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\(^{8} \mathrm{Li}\) is an isotope that has a lifetime of less than one second. Its mass is \(8.022485 \mathrm{u} .\) Calculate its binding energy in \(\mathrm{MeV}\).

Short Answer

Expert verified
Based on the step-by-step solution provided, provide a short answer to the question: To calculate the binding energy of Lithium-8, first determine the number of protons and neutrons, which are 3 and 5, respectively. Then, find the total mass of protons and neutrons using their respective masses in atomic mass units. Next, calculate the mass defect by subtracting the mass of the Lithium-8 nucleus from the sum of the masses of its individual protons and neutrons. Finally, convert the mass defect to energy using the mass-energy equivalence formula E=mc² and the conversion factor 1u = 931.4941 MeV/c². This will give you the binding energy of Lithium-8 in MeV.

Step by step solution

01

Determine the number of protons and neutrons

As we are given Lithium-8, it has 8 nucleons in total, with 3 protons (since Lithium has an atomic number of 3) and the rest being neutrons. So there are 5 neutrons in the nucleus.
02

Calculate the mass of individual protons and neutrons

Knowing the atomic mass unit, which is approximately \(1.660539 \times 10^{-27} \mathrm{kg}\) and using the mass of a proton (\(1.007276 \mathrm{u}\)) and the mass of a neutron (\(1.008665 \mathrm{u}\)), we can calculate the mass of 3 protons and 5 neutrons. Total mass of protons: \(3 \times 1.007276 \mathrm{u}\) Total mass of neutrons: \(5 \times 1.008665 \mathrm{u}\)
03

Calculate the mass defect

Now, we will find the mass defect by subtracting the mass of the Lithium-8 nucleus from the sum of the masses of its individual protons and neutrons. Mass defect = (Total mass of protons + Total mass of neutrons) - Mass of Lithium-8 Mass defect = \((3 \times 1.007276 \mathrm{u} + 5 \times 1.008665 \mathrm{u}) - 8.022485 \mathrm{u}\)
04

Convert mass defect to energy

Using the mass-energy equivalence formula, E=mc², where 'E' is energy, 'm' is the mass defect, and 'c' is the speed of light (\(2.99792458 \times 10^8 \mathrm{m/s}\)). We will also use the conversion factor \(1 \mathrm{u} = 931.4941 \mathrm{MeV/c^2}\) in our calculation. Energy = Mass defect × c² Energy = Mass defect × (\(931.4941 \mathrm{MeV/c^2}\))
05

Calculate the binding energy

Now, plug in the mass defect from step 3 into the energy equation from step 4 and solve for the binding energy. Binding energy = (Mass defect) × (931.4941 MeV/c²) Make sure to include units in the final answer.

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Key Concepts

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

Isotope
An isotope refers to any of the different forms of atoms of the same element. These forms have the same number of protons but different numbers of neutrons.

For instance, take Lithium, which normally has 3 protons, as its atomic number is 3. An isotope like \(^8\text{Li}\) is distinguished by its total of 8 nucleons. When we subtract the number of protons, we find that there are 5 neutrons in \(^8\text{Li}\). Different isotopes of the same element behave differently in nuclear calculations, such as binding energy analysis.

Isotopes are crucial because they allow scientists to study atomic behavior in different nuclear and chemical reactions, highlighting the versatility and depth of chemical elements.
Mass Defect
The mass defect is the difference in mass between the measured atomic mass of an isotope and the sum of the individual masses of its constituent protons, neutrons, and electrons. It provides insight into how energy is released during nuclear reactions.

Calculating the mass defect involves summing the masses of protons and neutrons individually and subtracting the actual isotopic mass from this total.
  • The mass of 3 protons: \(3 \times 1.007276 \, \text{u}\)

  • The mass of 5 neutrons: \(5 \times 1.008665 \, \text{u}\)

The difference yields the mass defect, showing the mass converted into binding energy that helps hold the nucleus together. The mass defect is a clear indicator of the stability and binding within an atomic nucleus.
Mass-Energy Equivalence
Mass-energy equivalence is a fundamental concept introduced by Albert Einstein, described by the famous equation \(E=mc^2\). This equation tells us that mass can be converted into energy and vice versa.

In the context of binding energy calculations, the mass defect calculated from the isotopes can be used to find the energy released or required to hold that nucleus together.

To convert the mass defect into energy:
  • Multiply the mass defect by the speed of light squared (\(c^2\)).

This results in energy typically expressed in MeV (mega-electronvolts), a convenient unit in nuclear physics. Understanding mass-energy equivalence is crucial in explaining why nuclear reactions release such large amounts of energy.
Atomic Mass Unit
An atomic mass unit (u) is a unit of mass used to measure atomic and molecular weights, approximately equal to \(1.660539 \times 10^{-27} \, \text{kg}\).

It provides a more practical scale for measuring the masses of atoms and subatomic particles.

The atomic mass unit is a standard in physics and chemistry for denoting the mass on an atomic scale. It is essential in calculating the mass defect and subsequently the binding energy.

Key aspects include:
  • 1 u = 1/12 of the mass of a carbon-12 atom

  • Conveniently used to express mass on a nuclear scale

By using atomic mass units, scientists can readily calculate the differences and transformations in nuclear reactions with ease.

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Most popular questions from this chapter

The mass of an atom (atomic mass) is equal to a) the sum of the masses of the protons. b) the sum of the masses of protons and neutrons. c) the sum of the masses of protons, neutrons and electrons. d) the sum of the masses of protons, neutrons, and electrons minus the atom's binding energy.

Write down equations to describe the alpha decay of the following atoms: a) \(^{212} \mathrm{Rn}\) b) \({ }^{241} \mathrm{Am}\)

Using the table of isotopes in Appendix B, calculate the binding energies of the following nuclei. a) \({ }^{7} \mathrm{Li}\) b) \({ }^{12} \mathrm{C}\) c) \({ }^{56} \mathrm{Fe}\) d) \({ }^{85} \mathrm{Rb}\)

a) What is the energy released in the fusion reaction \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+Q ?\) b) The oceans have a total mass of water of \(1.50 \cdot 10^{16} \mathrm{~kg}\), and \(0.0300 \%\) of this quantity is deuterium, \({ }_{1}^{2} \mathrm{H} .\) If all the deuterium in the oceans were fused by controlled fusion into \({ }_{2}^{4} \mathrm{He},\) how many joules of energy would be released? c) World power consumption is about \(1.00 \cdot 10^{13} \mathrm{~W}\). If consumption were to stay constant and all problems arising from ocean water consumption (including those of political, meteorological, and ecological nature) could be avoided, how many years would the energy calculated in part (b) last?

Refer to the subsection "Terrestrial Fusion" in Section 40.4 to see how achieving controlled fusion would be the solution to mankind's energy problems, and how difficult it is to do. Why is it so hard? The Sun does it all the time (see the previous subsection, "Stellar Fusion"). Do we need to understand better how the Sun works to build a useful nuclear fusion reactor?

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