The most stable nucleus in terms of binding energy per nucleon is \({ }^{56} \mathrm{Fe}\). If the atomic mass of \({ }^{56} \mathrm{Fe}\) is \(55.9349 \mathrm{amu}\), 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 \(1.36948\times10^{-12}\,\mathrm{J}\).

Step by step solution

01

Calculate the mass defect

To find the mass defect of \({}^{56}\mathrm{Fe}\), we first need to find out the total mass of the individual nucleons. In this case, we have 26 protons and 30 neutrons. For protons, we'll use the mass of \({}^1\mathrm{H}\) because they are essentially the same as the hydrogen atom. The mass of a proton is approximately \(1.007276\,\mathrm{amu}\). For neutrons, we'll approximate their mass to be \(1.008665\,\mathrm{amu}\). Total mass of protons = number of protons × mass of a proton Total mass of protons = 26 × \(1.007276\,\mathrm{amu}\) = \(26.189176\,\mathrm{amu}\) Total mass of neutrons = number of neutrons × mass of a neutron Total mass of neutrons = 30 × \(1.008665\,\mathrm{amu}\) = \(30.25995\,\mathrm{amu}\) Mass defect = (Total mass of protons + Total mass of neutrons) - Atomic mass of \({}^{56}\mathrm{Fe}\) Mass defect = (\(26.189176\,\mathrm{amu}\) + \(30.25995\,\mathrm{amu}\)) - \(55.9349\,\mathrm{amu}\) = \(0.514226\,\mathrm{amu}\)
02

Calculate the total binding energy

We'll use Einstein's famous mass-energy equivalence formula, \(E=mc^2\), to find the total binding energy. To do this, we need to convert the mass defect (\(0.514226\,\mathrm{amu}\)) into kilograms, and the speed of light, \(c\), is \(2.998\times10^8\,\mathrm{m/s}\). Mass defect in kg = \(0.514226\,\mathrm{amu}\) × \(1.66054\times10^{-27}\,\mathrm{kg/amu}\) = \(8.54254\times10^{-28}\,\mathrm{kg}\) Total binding energy = mass defect × \(c^2\) Total binding energy = \(8.54254\times10^{-28}\,\mathrm{kg}\) × \((2.998\times10^8\,\mathrm{m/s})^2\) = \(7.66513\times10^{-11}\,\mathrm{J}\)
03

Calculate the binding energy per nucleon

Finally, we'll divide the total binding energy by the number of nucleons (\({ }^{56}\mathrm{Fe}\) has 56 nucleons). Binding energy per nucleon = Total binding energy / Number of nucleons Binding energy per nucleon = \(7.66513\times10^{-11}\,\mathrm{J}\) / 56 = \(1.36948\times10^{-12}\,\mathrm{J}\) Thus, the binding energy per nucleon for \({ }^{56}\mathrm{Fe}\) is approximately \(1.36948\times10^{-12}\,\mathrm{J}\).

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Key Concepts

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

Nuclear Stability
The concept of nuclear stability is fundamental to understanding why certain nuclei are more tightly bound than others. A stable nucleus implies that it has a high binding energy per nucleon; this means it would require a significant amount of energy to break the nucleus apart into individual protons and neutrons.

For instance, ^{56}Fe, which is the most stable nucleus in terms of binding energy per nucleon, can be taken as a benchmark for stability. Nuclear stability is governed by a delicate balance between the strong nuclear force, which works to hold the nucleons together, and the electrostatic repulsion between protons. Nuclei are most stable when they have an optimal ratio of neutrons to protons, which for light elements is approximately 1:1, but shifts towards a higher proportion of neutrons in heavier elements. ^{56}Fe's stability comes from having an almost ideal number of neutrons and protons, which minimizes the repulsive forces while maximizing the strong nuclear force.
Mass Defect Calculation
Mass defect refers to the difference in mass between the sum of the separate nucleons (protons and neutrons) and the actual mass of the nucleus. It's a manifestation of the energy released when a nucleus is formed from its constituent particles.

To calculate this, you must sum the masses of the protons and neutrons when outside the nucleus and then subtract the measured atomic mass of the nucleus, just as seen in the solution for ^{56}Fe. Here's the formula simplified:
Mass Defect = (Total mass of protons + Total mass of neutrons) - Atomic mass of the nuclide


This calculation is important because the mass defect directly relates to the binding energy—the energy required to split the nucleus into its individual nucleons. In teaching the notion of mass defect, highlighting the accuracy of measured atomic masses is critical, as they can affect the resulting calculation of nuclear stability.
Einstein's Mass-Energy Equivalence
Einstein's famous equation,
E=mc^2
, known as the mass-energy equivalence, is pivotal in nuclear physics. This equation tells us that mass can be converted into energy and vice versa, implying that the energy which binds the nucleus together is actually derived from the mass defect.

When calculating binding energy, like in the provided solution for ^{56}Fe, this equation is used to convert the mass defect from atomic mass units into energy (joules). A high binding energy signifies a strong binding force between nucleons, thereby contributing to a more stable nucleus. It's these interactions at the subatomic level that illustrate the profound link between mass and energy, a cornerstone of modern physics that revolutionized scientific understanding.

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Most popular questions from this chapter

Radioactive cobalt- 60 is used to study defects in vitamin \(\mathrm{B}_{12}\) absorption because cobalt is the metallic atom at the center of the vitamin \(\mathrm{B}_{12}\) molecule. The nuclear synthesis of this cobalt isotope involves a three-step process. The overall reaction is iron58 reacting with two neutrons to produce cobalt-60 along with the emission of another particle. What particle is emitted in this nuclear synthesis? What is the binding energy in J per nucleon for the cobalt-60 nucleus (atomic masses: \({ }^{60} \mathrm{Co}=59.9338\) amu; \({ }^{1} \mathrm{H}=1.00782\) amu \()\) ? What is the de Broglie wavelength of the emitted particle if it has a velocity equal to \(0.90 c\) where \(c\) is the speed of light?

A proposed system for storing nuclear wastes involves storing the radioactive material in caves or deep mine shafts. One of the most toxic nuclides that must be disposed of is plutonium- 239, which is produced in breeder reactors and has a half-life of 24,100 years. A suitable storage place must be geologically stable long enough for the activity of plutonium- 239 to decrease to \(0.1 \%\) of its original value. How long is this for plutonium- \(239 ?\)

Radioactive copper-64 decays with a half-life of \(12.8\) days. a. What is the value of \(k\) in \(\mathrm{s}^{-1}\) ? b. A sample contains \(28.0 \mathrm{mg}^{64} \mathrm{Cu}\). How many decay events will be produced in the first second? Assume the atomic mass of \({ }^{64} \mathrm{Cu}\) is \(64.0 .\) c. A chemist obtains a fresh sample of \({ }^{64} \mathrm{Cu}\) and measures its radioactivity. She then determines that to do an experiment, the radioactivity cannot fall below \(25 \%\) of the initial measured value. How long does she have to do the experiment?

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

Americium- 241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of \({ }^{241}\) Am is 433 years, and it decays by emitting alpha particles. How many alpha particles are emitted each second by a \(5.00-\mathrm{g}\) sample of \({ }^{241} \mathrm{Am} ?\)

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