Video Transcript
The given plot shows the number of
protons and neutrons for all the stable nuclei known to exist. What name is given to the area of
the graph within which all stable nuclei are found? (A) Belt of elements, (B) magic
number, (C) valley of decay, (D) strong nuclear zone, or (E) band of stability.
The nucleus of an atom contains
protons and neutrons. If the nucleus contains too much
energy, too many protons, or too few neutrons, then it can become unstable. So the number of protons and the
number of neutrons relates to the stability of a nucleus, which is why in the plot
we’ve been given of stable nuclei the 𝑥- and 𝑦-axes are labeled with the number of
protons and the number of neutrons. More specifically, it’s the
neutron-to-proton or 𝑁 𝑍 ratio that directly relates to the stability of a
nucleus.
The smallest atomic nuclei are
stable when they have a neutron-to-proton number ratio of exactly or approximately
one. We can see this in the plot
given. As the blue line has a slope of
one, therefore it has a neutron-to-proton ratio of one. The smallest stable nuclei can be
found on or close to this line. Medium-sized atomic nuclei tend to
be stable when the 𝑁 𝑍 value is approximately equal to 1.25. As the neutron-to-proton number
ratio increases, the data points move closer to the axis labeled number of
neutrons. There is evidence for this in the
plot given. The data points in the middle are
above the blue line and closer to the axis labeled number of neutrons.
Stable large nuclei tend to have a
neutron-to-proton number ratio of approximately 1.5. We can see this in the plot
given. The data points are even further
away from the 𝑁 𝑍 equals one line. If we draw the line 𝑁 𝑍 equals
1.5, we can see that the data points for the large stable nuclei are closer to this
line than they are to the line 𝑁 𝑍 equals one. As the plot is of stable nuclei,
the answer to the question “What name is given to the area of the graph within which
all stable nuclei are found?” is (E) the band of stability.
The orange circle on the plot
represents the unstable isotope 138 55 Cs. How might this isotope decay to
form a more stable nucleus? (A) 𝛽 minus decay, (B) 𝛼 decay,
(C) electron capture, (D) 𝛽 plus decay, or (E) gamma emission.
This orange circle represents this
isotope. Cs is the chemical symbol for
cesium. This isotope has a total of 138
protons and neutrons. So we can call it cesium-138. As this isotope is unstable, it can
undergo radioactive decay. Radioactive decay is the
spontaneous emission of radiation from an unstable nucleus. There are different types of
radioactive decay. We have been given five different
types. Unstable atomic nuclei will tend to
undergo the type of radioactive decay that gives them the 𝑁 𝑍 ratio of a stable
atomic nucleus. So different nuclei will release
different types of radiation.
Let’s look at the options given in
turn, starting with (A) 𝛽 minus decay. During 𝛽 minus decay, a neutron is
converted to a proton. So this type of decay increases the
number of protons. It tends to occur for species that
have a lot of neutrons, so species that are neutron rich. Option (B) is 𝛼 decay. During 𝛼 decay, an 𝛼 particle,
which is identical to a helium nucleus, containing two protons and two neutrons, is
released. So the number of protons and
neutrons in the nucleus will decrease. This type of decay happens for the
nuclei of very heavy atoms, which are the ones that have a lot of protons and a lot
of neutrons.
Option (C) is electron capture, and
option (D) is 𝛽 plus decay. Both of these processes involve a
proton decaying into a neutron. But electron capture typically
occurs for heavier nuclei that are neutron deficient but do not have enough energy
to undergo 𝛽 plus decay. As these processes decrease the
number of protons and increase the number of neutrons, they occur for nuclei which
are proton rich.
Option (E) is gamma emission. Gamma rays are emitted from a
nucleus with excess energy. We only know about the number of
neutrons and protons, not about how much energy is within the nucleus. So we can rule out option (E) gamma
emission.
Let’s now look at the plot and see
which neutrons are neutron rich, proton rich, and come from heavy atoms. Nuclei from heavy atoms will have a
large number of neutrons and a large number of protons. So data points from this section of
the graph will represent nuclei from heavy atoms. If those nuclei are unstable, they
are most likely to undergo 𝛼 decay. Nuclei that are proton rich will be
closer to the 𝑥-axis, as the 𝑥-axis is the number of protons. So nuclei in this section of the
graph will undergo 𝛽 plus decay or electron capture if they’re unstable.
Neutron-rich nuclei will have data
points that are closer to the 𝑦-axis. So unstable nuclei in this section
of the graph are most likely to undergo 𝛽 minus decay. The orange circle is in the section
of the plot where the unstable nuclei are most likely to undergo 𝛽 minus decay. So the answer to the question “How
might this isotope decay to form a more stable nucleus?” is (A) 𝛽 minus decay.
But it’s worth noting that although
we used the plot to find out the answer to this question, we could have used the
periodic table. If we find cesium in the periodic
table, we see that it has an atomic number of 55, which means that it has 55
protons. It also has an atomic mass of
132.91. This mass is calculated using all
known isotopes and their relative abundances. It tells us the total number of
protons and neutrons for an average atom of cesium.
The isotope that we’ve been given
is cesium-138. As 138 is considerably larger than
132.9, then we can deduce that the unstable isotope we’ve been given is neutron
rich. Thus, this unstable isotope is most
likely to undergo 𝛽 minus decay to form a more stable nucleus.
