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Calculate the binding energy per nucleon of , the particle.
Strategy
To find , we first find BE using the Equation and then divide by . This is straightforward once we have looked up the appropriate atomic masses in Appendix A.
Solution
The binding energy for a nucleus is given by the equation
For , we have ; thus,
Appendix A gives these masses as , , and . Thus,
Noting that , we find
Since , we see that is this number divided by 4, or
Discussion
This is a large binding energy per nucleon compared with those for other low-mass nuclei, which have . This indicates that is tightly bound compared with its neighbors on the chart of the nuclides. You can see the spike representing this value of for on the graph in [link] . This is why is stable. Since is tightly bound, it has less mass than other nuclei and, therefore, cannot spontaneously decay into them. The large binding energy also helps to explain why some nuclei undergo decay. Smaller mass in the decay products can mean energy release, and such decays can be spontaneous. Further, it can happen that two protons and two neutrons in a nucleus can randomly find themselves together, experience the exceptionally large nuclear force that binds this combination, and act as a unit within the nucleus, at least for a while. In some cases, the escapes, and decay has then taken place.
There is more to be learned from nuclear binding energies. The general trend in is fundamental to energy production in stars, and to fusion and fission energy sources on Earth, for example. The abundance of elements on Earth, in stars, and in the universe as a whole is related to the binding energy of nuclei and has implications for the continued expansion of the universe.
Why is the number of neutrons greater than the number of protons in stable nuclei having greater than about 40, and why is this effect more pronounced for the heaviest nuclei?
is a loosely bound isotope of hydrogen. Called deuterium or heavy hydrogen, it is stable but relatively rare—it is 0.015% of natural hydrogen. Note that deuterium has , which should tend to make it more tightly bound, but both are odd numbers. Calculate , the binding energy per nucleon, for and compare it with the approximate value obtained from the graph in [link] .
1.112 MeV, consistent with graph
is among the most tightly bound of all nuclides. It is more than 90% of natural iron. Note that has even numbers of both protons and neutrons. Calculate , the binding energy per nucleon, for and compare it with the approximate value obtained from the graph in [link] .
is the heaviest stable nuclide, and its is low compared with medium-mass nuclides. Calculate , the binding energy per nucleon, for and compare it with the approximate value obtained from the graph in [link] .
7.848 MeV, consistent with graph
(a) Calculate for , the rarer of the two most common uranium isotopes. (b) Calculate for . (Most of uranium is .) Note that has even numbers of both protons and neutrons. Is the of significantly different from that of ?
(a) Calculate for . Stable and relatively tightly bound, this nuclide is most of natural carbon. (b) Calculate for . Is the difference in between and significant? One is stable and common, and the other is unstable and rare.
(a) 7.680 MeV, consistent with graph
(b) 7.520 MeV, consistent with graph. Not significantly different from value for , but sufficiently lower to allow decay into another nuclide that is more tightly bound.
The fact that is greatest for near 60 implies that the range of the nuclear force is about the diameter of such nuclides. (a) Calculate the diameter of an nucleus. (b) Compare for and . The first is one of the most tightly bound nuclides, while the second is larger and less tightly bound.
The purpose of this problem is to show in three ways that the binding energy of the electron in a hydrogen atom is negligible compared with the masses of the proton and electron. (a) Calculate the mass equivalent in u of the 13.6-eV binding energy of an electron in a hydrogen atom, and compare this with the mass of the hydrogen atom obtained from Appendix A. (b) Subtract the mass of the proton given in "Substructure of the Nucleus" from the mass of the hydrogen atom given in Appendix A. You will find the difference is equal to the electron’s mass to three digits, implying the binding energy is small in comparison. (c) Take the ratio of the binding energy of the electron (13.6 eV) to the energy equivalent of the electron’s mass (0.511 MeV). (d) Discuss how your answers confirm the stated purpose of this problem.
(a) vs. 1.007825 u for
(b) 0.000549 u
(c)
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