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Phet explorations: nuclear fission

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Nuclear Fission

Section summary

  • The binding energy (BE) of a nucleus is the energy needed to separate it into individual protons and neutrons. In terms of atomic masses,
    BE = { [ Zm ( 1 H ) + Nm n ] m ( A X ) } c 2 ,
    where m 1 H size 12{m left ("" lSup { size 8{1} } H right )} {} is the mass of a hydrogen atom, m A X is the atomic mass of the nuclide, and m n is the mass of a neutron. Patterns in the binding energy per nucleon, BE / A , reveal details of the nuclear force. The larger the BE / A size 12{"BE"/A} {} , the more stable the nucleus.

Conceptual questions

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?

Problems&Exercises

2 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 size 12{Z=N} {} , which should tend to make it more tightly bound, but both are odd numbers. Calculate BE/ A , the binding energy per nucleon, for 2 H and compare it with the approximate value obtained from the graph in [link] .

1.112 MeV, consistent with graph

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

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

7.848 MeV, consistent with graph

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

(a) Calculate BE / A for 12 C . Stable and relatively tightly bound, this nuclide is most of natural carbon. (b) Calculate BE / A for 14 C . Is the difference in BE / A between 12 C and 14 C 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 12 C , but sufficiently lower to allow decay into another nuclide that is more tightly bound.

The fact that 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 = 60 nucleus. (b) Compare BE / A for 58 Ni and 90 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 [link] 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) 1 . 46 × 10 8 u vs. 1.007825 u for 1 H

(b) 0.000549 u

(c) 2 . 66 × 10 5 size 12{2 "." "66" times "10" rSup { size 8{ - 5} } } {}

Unreasonable Results

A particle physicist discovers a neutral particle with a mass of 2.02733 u that he assumes is two neutrons bound together. (a) Find the binding energy. (b) What is unreasonable about this result? (c) What assumptions are unreasonable or inconsistent?

(a) –9.315 MeV

(b) The negative binding energy implies an unbound system.

(c) This assumption that it is two bound neutrons is incorrect.

Practice Key Terms 2

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Source:  OpenStax, Physics 101. OpenStax CNX. Jan 07, 2013 Download for free at http://legacy.cnx.org/content/col11479/1.1
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