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The atomic masses of nitrogen-14, titanium-48, and xenon-129 are \(13.999234\) amu, \(47.935878\) amu, and \(128.904779\) amu, respectively. For each isotope, calculate (a) the nuclear mass, (b) the nuclear binding energy, (c) the nuclear binding energy per nucleon.

Short Answer

Expert verified
The nuclear masses for Nitrogen-14, Titanium-48, and Xenon-129 are 13.9954 amu, 47.9299 amu, and 128.8995 amu, respectively. The nuclear binding energies for these isotopes are 127.188 MeV, 275.883 MeV, and 1054.269 MeV, respectively. The nuclear binding energy per nucleon for each isotope is 9.084 MeV (N-14), 5.747 MeV (Ti-48), and 8.173 MeV (Xe-129).

Step by step solution

01

Determine the number of protons and neutrons

Given the isotope notation, we know Nitrogen-14 has 7 protons and 7 neutrons.
02

Calculate the mass of electrons

As there are 7 protons and hence 7 electrons, the mass of electrons in the Nitrogen-14 atom is 7 * (mass of one electron = 0.0005485799 amu) = 0.0038405593 amu.
03

Calculate the nuclear mass

Subtract the total mass of electrons from the atomic mass. In this case, nuclear_mass(N-14) = 13.999234 amu - 0.0038405593 amu = 13.9953934407 amu.
04

Calculate the mass defect

Determine the sum of the individual masses of the protons and neutrons. Then, subtract the nuclear mass from the sum: mass_defect(N-14) = (7 * 1.007276 amu) + (7 * 1.008665 amu) - 13.9953934407 amu = 0.1365585593 amu
05

Calculate the nuclear binding energy

Using the mass defect and the E = mc^2 equation (where c = 931.494 MeV/amu), we can find the nuclear binding energy: binding_energy(N-14) = 0.1365585593 amu * 931.494 MeV/amu = 127.188 MeV
06

Calculate the nuclear binding energy per nucleon

Divide the nuclear binding energy by the total number of nucleons (protons and neutrons): binding_energy_per_nucleon(N-14) = 127.188 MeV / 14 = 9.084 MeV For Titanium-48 and Xenon-129, we can repeat the same process. This involves changing the number of protons, neutrons, and the atomic mass for each isotope. The final answers are as follows: For Titanium-48: Nuclear mass: 47.9298836936 amu Mass defect: 0.2961403064 amu Nuclear binding energy: 275.883 MeV Nuclear binding energy per nucleon: 5.747 MeV For Xenon-129: Nuclear mass: 128.8995461809 amu Mass defect: 1.1314168191 amu Nuclear binding energy: 1054.269 MeV Nuclear binding energy per nucleon: 8.173 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 fundamental concept in nuclear physics. It refers to the difference between the sum of the masses of a system's individual protons and neutrons and the actual mass of the nucleus.
When protons and neutrons come together to form a nucleus, some of their mass is converted into energy, which binds them together. This phenomenon is a result of binding energy, which stabilizes the nucleus. The difference in mass is the mass defect.
  • To calculate the mass defect, first identify the number of protons and neutrons in the nucleus.
  • Next, find their individual masses. For protons, this is approximately 1.007276 amu, and for neutrons, it's about 1.008665 amu.
  • Sum these masses and subtract the nuclear mass from this sum. This gives you the mass defect in atomic mass units (amu).
Understanding mass defect helps explain why the binding energy exists and how much energy would be required to break a nucleus apart, making it a crucial part of nuclear science.
Nuclear Mass
Nuclear mass is the actual mass of an atom's nucleus. It is slightly less than the sum of the masses of the individual protons and neutrons due to the mass defect.
Calculating the nuclear mass requires a few steps:
  • Start with the atomic mass of the atom, which includes the mass of the electrons.
  • Subtract the total mass of the electrons (obtained by multiplying the number of electrons by the mass of a single electron, approximately 0.0005485799 amu).
  • The result is the nuclear mass, which provides insight into the stability and energy characteristics of the nucleus.
Knowing the nuclear mass is essential for calculating the mass defect and, subsequently, the nuclear binding energy.
Atomic Mass
Atomic mass is the total mass of an atom, expressed in atomic mass units (amu). It accounts for the mass of protons, neutrons, and electrons. The atomic mass is significant as it serves as the starting point for determining other important nuclear properties.
It's useful to note:
  • Atomic mass is not always a whole number due to the presence of isotopes, which are different forms of elements with varying neutron numbers.
  • This average mass includes the contributions from all isotopes of an element, weighted by their abundance.
  • To differentiate from nuclear mass, remember that atomic mass includes electron mass as well, albeit it is relatively negligible compared to the total mass of protons and neutrons.
Atomic mass provides a comprehensive picture of the atom's size and can significantly impact calculations related to nuclear chemistry and physics.

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Most popular questions from this chapter

A \(26.00-\mathrm{g}\) sample of water containing tritium, \({ }^{3} \mathrm{H}\), emits \(1.50 \times 10^{3}\) beta particles per second. Tritium is a weak beta emitter with a half-life of \(12.3 \mathrm{yr}\). What fraction of all the hydrogen in the water sample is tritium?

Iodine-131 is a convenient radioisotope to monitor thyroid activity in humans. It is a beta emitter with a half-life of \(8.02\) days. The thyroid is the only gland in the body that uses iodine. A person undergoing a test of thyroid activity drinks a solution of Nal, in which only a small fraction of the iodide is radioactive. (a) Why is NaI a good choice for the source of iodine? (b) If a Geiger counter is placed near the person's thyroid (which is near the neck) right after the sodium iodide solution is taken, what will the data look like as a function of time? (c) A normal thyroid will take up about \(12 \%\) of the ingested iodide in a few hours. How long will it take for the radioactive iodide taken up and held by the thyroid to decay to \(0.01 \%\) of the original amount?

(a) Which of the following statements about the uranium used in nuclear reactors is or are true? (i) Natural uranium has too little \({ }^{295} \mathrm{U}\) to be used as a fuel. (ii) \({ }^{24} \mathrm{U}\) cannot be used as a fucl because it forms a supereritical mass too casily. (iii) To be used as fuel, uranium must be enriched so that it is more than \(50 \%{ }^{2.35} \mathrm{U}\) in composition. (iv) The neutron-induced fission of \({ }^{235} \mathrm{U}\) releases more neutrons per nucleus than fission of \({ }^{2.85} \mathrm{U}\). (b) Which of the following statements about the plutonium shown in the chapter-opening photograph explains why it cannot be used for nuclear power plants or nuclear weapons? (i) None of the isotopes of Pu possess the characteristics needed to support nuclear fission chain reactions. (ii) The orange glow indicates that the only radioactive decay products are heat and visible light. (iii) The particular isotope of plutonium used for RTGs is incapable of sustaining a chain reaction. (iv) Plutonium can be used as a fuel, but only atter it decays to uranium.

Write balanced equations for (a) \({ }_{92}^{238} \mathrm{U}(\alpha, \mathrm{n})_{{ }_{94}^{24}}^{24} \mathrm{Pu}\), (b) \({ }_{7}^{14} \mathrm{~N}(\alpha, \mathrm{p})_{8}^{17} \mathrm{O},(\mathrm{c}){ }_{26}^{56} \mathrm{Fe}\left(\alpha, \beta^{-}\right)_{29}^{60} \mathrm{Cu}\).

One of the nuclides in each of the following pairs is radioactive. Predict which is radioactive and which is stable: (a) \({ }_{19}^{39} \mathrm{~K}\) and \({ }_{19}^{40} \mathrm{~K}\), (b) \({ }^{209} \mathrm{Bi}\) and \({ }^{208} \mathrm{Bi}\), (c) nickel-58 and nickel-65.

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