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Chapter 18: 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 the energy changes for chemical and physical processes because nuclear processes involve the strong nuclear force, which is many orders of magnitude stronger than the electromagnetic force responsible for chemical and physical processes. Nuclear processes, such as fission and fusion, involve changes in the binding energy of atomic nuclei, leading to energy release or absorption measured in mega-electron volts (MeV). In contrast, chemical and physical processes involve interactions between electrons and weaker forces between molecules, resulting in smaller energy changes measured in electron volts (eV).

Step by step solution

01

Understanding Nuclear Processes

Nuclear processes refer to changes that involve atomic nuclei. Nuclear reactions, such as fission and fusion, involve the release or absorption of energy when atomic nuclei interact and change. The energy is released or absorbed depending on the binding energy of the nucleons (protons and neutrons).
02

Understanding Physical and Chemical Processes

Physical processes typically refer to changes in the state of matter and do not involve changes in the identity of the substances involved. Chemical processes involve the creation and breaking of chemical bonds in molecules, which are the result of interactions between electrons.
03

Energy changes in Nuclear Processes

The energy changes in nuclear processes stem from the strong nuclear force holding the nucleus together. Nuclei often rearrange in ways that lead to lower binding energies, and the "excess" energy is released. The energy scale of nuclear processes is measured in mega-electron volts (MeV), much higher than chemical and physical processes, which are measured in electron volts (eV).
04

Energy changes in Chemical and Physical Processes

The energy changes in chemical and physical processes come from the interactions between electrons and, to a lesser extent, the forces between the molecules (such as Van der Waals forces). These interactions are much weaker than the strong nuclear force and their energy scale is smaller.
05

Compare Energy Changes

The strong nuclear force in nuclear processes is many orders of magnitude stronger than the electromagnetic force in chemical and physical processes. As a result, the energy changes in nuclear processes are significantly higher than those in chemical and physical processes. This is the main reason why observed energy changes for nuclear processes are much larger than the energy changes for chemical and physical processes.

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Most popular questions from this chapter

The most stable nucleus in terms of binding energy per nucleon is \(^{56} \mathrm{Fe}\). If the atomic mass of \(^{56} \mathrm{Fe}\) is \(55.9349 \mathrm{u},\) calculate the binding energy per nucleon for \(^{56} \mathrm{Fe}\).

Consider the following reaction to produce methyl acetate: When this reaction is carried out with \(\mathrm{CH}_{3} \mathrm{OH}\) containing oxygen-18, the water produced does not contain oxygen-18. Explain.

Uranium- 235 undergoes many different fission reactions. For one such reaction, when \(^{235} \mathrm{U}\) is struck with a neutron, \(^{144} \mathrm{Ce}\) and \(^{90}\) Sr are produced along with some neutrons and electrons. How many neutrons and \(\beta\) -particles are produced in this fission reaction?

Calculate the amount of energy released per gram of hydrogen nuclei reacted for the following reaction. The atomic masses are \(^{1}_{1}{H}, 1.00782 \mathrm{u} ; \frac{2}{1} \mathrm{H}, 2.01410 \mathrm{u} ;\) and an electron, \(5.4858 \times\) \(10^{-4}\) u. (Hint: Think carefully about how to account for the electron mass.)$$\mathrm{i} \mathrm{H}+\mathrm{i} \mathrm{H} \longrightarrow_{\mathrm{i}}^{2} \mathrm{H}+_{+\mathrm{i}}^{0}$$

Natural uranium is mostly nonfissionable \(^{238} \mathrm{U} ;\) it contains only about \(0.7 \%\) of fissionable \(^{235}\) U. For uranium to be useful as a nuclear fuel, the relative amount of \(^{235}\) U must be increased to about \(3 \% .\) This is accomplished through a gas diffusion process. In the diffusion process, natural uranium reacts with fluorine to form a mixture of \(^{238} \mathrm{UF}_{6}(g)\) and \(^{235} \mathrm{UF}_{6}(g) .\) The fluoride mixture is then enriched through a multistage diffusion process to produce a \(3 \%^{235} \mathrm{U}\) nuclear fuel. The diffusion process utilizes Graham's law of effusion (see Chapter \(8,\) Section \(8-7\) ). Explain how Graham's law of effusion allows natural uranium to be enriched by the gaseous diffusion process.
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