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Chapter 19: Problem 7

Why are the observed energy changes for nuclear processes so much larger than the energy changes for chemical and physical processes?

Short Answer

Expert verified
The observed energy changes for nuclear processes are much larger than those for chemical and physical processes because nuclear processes involve the strong nuclear force, which is much stronger than the electromagnetic force that governs chemical and physical processes. This leads to a larger amount of energy stored in atomic nuclei, resulting in significantly larger energy changes during nuclear reactions compared to those in chemical and physical processes. Typical energy changes in nuclear processes range up to a billion electron volts (MeV), while energy changes in chemical and physical processes only go up to about 10 electron volts (eV).

Step by step solution

01

Nuclear processes involve changes in the nuclei of atoms, including nuclear reactions such as fission (splitting of a nucleus), fusion (combining of two or more nuclei), and radioactive decay (spontaneous disintegration of a nucleus). The energy changes involved in these processes come from the strong nuclear force that binds protons and neutrons together in the nucleus. Typical energy changes in nuclear processes range from a million to a billion electron volts (MeV) per reaction. #Step 2: Introduction to chemical and physical processes#

Chemical and physical processes involve changes in the arrangement of atoms in molecules or changes in the physical state of a substance. These processes are governed by electromagnetic forces between charged particles (such as electrons and atomic nuclei), like the electrostatic force and the covalent bonding. Typical energy changes in chemical reactions are in the range of 1 to 10 electron volts (eV) per reaction, while physical processes like phase changes have energy changes on the order of 0.01 to 1 eV per atom or molecule. #Step 3: Comparing the energy scales and explaining the difference#
02

Comparing the energy scales of nuclear processes and chemical or physical processes shows that nuclear processes involve energy changes that are several orders of magnitude larger. This is because the strong nuclear force, responsible for binding protons and neutrons in nuclei, is much stronger than the electromagnetic force responsible for holding atoms together in molecules. As a result, the energy stored in nuclei is much greater than the energy stored in chemical bonds or the energy associated with physical state changes. When a nuclear process occurs, a larger amount of energy is released or absorbed compared to the energy changes in a chemical or physical process. #Step 4: Conclusion#

In conclusion, the observed energy changes for nuclear processes are much larger than the energy changes for chemical and physical processes because nuclear processes involve changes in the strong nuclear force, which is much stronger than the electromagnetic force governing the chemical and physical processes. Consequently, the energy stored in atomic nuclei is significantly higher than the energy stored in chemical bonds or associated with physical state changes, resulting in larger energy changes when nuclear processes occur.

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Most popular questions from this chapter

A rock contains \(0.688 \mathrm{mg}^{206} \mathrm{~Pb}\) for every \(1.000 \mathrm{mg}{ }^{238} \mathrm{U}\) present. Assuming that no lead was originally present, that all the \({ }^{206} \mathrm{~Pb}\) formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between \({ }^{238} \mathrm{U}\) and \({ }^{206} \mathrm{~Pb}\) is negligible, calculate the age of the rock. (For \({ }^{238} \mathrm{U}\), \(t_{1 / 2}=4.5 \times 10^{9}\) years. \()\)

Breeder reactors are used to convert the nonfissionable nuclide \({ }_{92}^{238} \mathrm{U}\) to a fissionable product. Neutron capture of the \({ }_{92}^{238} \mathrm{U}\) is followed by two successive beta decays. What is the final fissionable product?

The easiest fusion reaction to initiate is $${ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}$$ Calculate the energy released per \({ }_{2}^{4} \mathrm{He}\) nucleus produced and per mole of \({ }_{2}^{4}\) He produced. The atomic masses are \({ }_{1}^{2} \mathrm{H}, 2.01410 ;{ }_{1}^{3} \mathrm{H}\), \(3.01605\); and \({ }_{2}^{4}\) He, \(4.00260\). The masses of the electron and neutron are \(5.4858 \times 10^{-4}\) amu and \(1.00866\) amu, respectively.

Each of the following isotopes has been used medically for the purpose indicated. Suggest reasons why the particular element might have been chosen for this purpose. a. cobalt-57, for study of the body's use of vitamin \(\mathrm{B}_{12}\) b. calcium- 47 , for study of bone metabolism c. iron- 59 , for study of red blood cell function

Which type of radioactive decay has the net effect of changing a neutron into a proton? Which type of decay has the net effect of turning a proton into a neutron?
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