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Chapter 19: Problem 46

Calculate the binding energy per nucleon for \({ }_{1}^{2} \mathrm{H}\) and \({ }_{1}^{3} \mathrm{H}\). The atomic masses are \({ }_{1}^{2} \mathrm{H}, 2.01410 \mathrm{u} ;\) and \({ }_{1}^{3} \mathrm{H}, 3.01605 \mathrm{u}\).

Short Answer

Expert verified
The binding energy per nucleon for \({ }_{1}^{2} \mathrm{H}\) is approximately 0.862 MeV/nucleon, and for \({ }_{1}^{3} \mathrm{H}\), it is approximately 2.661 MeV/nucleon.

Step by step solution

01

Mass defect of \({ }_{1}^{2} \mathrm{H}\)

The mass defect is the difference between the sum of the individual masses of the protons and neutrons and the mass of the nucleus. For \({ }_{1}^{2} \mathrm{H}\), we have 1 proton and 1 neutron. The mass of a proton is 1.00728 u, and the mass of a neutron is 1.00867 u. Mass defect = (mass of proton + mass of neutron) - atomic mass = (1.00728 u + 1.00867 u) - 2.01410 u = 0.00185 u
02

Mass defect of \({ }_{1}^{3} \mathrm{H}\)

For \({ }_{1}^{3} \mathrm{H}\), we have 1 proton and 2 neutrons. Mass defect = (mass of proton + 2 * mass of neutron) - atomic mass = (1.00728 u + 2 * 1.00867 u) - 3.01605 u = 0.00857 u #Step 2: Convert mass defect to energy#
03

Energy of \({ }_{1}^{2} \mathrm{H}\)

Using Einstein's mass-energy equivalence formula \(E=mc^2\), we convert the mass defect from u to MeV. 1 u = 931.5 MeV/c² Energy = 0.00185 u * 931.5 MeV/u = 1.723 MeV
04

Energy of \({ }_{1}^{3} \mathrm{H}\)

Similarly, for \({ }_{1}^{3} \mathrm{H}\), Energy = 0.00857 u * 931.5 MeV/u ≈ 7.982 MeV #Step 3: Calculate the binding energy per nucleon#
05

Binding energy per nucleon for \({ }_{1}^{2} \mathrm{H}\)

Binding energy per nucleon = Total binding energy / Number of nucleons = 1.723 MeV / 2 ≈ 0.862 MeV/nucleon
06

Binding energy per nucleon for \({ }_{1}^{3} \mathrm{H}\)

Binding energy per nucleon = Total binding energy / Number of nucleons = 7.982 MeV / 3 ≈ 2.661 MeV/nucleon So, the binding energy per nucleon for \({ }_{1}^{2} \mathrm{H}\) is approximately 0.862 MeV/nucleon, and for \({ }_{1}^{3} \mathrm{H}\), it is approximately 2.661 MeV/nucleon.

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Most popular questions from this chapter

Natural uranium is mostly nonfissionable \({ }^{238} \mathrm{U} ;\) it contains only about \(0.7 \%\) of fissionable \({ }^{235} \mathrm{U}\). For uranium to be useful as a nuclear fuel, the relative amount of \({ }^{235} \mathrm{U}\) must be increased to about \(3 \%\). This is accomplished through a gas diffusion process. In the diffusion process, natural uranium reacts with fluorine to form a mixture of \({ }^{238} \mathrm{UF}_{6}(g)\) and \({ }^{235} \mathrm{UF}_{6}(g)\). The fluoride mixture is then enriched through a multistage diffusion process to produce a \(3 \%^{235} \mathrm{U}\) nuclear fuel. The diffusion process utilizes Graham's law of effusion (see Chapter 5, Section 5.7). Explain how Graham's law of effusion allows natural uranium to be enriched by the gaseous diffusion process.

In addition to the process described in the text, a second process called the carbon-nitrogen cycle occurs in the sun: $$ \begin{aligned} { }_{1}^{1} \mathrm{H}+{ }_{6}^{12} \mathrm{C} \longrightarrow{ }_{7}^{13} \mathrm{~N}+{ }_{0}^{0} \gamma \\ { }_{7}^{13} \mathrm{~N} & \longrightarrow{ }_{6}^{13} \mathrm{C}+{ }_{+1}^{0} \mathrm{e} \\ { }_{1}^{1} \mathrm{H}+{ }_{6}^{13} \mathrm{C} &{ }_{7}^{14} \mathrm{~N}+{ }_{0}^{0} \gamma \\ { }_{1}^{1} \mathrm{H}+{ }_{7}^{14} \mathrm{~N} \longrightarrow &{ }_{8}^{15} \mathrm{O}+{ }_{0}^{0} \gamma \\ { }_{8}^{15} \mathrm{O} \longrightarrow{ }_{7}^{15} \mathrm{~N}+{ }_{+1}^{0} \mathrm{e} \\ { }_{1}^{1} \mathrm{H}+{ }_{7}^{15} \mathrm{~N} \longrightarrow{ }_{6}^{12} \mathrm{C}+{ }_{2}^{4} \mathrm{He}+{ }_{0}^{0} \gamma \\ \hline \end{aligned} $$ reaction: \(\quad 4{ }_{1}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+2{ }_{+1}^{0} \mathrm{e}\) a. What is the catalyst in this process? b. What nucleons are intermediates? c. How much energy is released per mole of hydrogen nuclei in the overall reaction? (The atomic masses of \({ }_{1} \mathrm{H}\) and \({ }_{2}^{4} \mathrm{He}\) are \(1.00782 \mathrm{u}\) and \(4.00260 \mathrm{u}\), respectively. \()\)

The radioactive isotope \({ }^{247} \mathrm{Bk}\) decays by a series of \(\alpha\) -particle and \(\beta\) -particle productions, taking \({ }^{247} \mathrm{Bk}\) through many transformations to end up as \({ }^{207} \mathrm{~Pb}\). In the complete decay series, how many \(\alpha\) particles and \(\beta\) particles are produced?

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

Iodine-131 is used in the diagnosis and treatment of thyroid disease and has a half-life of \(8.0\) days. If a patient with thyroid disease consumes a sample of \(\mathrm{Na}^{131} \mathrm{I}\) containing \(10 . \mu \mathrm{g}{ }^{131} \mathrm{I}\), how long will it take for the amount of \({ }^{131} \mathrm{I}\) to decrease to \(1 / 100\) of the original amount?
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