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Of the following nuclides, the highest nuclear binding energy per nucleon is found in (a) \(_{1}^{3} \mathrm{H} ;\) (b) \(_{8}^{16} \mathrm{O} ;\) (c) \(_{26}^{56} \mathrm{Fe}\); (d) \(_{92}^{235} \mathrm{U}\).

Short Answer

Expert verified
The nuclide with the highest nuclear binding energy per nucleon is \(_{26}^{56} \mathrm{Fe}\).

Step by step solution

01

Understanding the information

The task gives us four nuclides: Tritium \(_{1}^{3} \mathrm{H}\), Oxygen \(_{8}^{16} \mathrm{O}\), Iron \(_{26}^{56} \mathrm{Fe}\), and Uranium \(_{92}^{235} \mathrm{U}\). We need to determine which one of these has the highest nuclear binding energy per nucleon.
02

Refer to the binding energy curve

The binding energy curve charts binding energy per nucleon against atomic number. First, it increases, reaching a maximum at Iron-56 (\(_{26}^{56} \mathrm{Fe}\)), then it decreases for elements with higher atomic numbers.
03

Apply knowledge of the binding energy curve to given nuclides

According to the curve, the binding energy per nucleon is highest for Iron-56 (\(_{26}^{56} \mathrm{Fe}\)), lower for elements lighter than iron like Tritium (\(_{1}^{3} \mathrm{H}\)) and Oxygen (\(_{8}^{16} \mathrm{O}\)), and even lower for heavier elements like Uranium (\(_{92}^{235} \mathrm{U}\)).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nuclides
When discussing nuclear physics, the term "nuclides" comes up frequently. A nuclide refers to a specific type of atomic nucleus, defined by its number of protons (atomic number) and neutrons. This means that each nuclide is characterized by a unique combination of these two subatomic particles. Understanding nuclides is crucial because they form the foundation of nuclear chemistry and physics. They help in identifying the stability and energy characteristics of atoms.

Nuclides are often represented using the notation \(_{Z}^{A}X\), where:
  • \(Z\) is the atomic number (number of protons)
  • \(A\) is the mass number (total number of protons and neutrons)
  • \(X\) is the chemical symbol of the element
This notation gives clarity on the composition of the nucleus of each element, making it easier to compare and understand their properties, especially when discussing concepts like nuclear binding energy.
Iron-56
Iron-56 is a special nuclide often referenced in the study of nuclear physics and chemistry. It is denoted as \(_{26}^{56} \, \mathrm{Fe}\), where it consists of 26 protons and 30 neutrons. This particular nuclide is significant because it has one of the highest binding energies per nucleon. Binding energy per nucleon is an indicator of the stability of a nucleus - the higher it is, the more stable the nucleus.

In fact, Iron-56 is considered to be the most stable nuclide in existence, at least in terms of nuclear binding energy. This results from the optimal ratio and distribution of protons and neutrons within its nucleus. Due to its stability, it is prevalent in the universe and commonly discussed in the context of stellar nuclear processes, where many elements form and degrade based on their stability relative to Iron-56.
Binding Energy Curve
The concept of the "Binding Energy Curve" is essential for understanding the stability of different nuclides. This curve represents the binding energy per nucleon plotted against the atomic number of different elements. Initially, as the atomic number increases, so does the binding energy per nucleon. This trend reaches its peak at Iron-56, which holds the maximum value.
  • This indicates that Iron-56 is the most stable, with a strong nuclear force holding together its protons and neutrons.
  • After Iron-56, the binding energy per nucleon begins to decrease as elements become heavier.
This decrease implies that the nuclei of heavier elements, such as Uranium-235, are less stable due to the excess of protons and higher repulsive forces. Understanding the binding energy curve is crucial for comprehending why certain elements undergo processes like nuclear fission while others, like Iron-56, do not actively engage in such reactions.

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Most popular questions from this chapter

Write a plausible equation for the decay of tritium, 3 \(\mathrm{H}\), the radioactive isotope of hydrogen. 1 \(\textrm{ }\).

The packing fraction of a nuclide is related to the fraction of the total mass of a nuclide that is converted to nuclear binding energy. It is defined as the fraction \((M-A) / A,\) where \(M\) is the actual nuclidic mass and \(A\) is the mass number. Use data from a handbook (such as the Handbook of Chemistry and Physics, published by the CRC Press) to determine the packing fractions of some representative nuclides. Plot a graph of packing fraction versus mass number, and compare it with Figure \(25-6 .\) Explain the relationship between the two.

Of the following nuclides, the one most likely to be radioactive is \((a)^{31} P ;(b)^{66} Z n ;(c)^{35} C l ;(d)^{108} A g\).

Write nuclear equations to represent (a) the decay of \(^{214} \mathrm{Ra}\) by \(\alpha\) -particle emission (b) the decay of \(^{205}\) At by positron emission (c) the decay of \(^{212} \mathrm{Fr}\) by electron capture (d) the reaction of two deuterium nuclei (deuterons) to produce a nucleus of \(\frac{3}{2} \mathrm{He}\). (e) the production of \({243}_{97} \mathrm{Bk}\) get by the \(\alpha\) -particle bombardment of\({241}_{95} \mathrm{Am}\) (f) a nuclear reaction in which thorium-232 is bombarded with \(\alpha\) particles, producing a new nuclide and four neutrons.

For medical uses, radon-222 formed in the radioactive decay of radium-226 is allowed to collect over the radium metal. Then, the gas is withdrawn and sealed into a glass vial. Following this, the radium is allowed to disintegrate for another period, when a new sample of radon- 222 can be withdrawn. The procedure can be continued indefinitely. The process is somewhat complicated by the fact that radon-222 itself undergoes radioactive decay to polonium- 218 , and so on. The half-lives of radium-226 and radon-222 are \(1.60 \times 10^{3}\) years and 3.82 days, respectively.(a) Beginning with pure radium- \(226,\) the number of radon-222 atoms present starts at zero, increases for a time, and then falls off again. Explain this behavior. That is, because the half-life of radon-222 is so much shorter than that of radium- \(226,\) why doesn't the radon-222 simply decay as fast as it is produced, without ever building up to a maximum concentration?(b) Write an expression for the rate of change \((d \mathrm{D} / d t)\) in the number of atoms (D) of the radon- 222 daughter in terms of the number of radium- 226 atoms present initially ( \(\mathrm{P}_{0}\) ) and the decay constants of the parent \(\left(\lambda_{\mathrm{p}}\right)\) and daughter \(\left(\lambda_{\mathrm{d}}\right)\) (c) Integration of the expression obtained in part (b) yields the following expression for the number of atoms of the radon-222 daughter (D) present at a time \(t\).$$\mathrm{D}=\frac{\mathrm{P}_{0} \lambda_{\mathrm{p}}\left(\mathrm{e}^{-\lambda_{\mathrm{p}} \times t}-\mathrm{e}^{-\lambda_{\mathrm{d}} \times t}\right)}{\lambda_{\mathrm{d}}-\lambda_{\mathrm{p}}}$$,Starting with \(1.00 \mathrm{g}\) of pure radium- \(226,\) approximately how long will it take for the amount of radon222 to reach its maximum value: one day, one week, one year, one century, or one millennium?

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