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Chapter 43: Problem 8

43.8. Calculate (a) the total binding energy and (b) the binding energy per nucleon of \(^{12} \mathrm{C}\) (c) What percent of the rest mass of this nucleus is its total binding energy?

Short Answer

Expert verified
(a) 89.065 MeV; (b) 7.422 MeV/nucleon; (c) 0.797%.

Step by step solution

01

Determine the Mass Defect

The mass defect, \(\Delta m\), is the difference between the mass of the individual nucleons and the mass of the nucleus. For \(^{12} \text{C}\):- Proton mass \(m_p = 1.00728 \, \text{u}\)- Neutron mass \(m_n = 1.00866 \, \text{u}\).- Atomic mass of \(^{12} \text{C} = 12 \text{ u}\).Calculate the mass of 6 protons and 6 neutrons:\[m_{total} = 6m_p + 6m_n = 6(1.00728) + 6(1.00866) = 12.09564 \, \text{u}\]Mass defect: \[\Delta m = m_{total} - m_{\text{C}} = 12.09564 - 12 = 0.09564 \, \text{u}\]
02

Convert Mass Defect to Energy

To find the total binding energy \(E_b\), convert the mass defect from atomic mass units to energy (MeV) using the equation \(E = \Delta m \cdot c^2\) and knowing \(1 \text{ u} = 931.5 \, \text{MeV/c}^{2}\):\[E_b = \Delta m \times 931.5 \, \text{MeV/u} = 0.09564 \, \text{u} \times 931.5 \, \text{MeV/u} = 89.06546 \, \text{MeV}\]
03

Calculate the Binding Energy Per Nucleon

For \(^{12} \text{C}\), there are 12 nucleons. Calculate the binding energy per nucleon by dividing the total binding energy by the number of nucleons:\[E_{b_{\text{per nucleon}}} = \frac{E_b}{12} = \frac{89.06546}{12} = 7.4221 \, \text{MeV}\]
04

Calculate the Percent of Rest Mass as Binding Energy

To find what percent of the rest mass is the total binding energy, use the equation:\[\text{Percent} = \left(\frac{E_b}{m_{\text{C}} \times c^2}\right) \times 100\]Substitute the known values, remembering that \(m_{\text{C}} = 12 \, \text{u}\) and \(\text{converted to MeV} = 12 \times 931.5 = 11178 \, \text{MeV}\):\[\text{Percent} = \left(\frac{89.06546}{11178}\right) \times 100 = 0.7967\%\]

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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 term "mass defect" refers to the difference between the summed masses of the individual nucleons (protons and neutrons) and the actual mass of the nucleus itself. This can be expressed as
  • Mass of protons + Mass of neutrons - Mass of nucleus = Mass defect.
The mass defect arises because some of the mass is converted into binding energy. This energy is what keeps the nucleus from falling apart due to the repulsive forces between positively charged protons. For example, in the carbon-12 nucleus, we calculated the mass defect by finding the difference between the mass of six protons and six neutrons, and the measured mass of the carbon nucleus.
Nucleus
The nucleus is the compact center of an atom, composed of protons and neutrons.
  • Protons are positively charged particles.
  • Neutrons have no charge (are neutral).
This composition is why the nucleus is dense and holds almost all the atomic mass. Although consisting of charged protons, the nucleus stays intact due to the binding energy, which offsets the repulsion between protons. The strength of this energy illustrates why splitting or fusing nuclei can release a staggering amount of energy per unit mass.
Atomic Mass
Atomic mass is the average mass of an atom, typically measured in atomic mass units (u or amu). It includes the masses of protons, neutrons, and electrons. However, electrons contribute negligibly, so this measurement focuses primarily on the nucleus.
  • The nucleus accounts for nearly all the atomic mass.
  • Atomic mass is close to but not exactly equal to the sum of the numbers of protons and neutrons because of binding energy.
In this specific exercise, the carbon-12 isotope has an atomic mass exactly 12 u, highlighting its role as a standard for measuring atomic masses across elements.
Nucleons
Nucleons are the particles that make up the nucleus: protons and neutrons. Each type of nucleon plays a fundamental role in defining an atom's identity and stability.
  • Protons determine the element's atomic number.
  • Neutrons contribute to the isotope variation of elements.
The force that holds nucleons together is incredibly strong. In a nucleus such as carbon-12, nucleons must overcome the electromagnetic repulsion of protons, and they do so thanks to the strong nuclear force, which is significantly stronger than the electromagnetic force at very short distances.

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Most popular questions from this chapter

43.80. Industrial Radioactivity. Radioisotopes are used in a variety of manufacturing and testing techniques. Wear measurements can be made using the following method. An automobile engine is produced using piston rings with a total mass of 100 \(\mathrm{g}\) , which includes 9.4\(\mu \mathrm{Ci}\) of \(^{59} \mathrm{Fe}\) whose half-life is 45 days. The engine is test-run for 1000 hours, after which the oil is drained and its activity is measured. If the activity of the engine oil is 84 decays/s, how much mass was worn from the piston rings per hour of operation?

43.67 Measurements indicate that 27.83\(\%\) of all rubidium atoms currently on the earth are the radioactive \(^{87} \mathrm{Rb}\) isotope. The rest are the stable \(^{85} \mathrm{Rb}\) isotope. The half-life of \(^{87} \mathrm{Rb}\) is 4.75 \(\times 10^{10} \mathrm{y}\) . Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were \(^{87} \mathrm{Rb}\) when our solar system was formed \(4.6 \times 10^{9} \mathrm{y}\) ago?

43.22. Radioactive isotopes used in cancer therapy have a "shelf- life," like pharmaccuticals used in chemotherapy. Just after it has been manufactured in a nuclear reactor, the activity of a sample of "Co is 5000 Ci. When its activity falls below 3500 \(\mathrm{Ci}\) , it is considered too weak a source to use in treatment. You work in the radiology department of a large hospital. One of these "Co sources in your inventory was manufactured on October \(6,2004\) . It is now April \(6,2007\) . Is the source still usable? The half-life of \(^{6}\) Co is 5.271 years.

43.49. Use conservation of mass-energy to show that the energy released in alpha decay is positive whenever the mass of the original neutral atom is greater than the sum of the masses of the final neutral atom and the neutral "He atom. (Hint: Let the parent nucleus have atomic number \(Z\) and nucleon number \(A\) . First write the reaction in terms of the nuclei and particles involved, and then \(n\) then \(n\) add \(Z\) electron masses to both sides of the reaction and allot them as needed to arrive at neutral atoms.)

43.26. As a health physicist, you are being consulted about a spill in a radiochemistry lab. The isotope spilled was 500\(\mu \mathrm{Ci}\) of \(^{131} \mathrm{Ba}\) , which has a half-life of 12 days. (a) What mass of 33 \(\mathrm{Ba}\) was spilled? (b) Your recommendation is to clear the lab until the radiation level has fallen 1.00\(\mu \mathrm{Ci}\) . How long will the lab have to be closed?
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