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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 \(^{56} \mathrm{Fe} .\)

Short Answer

Expert verified
The binding energy per nucleon for \(^{56}\mathrm{Fe}\) is approximately 8.56 MeV.

Step by step solution

01

Identify the number of protons and neutrons

Since the isotope is \(^{56}\mathrm{Fe}\), it has 56 nucleons in total. Iron (Fe) has an atomic number of 26, which means it has 26 protons. To find the number of neutrons, we subtract the number of protons from the total number of nucleons: \(56 - 26 = 30\) neutrons.
02

Calculate the mass defect

The mass defect is the difference between the mass of the individual protons and neutrons and the actual mass of \(^{56}\mathrm{Fe}\). The atomic mass of a proton is approximately 1.00728 u and the atomic mass of a neutron is approximately 1.00867 u. We can calculate the mass defect using the following equation: \[Mass\,Defect = (26 \times 1.00728\,u + 30 \times 1.00867\,u) - 55.9349\,u\] \[Mass\,Defect = (26.18928\,u + 30.2601\,u) - 55.9349\,u\] \[Mass\,Defect = 56.44938\,u - 55.9349\,u\] \[Mass\,Defect = 0.51448\,u\]
03

Convert mass defect to energy

Now we will use Einstein's equation to convert the mass defect to energy. We need to multiply the mass defect by the atomic mass unit constant (\(c^2 = 931.5\,\frac{MeV}{c^2 u}\)) to convert the mass defect from atomic mass units (u) to energy in mega electron volts (MeV). \[Total\,Binding \,Energy = 0.51448\,u \times 931.5\,\frac{MeV}{c^2 u}\] \[Total\,Binding \,Energy = 479.17\,MeV\]
04

Calculate binding energy per nucleon

Finally, we will divide the total binding energy by the number of nucleons (56) to find the binding energy per nucleon. \[Binding\,Energy\,Per\,Nucleon = \frac{479.17\,MeV}{56}\] \[Binding\,Energy\,Per\,Nucleon = 8.56\,MeV\] Thus, the binding energy per nucleon for \(^{56}\mathrm{Fe}\) is approximately 8.56 MeV.

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Key Concepts

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

Mass Defect
In nuclear physics, the mass defect is a crucial concept. It helps explain why the mass of a nucleus is less than the sum of the masses of its constituent protons and neutrons. This difference is called the mass defect.

To find the mass defect:
  • Calculate the expected mass by adding up the masses of all protons and neutrons in the nucleus.
  • Subtract the actual mass of the nucleus from this expected mass.
This difference arises because some mass is converted into energy, which binds the nucleons together within the nucleus.

For example, in the case of iron-56 ( {^{56}Fe}), the mass defect plays a pivotal role in determining its nuclear binding energy. Here, the calculated mass defect reflects the level of stability and binding energy within the atomic nucleus. This energy holds the nucleus together, preventing it from breaking apart. Eventually calculating binding energy per nucleon gives a measure of stability.
Einstein's Equation
Einstein's equation, often written as \(E=mc^2\), is a fundamental principle in physics. It shows the relationship between mass (\(m\)) and energy (\(E\)), with \(c\) representing the speed of light in a vacuum. This equation highlights that mass can be converted into energy and vice versa, a concept central to nuclear physics.

In the context of nuclear binding energy:
  • The mass defect, which is the missing mass found when comparing the predicted and actual mass of a nucleus, is converted into energy.
  • When you multiply the mass defect by the speed of light squared, you get the binding energy, showing how much energy was converted from mass to stabilize the nucleus.
This energy is crucial because it keeps the nucleus intact, overcoming the repelling electromagnetic forces between protons. Thus, Einstein’s equation is key to understanding how nuclear reactions, such as fission and fusion, release tremendous energy.
Atomic Mass Unit
An atomic mass unit (amu or \(u\)) is a standard unit of mass that quantifies mass on an atomic or molecular scale. By definition, one atomic mass unit is one-twelfth of the mass of a carbon-12 atom. Nearly all atoms are measured in atomic mass units, making it a convenient way to express atomic and molecular masses.

In nuclear processes:
  • Mass defect calculations use atomic mass units to express the minute changes in mass that occur during nuclear reactions.
  • This unit helps in conversions, particularly when changing the mass defect into energy using Einstein's equation, as these reactions are dealing with extremely small mass amounts.
When analyzing the iron-56 nucleus, the mass defect is calculated using amu to precisely measure how much mass has been "lost" during nuclear formation, effectively transforming it into binding energy.
Iron-56
Iron-56 ( {^{56}Fe}) is a significant isotope in studies of nuclear stability. It is considered to be one of the most stable isotopes owing to its high binding energy per nucleon. This stability is due to its balanced forces and efficiently bound nucleons within its nucleus.

Key points about iron-56:
  • It has 26 protons and 30 neutrons.
  • Iron-56 is located at the peak of the nuclear binding energy curve, meaning it has the highest binding energy per nucleon compared to other elements.
This makes iron-56 particularly stable, as evidenced by its relatively high binding energy per nucleon. As a result, nucleosynthesis in stars often stops at iron, as fusing lighter nuclei into iron releases energy, whereas forming heavier nuclei requires energy input, thus further contributing to its prevalence and stability in the universe.

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Most popular questions from this chapter

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?

Consider the following graph of binding energy per nucleon as a function of mass number a. What does this graph tell us about the relative half-lives of the nuclides? Explain your answer. b. Which nuclide shown is the most thermodynamically stable? Which is the least thermodynamically stable? c. What does this graph tell us about which nuclides undergo fusion and which undergo fission to become more stable? Support your answer.

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, Sg, in honor of Glenn T. Seaborg, discoverer of the transuranium elements. a. \(^{263}\) Sg was produced by the bombardment of \(^{249} \mathrm{Cf}\) with a beam of \(^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \(^{263}\) g decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \(^{263} \mathrm{Sg}\) ?

The number of radioactive nuclides in a sample decays from \(1.00 \times 10^{20}\) to \(2.50 \times 10^{19}\) in 10.0 minutes. What is the half-life of this radioactive species?

Predict whether each of the following nuclides is stable or unstable (radioactive). If the nuclide is unstable, predict the type of radioactivity you would expect it to exhibit. a. \(_{19}^{45} \mathrm{K}\) b. \(_{56}^{26} \mathrm{Fe}\) c. \(_{20}^{11} \mathrm{Na}\) d. \(_{194}^{81} \mathrm{TI}\)

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