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Accelerated Chemistry: Nuclear Stability Guided Inquiry
Teacher’s Notes
Nuclear chemistry is the study of the atomic nucleus and nuclear reactions.
A nuclide is an atom consisting of 3 subatomic particles: electrons, protons and neutrons.
A nucleon is the nucleus of an atom consisting of neutrons and protons.
Keeping the Nucleus Together
Two of the four known forces in the universe are at work in atomic nuclei, electrostatic force and the nuclear
strong force. (The other two forces are gravitational force and the nuclear weak force.) It is important to
understand how electrostatic and nuclear forces work, in order to understand nuclear stability.
Electrostatic Forces - Like charges repel each other and opposite charges attract each other. This attraction and
repulsion is caused by electrostatic forces. They cause electrons and protons to be attracted to each other; they
cause protons to repel each other; and they cause electrons to repel each other. Electrostatic forces have no
effect on interactions between neutrons, between neutrons and electrons, and between neutrons and protons.
Electrostatic forces work over relatively long distances – many times greater than the diameter of the atomic
Nuclear Forces - Nuclear forces (aka strong forces) work in opposition to electrostatic forces. They are created by
the exchange of mesons between neutrons and protons. This exchange produces a force that attracts neutrons to
neutrons, protons to protons, and neutrons to protons. Nuclear forces are stronger than electrostatic forces, but
they work over very short distances - less than the diameter of the atomic nucleus. (Mesons are a subatomic
particle that is smaller than neutrons and protons. We will not be studying them.)
Mass Defect - Individual protons, neutrons and electrons have uniform masses:
proton 1.007 726 amu neutron 1.008 665 amu electron 0.000 5486 amu
When these subatomic particles combine to form atoms some of their mass is lost. In other words, the sum of the
masses of the subatomic particles that make up an atom do not equal the mass of the atom. This mass that is lost
during the creation of an atom is called the mass defect.
1. Look at the sample calculation for oxygen-16 in the first column below and then calculate the mass defect for
the other two isotopes of oxygen.
Nuclear Binding Energy – The mass lost when protons, neutrons and electrons are combined to create an atom is
converted into energy. The binding energy is the energy released when a nucleon is formed. It can also be
thought of as the amount of energy needed to break up a nucleon. Therefore, the higher the binding energy the
more stable the nucleon. Einstein’s famous equation, E = mc2, allows us to calculate the amount of energy
created by the lost mass.
where m is the mass defect in kg, c is the speed of light in m/s, and 1 amu = 1.6605 x 10-27 kg
Look at the binding energy calculation for oxygen-16,
2. Calculate the binding energy for the other two isotopes of oxygen and complete the table on the previous
3. Which of the 3 oxygen isotopes is most stable?
4. How does increasing the number of neutrons affect stability?
Adding neutrons increases the mass defect, which in turn increases the
binding energy and creates a more stable nucleus.
Nuclear Stability
The ratio of neutrons to protons plays a role in
the stability of a nucleus. While there is no exact
ratio that indicates a stable nucleus for all
elements, there is a general rule of thumb for
nuclear stability. As the figure indicates, stable
nuclei have a neutron to proton ratio of
anywhere from 1:1 for smaller nucleons up to
1.5:1 for larger nucleons. This shifting ratio
makes sense when we consider electrostatic and
nuclear forces. When protons are close to each
other (A & B in the figure below), like they are in
small nuclei, the strong, but short-range nuclear
forces are enough to overcome the electrostatic
forces and hold the protons together. There isn’t
much need for additional binding energy. As the
nucleus increases in size the protons at the
opposite sides of the nucleus (A & C in the figure
below) are too far apart to be held together by
nuclear forces; however the long-distance,
electrostatic forces still act to repel the protons.
Therefore, more nuclear binding energy is
needed to overcome the electrostatic repulsive
forces in larger nuclei.
We know from calculating the binding energies for the oxygen
isotopes, that adding more neutrons to a nucleus increases the
nuclear binding energy. Note that eventually the nucleus gets
too large and no number of additional neutrons can stabilize the
nucleus. Bismuth, which has an atomic number of 83, is the
largest stable nucleus. Nuclei above this atomic number, both
naturally occurring and man-made, are unstable.
Applying the Concepts
1. The mass of a neon-20 atom is 19.992 44 amu. Calculate the atom’s mass defect.
2. Calculate the binding energies of the following two nuclei, and indicate which nucleus releases more energy
when formed.
a. atomic mass 34.988011 amu, potassium-35
b. atomic mass 22.989767 amu, sodium-23
3. Calculate the neutron-proton ratios for the following nuclides:
a. carbon-12
b. lead-206
c. hydrogen-3
d. uranium-233
4. Locate the nuclides in #3 on the neutron-to-proton ratio graph in the notes.
a. Which ones lie within the band of stability? Carbon-12 and lead-206
b. For the stable nuclides, which ones tend toward 1:1? Which ones tend toward 1.5:1? Why?
The smaller stable nuclei tend toward the 1:1 ratio, while the larger nuclei
that approach an atomic number of 82 approach the 1.5: ratio.
Lighter nuclei don’t have as many protons and less electrostatic repulsive
force; therefore they don’t need as many neutrons to produce enough
nuclear force to overcome the electrostatic force. Heavier nuclei have more
protons and more electrostatic repulsive force; therefore they need more
neutrons to overcome the electrostatic force. Regardless of size, adding too
many neutrons will create unstable nuclei because the size of the nucleus
becomes too large for the relatively short range of the nuclear forces.
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