationale for this peak in binding energy is the interplay between the
coulombic repulsion of the protons in the nucleus, because like charges
repel each other, and the strong nuclear force, or strong force. The 8/10 12/19/21, 1:03 PM Nuclear Reactions | Boundless Chemistry strong force is what holds protons and neutrons together at short distances. As the size of the nucleus increases, the strong nuclear force is
only felt between nucleons that are close together, while the coulombic
repulsion continues to be felt throughout the nucleus; this leads to instability and hence the radioactivity and fissile nature of the heavier
elements. EXAMPLE Calculate the average binding energy per mole of a U-235 isotope. Show your answer in kJ/mole.
First, you must calculate the mass defect. U-235 has 92 protons, 143 neutrons, and has an observed mass of 235.04393
amu.
Md = (mn + mp ) − mo Md = (92(1.00728 amu)+143(1.00867 amu)) – 235.04393 amu
Md = 1.86564 amu
Calculate the mass in kg:
1.86564 amu x 1 kg
6.02214×10 26 amu = 3.09797 x 10-27 kg Now calculate the energy:
E = mc2
E = 3.09797 x 10-27 kg x (2.99792458 x 108 m
s )2 E =2.7843 x 10-10 J
Now convert to kJ per mole:
2.7843 × 10 1011 −10 Joules
atom × 6.02×10 23 atoms mole × 1 kJ
1000 joules = 1.6762 x kJ
mole 9/10 12/19/21, 1:03 PM Nuclear Reactions | Boundless Chemistry Previous Next Privacy Policy 10/10 ...
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