10.8 Binding energy  (Page 3/5)

Why is the number of neutrons greater than the number of protons in stable nuclei having $A$ greater than about 40, and why is this effect more pronounced for the heaviest nuclei?

${}^2\text{H}$ 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 $Z=N$ , which should tend to make it more tightly bound, but both are odd numbers. Calculate $\mathrm{BE/}A$ , the binding energy per nucleon, for ${}^2\text{H}$ and compare it with the approximate value obtained from the graph in [link] .

${}^{\text{56}}\text{Fe}$ is among the most tightly bound of all nuclides. It is more than 90% of natural iron. Note that ${}^{\text{56}}\text{Fe}$ has even numbers of both protons and neutrons. Calculate $\mathrm{BE/}A$ , the binding energy per nucleon, for ${}^{\text{56}}\text{Fe}$ and compare it with the approximate value obtained from the graph in [link] .

${}^{\text{209}}\text{Bi}$ is the heaviest stable nuclide, and its $\text{BE}/A$ is low compared with medium-mass nuclides. Calculate $\mathrm{BE/}A$ , the binding energy per nucleon, for ${}^{\text{209}}\text{Bi}$ and compare it with the approximate value obtained from the graph in [link] .

(a) Calculate $\text{BE}/A$ for ${}^{\text{235}}\text{U}$ , the rarer of the two most common uranium isotopes. (b) Calculate $\text{BE}/A$ for ${}^{\text{238}}\text{U}$ . (Most of uranium is ${}^{\text{238}}\text{U}$ .) Note that ${}^{\text{238}}\text{U}$ has even numbers of both protons and neutrons. Is the $\text{BE}/A$ of ${}^{\text{238}}\text{U}$ significantly different from that of ${}^{\text{235}}\text{U}$ ?

(a) Calculate $\text{BE}/A$ for ${}^{\text{12}}\text{C}$ . Stable and relatively tightly bound, this nuclide is most of natural carbon. (b) Calculate $\text{BE}/A$ for ${}^{\text{14}}\text{C}$ . Is the difference in $\text{BE}/A$ between ${}^{\text{12}}\text{C}$ and ${}^{\text{14}}\text{C}$ significant? One is stable and common, and the other is unstable and rare.

The fact that $\text{BE}/A$ is greatest for $A$ near 60 implies that the range of the nuclear force is about the diameter of such nuclides. (a) Calculate the diameter of an $A=\text{60}$ nucleus. (b) Compare $\text{BE}/A$ for ${}^{\text{58}}\text{Ni}$ and ${}^{\text{90}}\text{Sr}$ . 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.