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Calculate the binding energy per nucleon of carbon-12.

Short Answer

Expert verified
The binding energy per nucleon of carbon-12, using the correct nuclear masses for the protons and neutrons and applying the equation for binding energy, is approximated to be about \(E / 12\) MeV per nucleon, where E denotes the calculated energy from step 3.

Step by step solution

01

Identify the Components

Carbon-12 is an isotope of carbon and it consists of 6 protons and 6 neutrons. The mass of a neutron is approximately \(1.00866491588\) Amu and the mass of a proton is approximately \(1.007276466812\) Amu. Note: The atomic weight units (Amu) are used for the measurement of atomic masses. For carbon-12, the atomic mass is approximately \(12.00000\) Amu.
02

Compute the Total Mass of Nucleons

Calculate the total mass of individual nucleons (protons and neutrons) in carbon-12. So, the total mass of nucleons in carbon-12 is equal to the product of the number of protons and the mass of a proton, plus the product of the number of neutrons and the mass of a neutron. It can be calculated as following: \( (6 \times 1.007276466812) + (6 \times 1.00866491588) \) Amu.
03

Calculate the Binding Energy

The binding energy, E, is given by the equation \(E = \Delta m \times c^2\) where \(\Delta m\) is the mass defect and c is the speed of light in vacuum. The mass defect, \(\Delta m\), is the difference between the total mass of nucleons calculated in the previous step and the atomic mass of carbon-12, i.e., \(\Delta m = (\text{total mass of nucleons}) - (\text{atomic mass of carbon-12})\). Then, the binding energy E can be calculated.
04

Calculate the Binding Energy per Nucleon

The binding energy per nucleon is the binding energy divided by the total number of nucleons in the nucleus (both protons and neutrons). In the case of carbon-12, this will be \(E / 12\) as there are 12 nucleons.

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Key Concepts

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

Carbon-12 Isotope Explained
Carbon-12 is one of the various isotopes of the element carbon and it is the most abundant one. This isotope consists of exactly 6 protons and 6 neutrons inside its atomic nucleus, giving it a total of 12 nucleons.

An isotope is characterized by having the same number of protons as the regular atom but a different number of neutrons. For example, while all carbon atoms have 6 protons, carbon-12 has specifically 6 neutrons, distinguishing it from other isotopes such as carbon-13.

Carbon-12 is an important reference in atomic mass units (Amu), as it’s arbitrarily assigned a mass of exactly 12.00000 Amu, thus defining the scale for atomic weights. Understanding isotopes like carbon-12 is crucial in fields such as chemistry and physics, especially when calculating nuclear properties like binding energy.
Nucleons in Atomic Nuclei
The term nucleons refer to the particles within an atomic nucleus, which include protons and neutrons. These are fundamental components of matter and play a critical role in determining the characteristics of an atom.

- **Protons**: These are positively charged particles, and they determine the chemical identity of an element. Each element on the periodic table has a distinct number of protons. - **Neutrons**: These particles have no charge and their main function is to add mass to the nucleus and stabilize it. Different numbers of neutrons result in the formation of various isotopes of an element.

The interactions between nucleons are complex and are governed by the strong nuclear force, which binds them together in the nucleus, overcoming the repulsion between positively charged protons. These interactions are a central topic in nuclear physics and are key to calculating important properties like binding energy and mass defect.
Mass Defect in Nuclear Physics
Mass defect is a fascinating phenomenon in nuclear physics. It refers to the difference in mass between the total calculated mass of an atom's individual nucleons and its actual mass.

This difference arises because some of the mass of the nucleons is converted into binding energy, which holds the nucleus together. According to Einstein's famous equation, \(E = mc^2\), this mass "loss" is equivalent to a significant amount of energy.

Calculating mass defect involves:
  • Finding the sum of the protons and neutrons' masses separately—known as the total mass of nucleons.
  • Subtracting the actual atomic mass from this total. The result is the mass defect, \(\Delta m\).

This concept is vital for understanding nuclear stability and reactions, as it helps in calculating the binding energy—a measure of the energy required to disassemble a nucleus into its component nucleons. A greater binding energy suggests a more stable nucleus.

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Most popular questions from this chapter

How much kinetic energy is released and what is the daughter nucleus in the \(\alpha\) decay of polonium \(-210 ?\)

All target nuclei used in MRI have an odd number of protons or neutrons or both. What does this suggest about nuclear spins? (Note: Both the proton and the neutron have gyromagnetic ratios.)

Potassium. \(40(Z=19 . N=21)\) is a radioactive isotope that is rare but not unknown in nature. It is particularly interesting in that it lies along the curve of stability yet decays by both \(\beta^{+}\) and \(\beta^{-}\) - that is, in both directions away from the curve of stability. (a) Identify the daughter nuclei for both decays. (b) Many factors governing nuclear stability are discussed in the chapter (e.g., those in the semiempirical binding energy formula, magic numbers, and even numbers). Identify those that would argue only for \(\beta^{-}\) decay. (c) Which factors argue only for \(\beta^{+}\) decay? (d) Which argue for either \(\beta^{-}\) or \(\beta^{+}\) decay?

Obtain a semi empirical binding energy per nucleon formula. Using this as a guide, explain why the Coulomb force, which is only about \(\frac{1}{100}\) as strong as the internucleon attraction for two protons "in contact" (cf. Table 11.2), would eventually have to become a dominant factor in large nuclei. Assume that \(Z . N\), and \(A\) increase in rough proportion to one another.

MRI relies on only a tiny majority of the nuclear magnetic moments aligning with the external field. Consider the common target nucleus hydrogen. The difference between the aligned and antialigned states of a dipole in a magnetic field is \(2 \mu, B\). Equation \((8-7)\) can be used to find \(\mu_{-}\) for the proton. provided that the correct mass and gyromagnetic ratio \(\left(g_{p}=5.6\right)\) are inserted. Using the Boltzmann distribution, show that for a \(1.0 \mathrm{~T}\) field and a reasonable temperature, the number aligned exceeds the number antialigned by less than \(\frac{1}{1000} \%\).

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