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Chapter 19: Problem 77

Given the following information: Mass of proton \(=1.00728 \mathrm{u}\) Mass of neutron \(=1.00866 \mathrm{u}\) Mass of electron \(=5.486 \times 10^{-4} \mathrm{u}\) Speed of light \(=2.9979 \times 10^{8} \mathrm{~m} / \mathrm{s}\) Calculate the nuclear binding energy of \({ }_{12}^{24} \mathrm{Mg}\), which has an atomic mass of \(23.9850 \mathrm{u}\).

Short Answer

Expert verified
The nuclear binding energy of \({}_{12}^{24} \mathrm{Mg}\) is \(3.0844 \times 10^{-11}\) joules.

Step by step solution

01

Determine the number of protons and neutrons

The symbol \({}_{12}^{24} \mathrm{Mg}\) represents magnesum-24 with 12 protons and 24 total nucleons (protons + neutrons). To find the number of neutrons, subtract the number of protons from the total number of nucleons: Number of neutrons = 24 - 12 = 12
02

Calculate the total mass of protons and neutrons

Multiply the number of protons (12) by the mass of a proton (1.00728 u) and the number of neutrons (12) by the mass of a neutron (1.00866 u) to determine the total mass: Total mass of protons = 12 * 1.00728 u = 12.08736 u Total mass of neutrons = 12 * 1.00866 u = 12.10392 u Sum of masses = 12.08736 u + 12.10392 u = 24.19128 u
03

Calculate mass defect

The mass defect is the difference between the total mass of protons and neutrons and the given atomic mass of \({}_{12}^{24} \mathrm{Mg}\): Mass defect = 24.19128 u - 23.9850 u = 0.20628 u
04

Convert mass defect to energy

Use Einstein's equation, \(E = mc^2\), to convert the mass defect to energy. First, we need to convert the mass defect from atomic mass units (u) to kilograms. 1 u = \(1.66054 \times 10^{-27}\) kg: Mass defect in kg = \(0.20628 \mathrm{u} \times 1.66054 \times 10^{-27}\) kg/u = \(3.4274 \times 10^{-28}\) kg Now, plug the mass defect in kg and the speed of light into the equation: Nuclear binding energy = \((3.4274 \times 10^{-28}\) kg\() \times \)(\(2.9979 \times 10^{8}\) m/s\()\)^2\) = \(3.0844 \times 10^{-11}\) J The nuclear binding energy of \({}_{12}^{24} \mathrm{Mg}\) is \(3.0844 \times 10^{-11}\) joules.

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Most popular questions from this chapter

There is a trend in the United States toward using coal-fired power plants to generate electricity rather than building new nuclear fission power plants. Is the use of coal-fired power plants without risk? Make a list of the risks to society from the use of each type of power plant.

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \({ }^{68} \mathrm{Ga}\) (electron capture) c. \({ }^{212} \mathrm{Fr}(\alpha)\) b. \({ }^{62} \mathrm{Cu}\) (positron) d. \({ }^{129} \mathrm{Sb}(\beta)\)

Iodine-131 has a half-life of \(8.0\) days. How many days will it take for \(174 \mathrm{~g}\) of \({ }^{131}\) I to decay to \(83 \mathrm{~g}\) of \({ }^{131} \mathrm{I}\) ?

A chemist studied the reaction mechanism for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ by reacting \(\mathrm{N}^{16} \mathrm{O}\) with \({ }^{18} \mathrm{O}_{2}\). If the reaction mechanism is $$ \begin{aligned} \mathrm{NO}+\mathrm{O}_{2} & \rightleftharpoons \mathrm{NO}_{3}(\text { fast equilibrium }) \\ \mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2}(\text { slow }) \end{aligned} $$ what distribution of \({ }^{18} \mathrm{O}\) would you expect in the \(\mathrm{NO}_{2}\) ? Assume that \(\mathrm{N}\) is the central atom in \(\mathrm{NO}_{3}\), assume only \(\mathrm{N}^{16} \mathrm{O}^{18} \mathrm{O}_{2}\) forms, and assume stoichiometric amounts of reactants are combined.

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?
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