Calculate the nuclear binding energy (in \(\mathrm{J}\) ) and the binding energy per nucleon of the following isotopes: (a) \({ }_{3}^{7} \mathrm{Li}(7.01600 \mathrm{amu}),(\mathrm{b}){ }_{17}^{35} \mathrm{Cl}(34.95952 \mathrm{amu})\).

Short Answer

Expert verified
The nuclear binding energy and binding energy per nucleon for isotope \(^{7}_3Li\) are calculated to be roughly 4.86 x 10^-19 J and 6.94 x 10^-20 J/nucleon, respectively. For isotope \(^{35}_{17}Cl\), they are calculated to be roughly 4.88 x 10^-18 J and 1.39 x 10^-19 J/nucleon, respectively.

Step by step solution

01

Calculate the mass of individual nucleons

First, find the individual mass of the protons and neutrons in the isotope. For \(^{7}_3Li\), this is \(3(1.007825 AMU)\) for protons and \(4(1.008665 AMU)\) for neutrons. For \(^{35}_{17}Cl\), this is \(17(1.007825 AMU)\) for protons and \(18(1.008665 AMU)\) for neutrons.
02

Calculate the mass defect

The mass defect is the difference between the mass of individual nucleons and the AMU of the isotope. For \(^{7}_3Li\), this is \((3(1.007825 AMU) + 4(1.008665 AMU)) - 7.01600 AMU\). For \(^{35}_{17}Cl\), this is \((17(1.007825 AMU) + 18(1.008665 AMU)) - 34.95952 AMU\). Convert the mass defect from AMU to kilograms by multiplying by \(1.66053906660 × 10^{-27}kg/AMU\).
03

Calculate the nuclear binding energy

The nuclear binding energy can be found using Einstein's mass-energy equivalence principle, \(E=mc^2\). Multiply the mass defect in kg (from Step 2) by speed of light squared (\(3.00 x 10^8 m/s\)^2). That gives the binding energy in joules.
04

Calculate the binding energy per nucleon

To find the binding energy per nucleon, divide the nuclear binding energy (from Step 3) by the number of nucleons (protons and neutrons combined). For \(^{7}_3Li\), divide by 7, and for \(^{35}_{17}Cl\), divide by 35. This gives the binding energy per nucleon in joules.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Mass Defect
The concept of 'mass defect' refers to the difference between the predicted mass of a nucleus if it were assembled from free protons and neutrons and its actual mass. This discrepancy arises because energy is released when nucleons (protons and neutrons) bind together to form a nucleus, leading to a loss of mass. According to the mass-energy equivalence principle, this energy has mass, and so the combined mass of separate nucleons is actually greater than the mass of the nucleus they form.

The calculation of the mass defect involves summing the masses of the individual protons and neutrons and then subtracting the actual mass of the nucleus. The resulting value is very small and measured in atomic mass units (amu), a specialized unit used to measure atomic mass which reflects the scale and precision needed for these calculations.
Mass-Energy Equivalence

E=mc2 Explained

Mass-energy equivalence is a fundamental concept introduced by Albert Einstein, suggesting that mass and energy are two forms of the same thing and can be converted into each other. This principle is best known by Einstein's famous equation, E=mc2, where E stands for energy, m for mass, and c for the speed of light in a vacuum. The equation shows that even a tiny amount of mass can be converted into a huge amount of energy, as the speed of light squared (c2) is an enormous number.

In the context of nuclear physics, mass-energy equivalence explains how the mass defect corresponds to the binding energy of a nucleus. The energy released when a nucleus forms is equal to its mass defect times the square of the speed of light. When calculating nuclear binding energy, we apply this concept to determine the energy stability of a nucleus.
Binding Energy per Nucleon
The 'binding energy per nucleon' gives us a measure of the stability of a nucleus. It is calculated by dividing the total nuclear binding energy by the number of nucleons (both protons and neutrons) in the nucleus.

A higher binding energy per nucleon generally indicates a more stable nucleus, as it implies that more energy would be required to break the nucleus apart. This concept is important because it helps explain why certain isotopes are more stable than others and why some nuclei are candidates for nuclear power and weapons. The binding energy per nucleon peaks at elements around Iron on the periodic table, which is the reason why heavier elements can release energy through fission and lighter elements through fusion.
Einstein's Equation

Core of Nuclear Physics

Einstein's equation, E=mc2, is the cornerstone of nuclear physics, as it quantifies the relationship between mass and energy. In practice, nuclear binding energy calculations involve this equation to find the energy equivalent of the mass defect. It explains the vast energy contained in atomic nuclei and why nuclear reactions can release so much energy compared to chemical reactions.

By calculating the nuclear binding energy using Einstein's equation, scientists can make crucial predictions about nuclear reactions, such as the energy released in nuclear power plants or the explosive power of nuclear weapons.
Atomic Mass Unit (amu)
The atomic mass unit (amu) is a standard unit of mass that quantifies the mass of atoms or molecules. It is defined as one twelfth of the mass of a carbon-12 atom, with one amu approximately equal to \(1.66053906660 \times 10^{-27}\) kilograms.

Atomic mass units are crucial in nuclear physics because they provide a convenient scale to measure and express very small masses, such as those of protons, neutrons, and entire nuclei. When calculating mass defects and nuclear binding energies, we convert from amu to kilograms to use international system (SI) units, which allows us to apply Einstein's equation accurately.
Isotopes
Isotopes are variants of a particular chemical element that have the same number of protons but different numbers of neutrons in their nuclei, resulting in different mass numbers. These differences in neutron number can significantly affect the stability of a nucleus.

For example, the isotopes of Lithium (\( _3^7Li \) ) and Chlorine (\( _{17}^{35}Cl \) ) mentioned in the exercise have a specific number of protons and neutrons, which dictates their unique mass and stability. The study of isotopes is essential in understanding the variety of nuclear behaviors in nature and in numerous applications ranging from medical imaging to archaeology.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The radioactive isotope \({ }^{238} \mathrm{Pu},\) used in pacemakers, decays by emitting an alpha particle with a half-life of 86 yr. (a) Write an equation for the decay process. (b) The energy of the emitted alpha particle is \(9.0 \times 10^{-13} \mathrm{~J},\) which is the energy per decay. Assuming that all the alpha particle energy is used to run the pacemaker, calculate the power output at \(t=0\) and \(t=10 \mathrm{yr} .\) Initially \(1.0 \mathrm{mg}\) of \({ }^{238} \mathrm{Pu}\) was present in the pacemaker. (Hint: After \(10 \mathrm{yr}\), the activity of the isotope decreases by 8.0 percent. Power is measured in watts or \(\mathrm{J} / \mathrm{s}\).

Consider the decay series $$\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} \longrightarrow \mathrm{D}$$ where \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are radioactive isotopes with halflives of \(4.50 \mathrm{~s}, 15.0\) days, and \(1.00 \mathrm{~s},\) respectively, and \(\mathrm{D}\) is nonradioactive. Starting with 1.00 mole of A, and none of \(\mathrm{B}, \mathrm{C},\) or \(\mathrm{D},\) calculate the number of moles of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) left after 30 days.

Discuss factors that lead to nuclear decay.

Why is it preferable to use nuclear binding energy per nucleon for a comparison of the stabilities of different nuclei?

As a result of being exposed to the radiation released during the Chernobyl nuclear accident, the dose of iodine-131 in a person's body is \(7.4 \mathrm{mCi}\) \(\left(1 \mathrm{mCi}=1 \times 10^{-3} \mathrm{Ci}\right) .\) Use the relationship rate \(=\lambda N\) to calculate the number of atoms of iodine- 131 to which this radioactivity corresponds. (The halflife of \({ }^{131} \mathrm{I}\) is 8.1 d. \()\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free