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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 \(\frac{24}{12} \mathrm{Mg},\) which has an atomic mass of 23.9850 u.

Short Answer

Expert verified
The nuclear binding energy of \(^{24}_{12}\mathrm{Mg}\) is approximately 3.07643 × 10\(^{-11}\) joules.

Step by step solution

01

Determine the number of protons, neutrons, and electrons in the atom

In \(^{24}_{12}\mathrm{Mg}\), the atomic number (number of protons) is 12, and the mass number (protons + neutrons) is 24. Using this information, we can find the number of neutrons and electrons in the atom. We have: Number of protons = 12 Number of neutrons = mass number - atomic number = 24 - 12 = 12 Number of electrons = atomic number (assuming a neutral atom) = 12
02

Calculate the mass of the nucleus

To calculate the total mass of the nucleus, we need to multiply the number of protons and neutrons by their respective masses and add them together. Mass of nucleus = (number of protons × mass of proton) + (number of neutrons × mass of neutron) Mass of nucleus = (12 × 1.00728\(\mathrm{u}\)) + (12 × 1.00866\(\mathrm{u}\)) Mass of nucleus = 24.19128\(\mathrm{u}\)
03

Calculate the mass defect

The mass defect is the difference between the mass of nucleus (which we calculated in Step 2) and the given atomic mass of \(^{24}_{12}\mathrm{Mg}\). Mass defect = mass of nucleus - atomic mass Mass defect = 24.19128\(\mathrm{u}\) - 23.9850\(\mathrm{u}\) Mass defect = 0.20628\(\mathrm{u}\)
04

Convert mass defect to kilograms

1 unified atomic mass unit (\(\mathrm{u}\)) is equal to 1.660539040 × 10\(^{-27}\) kilograms. We need to convert the mass defect from atomic mass units to kilograms. Mass defect (kg) = 0.20628\(\mathrm{u}\) × 1.660539040 × 10\(^{-27}\) kg/\(\mathrm{u}\) Mass defect (kg) = 3.42464 × 10\(^{-28}\) kg
05

Calculate the nuclear binding energy using Einstein's equation

We can now use Einstein's famous equation, \(E=mc^2\), to find the nuclear binding energy. Here, \(E\) is the energy, \(m\) is the mass defect in kilograms, and \(c\) is the speed of light in meters per second. Nuclear binding energy = (mass defect) × (speed of light)\(^2\) Nuclear binding energy = (3.42464 × 10\(^{-28}\) kg) × (2.9979 × 10\(^8\)\(\mathrm{m/s}\))\(^2\) Nuclear binding energy = 3.07643 × 10\(^{-11}\) J Therefore, the nuclear binding energy of \(^{24}_{12}\mathrm{Mg}\) is approximately 3.07643 × 10\(^{-11}\) joules.

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Key Concepts

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

Mass Defect
The term mass defect refers to the difference between the calculated mass of the separated nucleons (protons and neutrons) within an atom's nucleus and the actual mass of the nucleus itself. This discrepancy is vital in understanding the stability of atomic nuclei and is indicative of the binding energy holding the nucleus together.

To put it simply, when nucleons come together to form a nucleus, the mass of the resulting nucleus is less than the total mass of the individual protons and neutrons. This 'missing' mass hasn't vanished – it's been converted into energy during the nuclear binding process, according to the principles of mass-energy equivalence. In the context of our exercise, when calculating mass defect for the magnesium-24 isotope, we find that the total mass of the nucleons when separated is greater than the mass of the nucleus. This 'defect' signifies the amount of energy that would be released if the nucleus were to break apart or that was required to bring it together from its constituent nucleons. From a pedagogical standpoint, it is essential to make students understand that this mass defect is not an error in measurement but a physical occurrence that reflects the stability and energy dynamics of the atomic nucleus.
Atomic Mass Unit
An atomic mass unit (u) is a standard unit of mass that quantifies mass on an atomic or molecular scale. One atomic mass unit is defined as one twelfth the mass of a single carbon-12 atom, at rest and in its ground state. Its exact value is 1.660539040 × 10−27 kilograms.

The atomic mass unit is particularly useful when dealing with the extremely small masses of atoms and subatomic particles because using kilograms would result in impractically small numbers that are difficult to work with. Understanding the concept of atomic mass units is essential for comprehending various phenomena in nuclear chemistry and physics, including the calculation of mass defect and binding energy. When students are solving problems involving nuclear binding energy, such as in the example of magnesium-24, they will often need to convert the atomic mass unit measurements into kilograms to use Einstein's equation appropriately. Ensuring clarity on what an atomic mass unit represents and how to convert it into kilograms is key to avoiding confusion and errors in calculations involving nuclear processes.
Einstein's Equation E=mc^2
Perhaps one of the most famous equations in physics, Einstein's equation E=mc2, encapsulates the principle of mass-energy equivalence. This formula tells us that energy (E) and mass (m) are interchangeable; they are different forms of the same thing. The speed of light (c) squared is the conversion factor that tells us how much energy is contained in a unit of mass.

In relation to nuclear binding energy calculations, this equation is fundamentally important. It allows us to calculate the amount of energy corresponding to the mass defect of a nucleus. As shown in the provided solution for the atomic nucleus of magnesium-24, we use the mass defect (in kilograms) multiplied by the square of the speed of light to find the nuclear binding energy. This calculation tells us how much energy would be released if the nucleus were completely converted into energy, or, equivalently, the energy required to break the nucleus apart into its constituent particles. It's crucial for learners to grasp that while the values for mass defect may seem miniscule, the speeds involved are so vast (the speed of light squared) that the resulting energy is significant. This core principle explains why nuclear reactions release such tremendous amounts of energy – whether it be in stars, nuclear reactors, or atomic weaponry.

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Most popular questions from this chapter

Strontium-90 and radon-222 both pose serious health risks. \(^{90}\) \(\mathrm{Sr}\) decays by \(\beta\) -particle production and has a relatively long half-life (28.9 years). Radon-222 decays by \(\alpha\) -particle production and has a relatively short half-life (3.82 days). Explain why each decay process poses health risks.

Breeder reactors are used to convert the nonfissionable nuclide \(\frac{238}{92} \mathrm{U}\) to a fissionable product. Neutron capture of the \(\frac{238}{92} \mathrm{U}\) is followed by two successive beta decays. What is the final fissionable product?

Define "third-life" in a similar way to "half-life," and determine the "third- life" for a nuclide that has a half-life of 31.4 years.

Zirconium is one of the few metals that retains its structural integrity upon exposure to radiation. The fuel rods in most nuclear reactors therefore are often made of zirconium. Answer the following questions about the redox properties of zirconium based on the half-reaction $$\mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad \mathscr{C}^{\circ}=-2.36 \mathrm{V}$$a. Is zirconium metal capable of reducing water to form hydrogen gas at standard conditions? b. Write a balanced equation for the reduction of water by zirconium. c. Calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\) for the reduction of water by zirconium metal. d. The reduction of water by zirconium occurred during the accidents at Three Mile Island in \(1979 .\) The hydrogen produced was successfully vented and no chemical explosion occurred. If \(1.00 \times 10^{3} \mathrm{kg}\) Zr reacts, what mass of \(\mathrm{H}_{2}\) is produced? What volume of \(\mathrm{H}_{2}\) at 1.0 atm and \(1000 .^{\circ} \mathrm{C}\) is produced? e. At Chernobyl in \(1986,\) hydrogen was produced by the reaction of superheated steam with the graphite reactor core:$$\mathbf{C}(s)+\mathbf{H}_{2} \mathbf{O}(g) \longrightarrow \mathbf{C O}(g)+\mathbf{H}_{2}(g)$$ It was not possible to prevent a chemical explosion at Chernobyl. In light of this, do you think it was a correct decision to vent the hydrogen and other radioactive gases into the atmosphere at Three Mile Island? Explain.

To determine the \(K_{\mathrm{sp}}\) value of \(\mathrm{Hg}_{2} \mathrm{I}_{2},\) a chemist obtained a solid sample of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) in which some of the iodine is present as radioactive \(^{131}\) I. The count rate of the \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) sample is \(5.0 \times 10^{11}\) counts per minute per mole of I. An excess amount of \(\mathrm{Hg}_{2} \mathrm{I}_{2}(s)\) is placed into some water, and the solid is allowed to come to equilibrium with its respective ions. A 150.0 -mL sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute. From this information, calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) $$\mathrm{Hg}_{22}(s) \rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}$$

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