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Calculation of nuclear binding energy

Determine the binding energy for the nuclide 2 4 He in:

(a) joules per mole of nuclei

(b) joules per nucleus

(c) MeV per nucleus

Solution

The mass defect for a 2 4 He nucleus is 0.0305 amu, as shown previously. Determine the binding energy in joules per nuclide using the mass-energy equivalence equation. To accommodate the requested energy units, the mass defect must be expressed in kilograms (recall that 1 J = 1 kg m 2 /s 2 ).

(a) First, express the mass defect in g/mol. This is easily done considering the numerical equivalence of atomic mass (amu) and molar mass (g/mol) that results from the definitions of the amu and mole units (refer to the previous discussion in the chapter on atoms, molecules, and ions if needed). The mass defect is therefore 0.0305 g/mol. To accommodate the units of the other terms in the mass-energy equation, the mass must be expressed in kg, since 1 J = 1 kg m 2 /s 2 . Converting grams into kilograms yields a mass defect of 3.05 × 10 –5 kg/mol. Substituting this quantity into the mass-energy equivalence equation yields:

E = m c 2 = 3.05 × 10 −5 kg mol × ( 2.998 × 10 8 m s ) 2 = 2.74 × 10 12 kg m 2 s −2 mol −1 = 2.74 × 10 12 J mol −1 = 2.74 TJ mol −1

Note that this tremendous amount of energy is associated with the conversion of a very small amount of matter (about 30 mg, roughly the mass of typical drop of water).

(b) The binding energy for a single nucleus is computed from the molar binding energy using Avogadro’s number:

E = 2.74 × 10 12 J mol −1 × 1 mol 6.022 × 10 23 nuclei = 4.55 × 10 −12 J = 4.55 pJ

(c) Recall that 1 eV = 1.602 × 10 –19 J. Using the binding energy computed in part (b):

E = 4.55 × 10 −12 J × 1 eV 1.602 × 10 −19 J = 2.84 × 10 7 eV = 28.4 MeV

Check your learning

What is the binding energy for the nuclide 9 19 F (atomic mass: 18.9984 amu) in MeV per nucleus?

Answer:

148.4 MeV

Got questions? Get instant answers now!

Because the energy changes for breaking and forming bonds are so small compared to the energy changes for breaking or forming nuclei, the changes in mass during all ordinary chemical reactions are virtually undetectable. As described in the chapter on thermochemistry, the most energetic chemical reactions exhibit enthalpies on the order of thousands of kJ/mol, which is equivalent to mass differences in the nanogram range (10 –9 g). On the other hand, nuclear binding energies are typically on the order of billions of kJ/mol, corresponding to mass differences in the milligram range (10 –3 g).

Nuclear stability

A nucleus is stable if it cannot be transformed into another configuration without adding energy from the outside. Of the thousands of nuclides that exist, about 250 are stable. A plot of the number of neutrons versus the number of protons for stable nuclei reveals that the stable isotopes fall into a narrow band. This region is known as the band of stability    (also called the belt, zone, or valley of stability). The straight line in [link] represents nuclei that have a 1:1 ratio of protons to neutrons (n:p ratio). Note that the lighter stable nuclei, in general, have equal numbers of protons and neutrons. For example, nitrogen-14 has seven protons and seven neutrons. Heavier stable nuclei, however, have increasingly more neutrons than protons. For example: iron-56 has 30 neutrons and 26 protons, an n:p ratio of 1.15, whereas the stable nuclide lead-207 has 125 neutrons and 82 protons, an n:p ratio equal to 1.52. This is because larger nuclei have more proton-proton repulsions, and require larger numbers of neutrons to provide compensating strong forces to overcome these electrostatic repulsions and hold the nucleus together.

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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