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Chapter 11: Problem 10

If the mass defect of \(_{4} X^{9}\) is \(0.090\) amu, then binding energy per nucleon is \((1 \mathrm{amu},=921.5 \mathrm{MeV})\) a. \(83.0 \mathrm{MeV}\) b. \(9.315 \mathrm{MeV}\) c. \(8.38 \mathrm{MeV}\) d. \(931.5 \mathrm{MeV}\)

Short Answer

Expert verified
b. 9.315 MeV

Step by step solution

01

Understand Key Concepts

The mass defect is the difference between the sum of the masses of the separate nucleons and the actual mass of the nucleus. This mass defect is converted into binding energy, which holds the nucleus together.
02

Calculate Total Binding Energy

Use the mass defect to calculate the total binding energy. The mass defect given is \(0.090\) amu. Knowing that \(1\) amu corresponds to \(921.5\) MeV, calculate:\[\text{Total Binding Energy} = 0.090 \, \text{amu} \times 921.5 \, \text{MeV/amu} = 82.935 \, \text{MeV}\]
03

Count the Number of Nucleons

The notation \(_{4} X^{9}\) indicates an element with a mass number of \(9\), meaning it has \(9\) nucleons (protons + neutrons) total.
04

Calculate Binding Energy Per Nucleon

To find the binding energy per nucleon, divide the total binding energy by the number of nucleons:\[\text{Binding Energy Per Nucleon} = \frac{82.935 \, \text{MeV}}{9} \approx 9.215 \, \text{MeV}\]
05

Round to Match Options

Round the binding energy per nucleon to match one of the given options. The closest value to the calculated \(9.215 \, \text{MeV}\) is option (b) \(9.315 \, \text{MeV}\).

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Key Concepts

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

Mass Defect
When discussing nuclear physics, the term 'mass defect' is fundamental. It refers to the difference in mass between a nucleus and its individual nucleons, such as protons and neutrons. These nucleons, when bound together in a nucleus, have less mass than when they are separate. This missing mass is not really missing; instead, it has been converted into energy according to Einstein's famous equation, \(E=mc^2\), where \(E\) is energy, \(m\) is mass, and \(c\) is the speed of light. This energy is known as the nuclear binding energy.
  • Mass defect demonstrates how powerful nuclear forces are, as they tightly bind nucleons together.
  • It is a key concept in understanding nuclear reactions, such as fusion and fission.
  • The mass defect is crucial for calculating binding energy, and further, the stability of the nucleus.
The greater the mass defect, the greater the binding energy, indicating a more stable nucleus.
Nucleons
Nucleons refer to the particles in an atomic nucleus, consisting of protons and neutrons. Together, they are responsible for almost the entire mass of an atom. Understanding the role of nucleons is essential when calculating mass defect and binding energy, as these particles determine the nucleus's behavior and properties.
  • Protons are positively charged particles, while neutrons have no charge.
  • The number of protons in a nucleus defines the chemical element (e.g., carbon, oxygen).
  • The total number of nucleons is called the mass number, which is different from atomic mass.
  • In the notation \(_{4} X^{9}\), \(9\) represents the total nucleon count in the nucleus.
Counting nucleons allows for determining the binding energy per nucleon, which relates to the energy required to remove a nucleon from the nucleus.
MeV (Mega-electron Volts)
Mega-electron Volts (MeV) is a unit of energy commonly used in nuclear physics to quantify binding energies. One MeV is equivalent to one million electron Volts (eV), where an electron Volt is the amount of energy gained or lost by a single electron when moved across an electric potential difference of one volt.
  • MeV is a convenient unit because nuclear energies are typically very large.
  • Given: \(1\) amu = \(921.5\) MeV indicates how mass is transformed into energy.
  • Use of MeV helps to easily compare results with experimental and theoretical predictions in nuclear physics.
When analyzing nuclear reactions, such as the one involving mass defect in this exercise, calculations involving MeV highlight the incredible energy changes in nuclear transformations. It provides a clear measurement to assess how much energy binds nucleons together, indicating nuclear stability and energy changes during reactions.

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Most popular questions from this chapter

A piece of wood from an ancient artifact has a carbon-14 activity of \(11.7\) disintegrations per minute per gram of carbon. Current carbon- 14 activity in fresh samples is \(15.3\) disintegrations per minute per gram of carbon. The half - life of carbon is 5715 years. Which of the statements is true? a. The rate constant for the decay is \(1.21 \times 10^{4}\) years. b. The carbon \(-14\) activity in the freshly cut sample is presumed to be different than in carbon dioxide in the air when doing calculations involving carbon \(-14\) dating c. The artifact could from the age of 2200 years nearly d. The age of the sample is 1270 years.

Half life period of the radioactive element \(X\) is 10 hours. Amount of \(\mathrm{X}\) left of 11 the hour starting with one \(\operatorname{mol} X\) is a. \((1 / 2)^{1 / 10}\) b. \((1 / 2)^{11 / 10}\) c. \((1 / 2)^{12 / 11}\) d. \((1 / 2)^{1 / 11}\)

For a radioactive element with \(\lambda=2.2 \times 10^{-12} \mathrm{~s}^{-1}\), calculate the number of atoms that would show an activity of \(10 \mathrm{mC}\). \(\left[1 \mathrm{C}=3.7 \times 10^{10}\right.\) disintegrations per second] a. \(1.68 \times 10^{21}\) b. \(1.68 \times 10^{20}\) c. \(3.36 \times 10^{21}\) d. \(3.36 \times 10^{22}\)

Match the lists I and II and pick the correct matching from the codes given below. List I List II A. Isotope (p) \(_{38} R a^{228}\) and \({ }_{39} A c^{228}\) B. Isobar(q) (q) \({ }_{18} \mathrm{Ar}^{39}\) and \({ }_{19} \mathrm{~K}^{40}\) C. Isotone (r) \({ }_{1} \mathrm{H}^{2}\) and \({ }_{1} \mathrm{H}^{3}\) D. Isosters (s) \({ }_{92} \mathrm{U}^{235}\) and \({ }_{92} \mathrm{U}^{2 \text { ss }}\) (t) \(\mathrm{CO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}\)

A nucleus with 70 neutrons has a neutron-to proton ration less than one. How can it gain stability? a. Alpha decay b. Beta decay c. Electron capture d. Gamma emission
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