Video Transcript
In this video, we’ll take a look
inside the nucleus of an atom. We’ll learn how the nuclear force
and the neutron–proton ratio affect nuclear stability. We’ll also calculate the binding
energy for an atomic nucleus.
This is an atom of helium. There are two electrons in the
electron cloud, and the nucleus contains two protons and two neutrons. Like charges repel each other. So, why don’t the two positively
charged protons in the helium nucleus repel each other? We know the protons in the nucleus
don’t repel each other. In fact, the particles in the
nucleus are tightly bound. So, what holds these particles
together? To understand that, we need to zoom
in further and take a look at the particles that make up the proton.
These particles are called
quarks. Quarks are held together into
larger particles, like the proton and neutron, by the strong force. The strong force is one of the four
fundamental forces of nature. Another fundamental force is
gravity, which is responsible for the attraction between two objects with
masses. The electromagnetic force is
responsible for the repulsion between protons. The final fundamental force is the
weak force, which is responsible for the process of nuclear decay. As the name suggests, the strong
force is the strongest of the fundamental forces over short distances. The strong force is so strong that
it results in a force of attraction between the particles in the nucleus. This force is often called the
nuclear force.
The nuclear force occurs between
all pairings of protons and neutrons: protons and neutrons, protons and protons, and
neutrons and neutrons. This force is so strong that it’s
able to overcome the repulsion between the protons in the nucleus, which keeps the
nucleus together. But the nuclear force is only this
strong under very short distances. At distances of about one to 2.5
centimeters, the nuclear force is strong enough to overcome the repulsions between
protons. At distances larger than 2.5
centimeters, the nuclear force isn’t strong enough to overcome the repulsions.
Now we know that protons and
neutrons are held together by the nuclear force. This means breaking the nucleus up
into protons and neutrons would be working against the nuclear force, which would
require energy. This energy is called the binding
energy. If we went in the opposite
direction and formed an atomic nucleus from protons and neutrons, energy would be
released that has the same magnitude as the binding energy. This energy that’s released when a
nucleus is formed has an interesting impact on the mass of the nucleus. To investigate this, let’s continue
to use the helium atom as an example.
The mass of the helium nucleus has
been measured to be 4.00153 unified atomic mass units. We can also calculate the mass of
the helium nucleus using the mass of the proton and the mass of the neutron. The helium nucleus has two protons
and two neutrons, which gives a total mass of 4.03188 unified atomic mass units for
the calculated mass of the helium nucleus. We can see that the measured mass
of the nucleus and the calculated mass of the nucleus are not the same. The calculated mass is larger. In fact, the calculated mass is
larger than the mass measured experimentally by 0.03035 unified atomic mass
units. This difference in mass is called
the mass defect.
The mass defect is related to the
energy that’s released when the nucleus is formed. To understand why, we need to turn
to Einstein’s famous equation, 𝐸 equals 𝑚𝑐 squared. This equation tells us that energy
and mass are proportional to one another. When energy is released due to the
formation of the atomic nucleus, the mass of the nucleus decreases by a proportional
amount. We can use this information to
calculate the binding energy for helium. The binding energy is the energy
required to disassemble the atomic nucleus. It has the same magnitude as the
energy released when the nucleus is formed.
So, if we plug in the mass defect
for 𝑚 in Einstein’s equation and solve for 𝐸, the quantity of energy we calculate
will be the binding energy. Before we plug in any values, we
should think about units. The energy in this equation is in
units of joules. Joules are defined as units of
kilograms times meters squared per second squared. The speed of light has units of
meters per second, so the speed of light squared has units of meters squared per
second squared. So, the mass we plug into this
equation should be in units of kilograms. So, we need to convert the mass
defect from unified atomic mass units into kilograms.
One unified atomic mass unit is
equal to 1.66 times 10 to the negative 27 kilograms. So, we can convert from unified
atomic mass units into kilograms if we multiply by this value. This gives us 5.0381 times 10 to
the negative 29 for the mass defect in kilograms. Now that we have this value, we can
use Einstein’s equation to solve for the binding energy. We can plug in the mass defect and
the speed of light, which has a value of three times 10 to the eighth meters per
second. If we perform the calculation,
we’ll get 4.53429 times 10 to the negative 12 joules.
We can leave our answer how it is,
but it is common to use units of electron volts or mega-electron volts to express
tiny quantities of energy involved in atomic processes. One electron volt is equal to 1.602
times 10 to the negative 19 joules. So, we can convert our answer from
joules to electron volts if we divide by this value. This gives us 28,303,932 electron
volts. There are one million electron
volts in one mega-electron volt. We can also write one million as 10
raised to the power of six. So, we can convert from electron
volts to mega-electron volts if we divide by 10 to the sixth. This gives us about 28.3039
mega-electron volts, which we’ll round to 28.3. The binding energy of helium is
28.3 mega-electron volts.
The binding energy is a very large
quantity of energy for an atomic process. The ionization energy for hydrogen
is only 13.6 electron volts. That means it only takes 13.6
electron volts of energy to remove the electron from a hydrogen atom. The binding energy of helium is
about two million times larger than the ionization energy of hydrogen. This shows how much stronger the
nuclear force is than the electrostatic force. It takes a lot of energy to
overcome the attractions between the particles in the nucleus that are caused by the
nuclear force. It doesn’t take nearly as much
energy to overcome the attraction between the positively charged proton and the
negatively charged electron that’s caused by the electrostatic force.
In this video, we’ve referred quite
a bit to helium four, the isotope of helium, with two neutrons. There are other isotopes of helium,
like helium two and helium five. Helium four is the most common
stable isotope of helium. Helium two and helium five are both
radioactive. They are unstable and will decay
over time. The only difference between these
are the number of neutrons, so the number of neutrons must play some role in nuclear
stability. If there are more neutrons in the
nucleus, that means there will be more attractions caused by the nuclear force to
keep the nucleus stable. This explains why helium two is
unstable. There aren’t enough neutrons in the
nucleus.
Too few neutrons in the nucleus
mean there aren’t enough attractions to overcome the repulsions between the
protons. Helium five is unstable because
there are too many neutrons in the nucleus. Remember that the nuclear force is
very short range. If there are too many neutrons, the
nucleus will be too large to be held together by the nuclear force. If we plot the number of neutrons
versus the number of protons for all of the stable isotopes, we can start to see
some trends in nuclear stability. Lighter elements are stable with a
neutron–proton ratio of one to one. This is what we saw with helium
four, which had two neutrons and two protons.
Carbon 12 also has a one-to-one
neutron–proton ratio with six protons and six neutrons. Nitrogen 14 has a one-to-one
neutron–proton ratio, as does oxygen 16. Isotopes of heavy elements tend to
be stable with a neutron–proton ratio of 1.5 to one. For example, lead 207, one of the
stable isotopes of lead, has 125 neutrons and 82 protons. This corresponds to a
neutron–proton ratio of 1.52 to one. So, as we can see, the ratio of
neutrons to protons can help us predict nuclear stability. Lighter elements tend to be stable
with a neutron–proton ratio of one to one. Heavier elements need more neutrons
to stabilize the nucleus, so heavier elements tend to be stable with a
neutron–proton ratio of 1.5 to one.
This trend occurs because heavier
elements have more protons in the nucleus. In order to overcome the repulsions
between those protons, there must be more neutrons providing more attractions from
the nuclear force. These stable isotopes are referred
to as the band of stability. Any isotopes with a neutron–proton
ratio outside of the band of stability will be radioactive. These isotopes are unstable and
will decay over time into an isotope that is on the band of stability. We can use the information on this
graph to help us predict the kind of nuclear decay an isotope will undergo.
For example, this isotope has too
many neutrons to be stable. So, this isotope will undergo some
kind of decay that decreases the number of neutrons. 𝛽 minus decay decreases the number
of neutrons, so it’s likely that this isotope will undergo 𝛽 minus decay to become
stable. With that, we’ve reached the end of
what we need to learn about the atomic nucleus. We’ve learned about the nuclear
force and how it holds together the nucleus. We learned how to calculate the
binding energy and how to predict nuclear stability using the neutron–proton
ratio. Before we conclude this video with
the key points, let’s work a problem.
Which of the following relations is
correct? (A) To convert an amount of energy
from joules to mega-electron volts, we divide it by 1.6 times 10 to the negative
27. (B) To convert an amount of energy
from joules to mega-electron volts, we divide it by 1.6 times 10 to the negative
13. (C) To convert an amount of energy
from joules to mega-electron volts, we multiply it by 1.6 times 10 to the negative
27. (D) To convert an amount of energy
from joules to mega-electron volts, we multiply it by 1.6 times 10 to the negative
13.
An electron volt is the amount of
energy gained by an electron when it travels through a potential of one volt. This unit of energy is commonly
used for energy changes in atomic processes. For example, removing the electron
from a hydrogen atom requires 13.6 electron volts of energy. Breaking apart the helium nucleus
into protons and neutrons requires 28.3 mega-electron volts of energy. Mega-electron volts are the subject
of this question. We need to determine the
relationship between mega-electron volts and joules.
There are one million electron
volts in a mega-electron volt. We can also express one million as
10 to the sixth. So we can convert from
mega-electron volts to electron volts if we multiply by 10 raised to the power of
six. One electron volt is equivalent to
1.602 times 10 to the negative 19 joules. So we can convert from electron
volts to joules if we multiply by this number. The combination of these two
conversions gives us 1.602 times 10 to the negative 13 joules, which we’ll round to
one decimal place to match the answer choices. So one mega-electron volt is equal
to 1.6 times 10 to the negative 13 joules.
We can rule out answer choices (A)
and (C) as our conversion factor is not 1.6 times 10 to the negative 27. Now we need to know if we should
divide or multiply by this number. So if we had some amount of energy
in joules and we want to convert it into mega-electron volts, we would divide by our
conversion factor. We can see this way that the units
of joules cancel, leaving us in the correct units of mega-electron volts. So of the relations in this
problem, answer choice (B) was correct. To convert an amount of energy from
joules to mega-electron volts, we divide it by 1.6 times 10 to the negative 13.
Now let’s wrap this video up with
the key points. The nuclear force holds together
the protons and neutrons in an atomic nucleus. The binding energy is the energy
required to disassemble an atomic nucleus into the unbound protons and neutrons it’s
composed of. Neutrons stabilize the atomic
nucleus. Lighter elements are stable with a
neutron–proton ratio of one to one, while heavier elements are stable with a
neutron–proton ratio of 1.5 to one.
