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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 1.0827 MeV.

Step by step solution

01

Determine the number of protons and neutrons in the nucleus

First, we need to determine the number of protons and neutrons in the \(^{56}\mathrm{Fe}\) nucleus. Since it's an iron atom, it has 26 protons. To find the number of neutrons, subtract the number of protons from the mass number (56): \( 56 - 26 = 30 \) neutrons.
02

Calculate the total mass of individual nucleons

Next, we need to find the total mass of the individual protons and neutrons. The atomic mass unit (u) is approximately equal to the mass of a proton and a neutron, so: Total mass of protons = number of protons × mass of one proton Total mass of protons = 26 × 1u = 26u Total mass of neutrons = number of neutrons × mass of one neutron Total mass of neutrons = 30 × 1u = 30u Total mass of individual nucleons = Total mass of protons + Total mass of neutrons Total mass of individual nucleons = 26u + 30u = 56u
03

Find the mass defect

Now we can find the mass defect, which is the difference between the mass of the nucleus and the total mass of its individual protons and neutrons: Mass defect = Total mass of individual nucleons - Mass of nucleus Mass defect = 56u - 55.9349u = 0.0651u
04

Convert the mass defect into energy

Using Einstein's mass-energy equivalence formula, we can convert the mass defect into energy. The formula is: \(E = mc^2\) Here, E is the energy, m is the mass defect, and c is the speed of light in a vacuum. Since 1u is equal to 931.5 MeV/c², we can convert the mass defect into MeV and then calculate the energy: Mass defect in MeV = 0.0651u × 931.5 MeV/c²/u = 60.6128 MeV
05

Calculate the binding energy per nucleon

Finally, to find the binding energy per nucleon, simply divide the total binding energy by the number of nucleons: Binding energy per nucleon = Total binding energy / Number of nucleons Binding energy per nucleon = 60.6128 MeV / 56 = 1.0827 MeV So the binding energy per nucleon for \(^{56}\mathrm{Fe}\) is approximately 1.0827 MeV.

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Key Concepts

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

Mass Defect
Mass defect is a crucial concept in nuclear physics that helps us understand the binding energy within a nucleus. It refers to the difference between the total mass of the individual nucleons (protons and neutrons) and the actual mass of the nucleus. This difference arises because a portion of the mass is converted into binding energy, which holds the nucleons together. To find the mass defect, you start by calculating the mass of all the protons and neutrons separately, assuming each has a mass close to 1 atomic mass unit (u). For a nucleus like - **Iron-56**, you have: - 26 protons and 30 neutrons, giving a combined mass of 56u (since each nucleon is approximately 1u). Then, subtract the actual nuclear mass from this total to find the mass defect. For - **Iron-56**, if the nuclear mass is 55.9349u, the mass defect is 56u - 55.9349u = 0.0651u. This small but significant difference gives us insight into how much energy is tied up in binding the nucleus together.
Einstein's Mass-Energy Equivalence
Einstein's mass-energy equivalence is an indispensable principle that explains how mass can be transformed into energy and vice versa. This relationship is described by the famous equation: \[ E = mc^2 \]which states that energy (E) is equal to mass (m) multiplied by the speed of light (c) squared.In the context of nuclear physics, this equation helps us understand how mass defect translates into binding energy. The mass lost during nucleon combination (mass defect) is not disappeared but converted into energy, stabilizing the nucleus. **For example:**- The mass defect of Iron-56 is 0.0651u. By using Einstein's equation, this mass defect, when multiplied by the conversion factor (1u = 931.5 MeV/c²), provides a binding energy: - Energy = 0.0651u × 931.5 MeV/c²/u = 60.6128 MeV.This tells us the total energy released or required to hold the nucleus together, showcasing the power of mass-energy equivalence in nuclear stability.
Atomic Mass Unit
The atomic mass unit (u) is a standard unit of mass that is primarily used to express atomic and molecular weights. It is defined as one twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state, approximately equal to 1.66053906660 x 10^{-27} kg. In nuclear physics, the atomic mass unit serves as a convenient measure for the mass of protons and neutrons: - Both have a mass close to 1u each. With this unit, it becomes easier to measure small mass differences, known as mass defects, which are crucial for calculating nuclear binding energies. - **For Iron-56 nucleus computations,** we see that it involves determining the mass of 26 protons and 30 neutrons in atomic mass units, producing results in the framework convenient for high precision. If the nucleus has fewer atomic mass units than these nucleons combined, the binding energy of the nucleus can be computed using Einstein’s mass-energy equivalence. Thus, the atomic mass unit not only streamlines calculations involving tiny quantities but also acts as a bridge between atomic-scale measurements and universal constants like the speed of light in mass-energy conversions.

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Most popular questions from this chapter

Consider the following information: i. The layer of dead skin on our bodies is sufficient to protect us from most \(\alpha\) -particle radiation. ii. Plutonium is an \(\alpha\) -particle producer. iii. The chemistry of \(\mathrm{Pu}^{4+}\) is similar to that of \(\mathrm{Fe}^{3+}\). iv. Pu oxidizes readily to \(\mathrm{Pu}^{4+}\) Why is plutonium one of the most toxic substances known?

Much of the research on controlled fusion focuses on the problem of how to contain the reacting material. Magnetic fields appear to be the most promising mode of containment. Why is containment such a problem? Why must one resort to magnetic fields for containment?

The most significant source of natural radiation is radon-222. \(^{222} \mathrm{Rn},\) a decay product of \(^{238} \mathrm{U},\) is continuously generated in the earth's crust, allowing gaseous Rn to seep into the basements of buildings. Because \(^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of 3.82 days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \(^{238} \mathrm{U}\) decays to \(^{222} \mathrm{Rn} ?\) What nuclei are produced when \(^{222} \mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \(^{222} \mathrm{Rn} ?\) c. Another problem associated with \(^{222} \mathrm{Rn}\) is that the decay of \(^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer \(\left(t_{1 / 2}=\right.\) 3.11 min) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by \(\alpha\) particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that \(^{222}\) Rn levels not exceed 4 pCi per liter of air (1 \(\mathrm{Ci}=1\) curie \(=3.7 \times 10^{10}\) decay events per second; \(1 \mathrm{pCi}=1 \times 10^{-12} \mathrm{Ci}\). Convert \(4.0 \mathrm{pCi}\) per liter of air into concentrations units of \(^{222} \mathrm{Rn}\) atoms per liter of air and moles of \(^{222}\) Rn per liter of air.

Technetium-99 has been used as a radiographic agent in bone scans ( \(_{43}^{99}\) Tc is absorbed by bones). If \(\frac{99}{43}\) Tc has a half-life of 6.0 hours, what fraction of an administered dose of \(100 . \mu \mathrm{g}\) \(^{99}_{43}\) Tc remains in a patient's body after 2.0 days?

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. \(\frac{56}{26} \mathrm{Fe}\) c. \(\frac{20}{11} \mathrm{Na}\) d. \(^{194}_{81} \mathrm{TI}\)

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