Activity 47 Nuclear Chemistry: Binding Energy

Why? Binding energies of nuclides determine the energy released in nuclear reactions and provide insights regarding the relative stability of the elements and their isotopes. The binding energy of a nuclide, Eb, is the energy required to separate the nuclide into its component protons and neutrons. The binding energy per nucleon, Eb/A, is a measure of the force holding a nuclide together. To obtain a value for the binding energy per nucleon, just divide the total binding energy by the number of nucleons in the nuclide.

What Do You Think?

Centuries ago, a major goal of alchemists was to turn lead into gold. Do you think this is possible? Explain why or why not.

Learning Objective

• Understand the origin of energy produced in nuclear reactions

Success Criteria

• Explain the source of energy in a nuclear reaction • Correctly calculate nuclear binding energies and the energy released in nuclear reactions.

Prerequisite

• Activity 46: Nuclear Chemistry: Radioactivity

Activity 47 —Nuclear Chemistry: Binding Energy 319 Procedure

Calculation of the Binding Energy per Nucleon Since the advent of Albert Einstein’s Theory of Special Relativity, mass and energy are considered to be different manifestations of the same quantity. The equation E = mc2, where c is the speed of light, provides the conversion between mass, m, and energy, E. In most situations, e.g., chemical reactions, energy changes are so small that corresponding changes in mass can not be observed. The binding energies of nuclides are very much larger, so it is useful to view the binding enertgies as changes in mass. Consequently the binding energy of a nuclide, Eb, and the binding energy per nucleon, Eb/A, can be calculated by the following procedure, where A is the mass number. Step (5) in this procedure is very important because to see how strongly one nucleon is held in the nuclide, it is necessary to divide the total binding energy of the nuclide by the number of nucleons comprising it.

(1) Eb = Eseparated nucleons – Enuclide 2 2 (2) Eb = mseparated nucleons c – mnuclide c

(3) ∆m = mseparated nucleons – mnuclide 2 (4) Eb = ∆mc

Eb (5) Eb/A = A

Analysis

How to calculate binding energies 1. Different books disagree over which isotope of iron is the most stable. a) Examine the Procedure above, and describe in words what you need to do to calculate the binding energy per nucleon for an isotope of iron.

b) Identify the concepts or ideas that you need to use in this calculation and explain why you need to use them.

320 Foundations of Chemistry 2. Calculate the binding energy per nucleon for iron-56 and iron-58. Place these points, along with the values you calculated, on the graph in the model, and identify the isotope that is more stable. For this calculation you need precise values for the mass of a proton, a neutron, and the iron isotopes. Remember that 1 J = 1 kg m2/s2, and note that the units on the graph are 108 kJ/mol. proton mass = 1.007276 g/mol, neutron mass = 1.008665 g/mol, Fe-56 mass = 55.934994 g/mol, Fe-58 mass = 57.933275 g/mol

Fe-56: mass of nucleons = (1.007276)(26) + (1.008665)(30) = 56.449126 g/mol given mass of nuclide = 55.934994 g/mol ∆m = 56.449126 - 55.934994 = 0.514132 g/mol -4 8 2 13 Eb = (5.14132 × 10 kg/mol)(2.99792 × 10 m/s) = 4.62077 × 10 J/mol  4.62077 × 1013 J mol-1  1 kJ  Eb/A =     56 1000 J  8 Eb/A = 8.25138 × 10 kJ/mol

For Fe-58 mass of nucleons = (1.007276)(26) + (1.008665)(32) = 58.466456 g/mol given mass of nuclide = 57.933275 g/mol ∆m = 58.466456 - 57.9332754 = 0.533181 g/mol -4 8 2 13 Eb = (5.33181 × 10 kg/mol)(2.99792 × 10 m/s) = 4.79198 × 10 J/mol  4.79198 × 1013 J mol-1  1 kJ  Eb/A =     58 1000 J  8 Eb/A = 8.26203 × 10 kJ/mol

Fe-58 has the larger binding energy per nucleon, which means that it is more stable than Fe-56. A point is included in Figure 47.1 between mass numbers 55 and 60. It is difficult to judge from the scales, but this point appears to be one of the iron isotopes. If we assume that it is iron-58, then the point for iron-56 would go to the left and very slightly below that point.

Activity 47 —Nuclear Chemistry: Binding Energy 321 Model: Binding Energies of Nucleons Figure 47.1

Binding Energy per Nucleon

10.00 9.00 8.00 7.00 6.00 5.00 4.00 3.00 Eb/A (10^8 kJ/mol) 2.00 1.00 0.00 0 20 40 60 80 100 120 140 160 180 200 220 240 MassMass Number Number

Key Questions

1. Of all isotopes in the model, which has the smallest binding energy per nucleon?

2. Why do you think the binding energy per nucleon is small when the mass number is very small?

3. Using the ideas of a repulsive electromagnetic force and an attractive nuclear force, why do you think the binding energy per nucleon first increases and then decreases as the number of nucleons increases past some intermediate value?

4. Which nuclides are the more stable: those with a large binding energy per nucleon or those with a small binding energy per nucleon? Explain.

322 Foundations of Chemistry 5. Is the reaction that converts a less stable nuclide to a more stable nuclide exothermic or endothermic? Would you expect the free energy change for this reaction to be positive or negative? Explain.

6. In view of your answer to Key Question 5, which of those isotopes in the model would you expect to be most abundant at equilibrium? Explain.

7. Why is the distribution of isotopes not the equilibrium distribution, e.g., why do we still have hydrogen, helium, and uranium?

8. What reaction of nuclides (fission or fusion) with a small mass number could produce more stable nuclides? Explain.

9. What reaction of nuclides (fission or fusion) with a large mass number could produce more stable nuclides? Explain.

10. There are two kinds of atomic bombs: hydrogen and uranium. A hydrogen bomb is much more destructive. Why do you think more energy is released in a hydrogen bomb than in a uranium bomb?

Information

All elements are thermodynamically unstable with respect to iron because the free energy change required to produce iron from these elements is negative. All elements should spontaneously convert to iron but do not do so when the activation energy for such conversion is large. Light nuclides can come together to form more stable heavier nuclides. This process is called fusion. It is the source of energy in the sun. Heavy nuclides can split to form more stable lighter nuclides. This process is called fission. It is the source of energy in nuclear power generation stations.

Activity 47 —Nuclear Chemistry: Binding Energy 323 Exercises

1. Calculate the energy released when one mole of 2H is converted into 3He. The nuclear reaction equation is 2 2H 3He + neutron. The nuclide masses are 2H = 2.0141 g/mol, 3He = 3.0160 g/mol.

Calculate the change in mass between the reactants and the products, multiply by the speed of light squared to convert to energy.

∆m = (mp )- (m r ) = [ (1 mol)(3.0160 g/mol + 1.008665 g/mol)] -[(2 mol)(2.0141 g/mol)] = -0.0035 g E = (-3.5 × 10-6 kg)(2.998 × 108 m/s) 2 = -3.1 × 1011J 11 -3.1 × 10 J 11 E = = -1.6 × 10 J/mol deuterium 2 mol The negative sign indicates that the reaction is exothermic.

2. How much energy is released per mole of uranium in the following nuclear reaction? The nuclide masses in g/mol are U (235.0439), Mo (99.9076), and Sn (133.9125).

235U + n 100Mo + 134Sn + 2n

∆m = (mp )- (m r ) =  99.9076 g + 133.9125 g + ( 2.01733 - 235.0439 g) - 1.008665 g = - 0.2151 g E = - (2.151 × 10-4 kg)(2.998 × 108 m/s) 2 E = - 1.933 × 1013 J/mol

Note: The answers to Exercises 1 and 2 when compared appear to contradict the answer to Key Question 10, but the answer to KQ 10 is based on the binding energy per nucleon, while the answers to Exercises 1 and 2 are based on the change in binding energies for the total number of nucleons involved in the reaction.

What do you think now?

Is it possible to turn lead into gold? If no, explain why not. If yes, explain how it might be done. In your explanation mention whether gold is more or less stable than lead.

324 Foundations of Chemistry