Problem 85 - Chemistry

Question

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 \(\mathrm{u}\) .

Step-by-Step Solution

Answer

The nuclear binding energy of \(_{12}^{24} \mathrm{Mg}\) can be found by first calculating the mass defect, which is the difference between the sum of individual masses of protons, neutrons, and electrons and the actual atomic mass. Convert the mass defect to kg, then use Einstein's mass-energy equivalence equation, \(E=mc^2\), to find the nuclear binding energy in Joules. Finally, convert the nuclear binding energy to MeV by dividing the energy in Joules by 1.602 × 10^{-13} Joules/MeV.

Step 1: Find the number of protons, neutrons, and electrons

Since it is \(_{12}^{24} \mathrm{Mg}\), it has 12 protons, and as the atomic mass number is 24, the number of neutrons is 24 - 12 = 12. As it is a neutral atom, it also has 12 electrons.

Step 2: Calculate the sum of individual masses

Using the given values for the mass of proton, neutron, and electron, we can calculate the sum of the individual masses. Sum of individual masses = (12 protons × mass of proton) + (12 neutrons × mass of neutron) + (12 electrons × mass of electron) Sum of individual masses = (12 × 1.00728 u) + (12 × 1.00866 u) + (12 × 5.486 × 10^{-4} u)

Step 3: Calculate the mass defect

Subtract the given atomic mass of \(_{12}^{24} \mathrm{Mg}\) from the sum of individual masses to find the mass defect. Mass defect = Sum of individual masses - Atomic mass

Step 4: Convert the mass defect in kilograms

We need to convert the mass defect from atomic mass units (u) to kilograms (kg) before we apply Einstein's mass-energy equivalence equation. One atomic mass unit (u) is equal to \(1.660539 \times 10^{-27}\) kg. Mass defect (kg) = Mass defect (u) × 1.660539 × 10^{-27} kg/u

Step 5: Calculate the nuclear binding energy

Now, we have the mass defect in kilograms, and we can apply Einstein's mass-energy equivalence equation, \(E=mc^2\), to find the nuclear binding energy. Nuclear binding energy = Mass defect (kg) × (Speed of light)^2 Nuclear binding energy (Joules) = Mass defect (kg) × (2.9979 × 10^8 m/s)^2

Step 6: Convert nuclear binding energy to MeV

Finally, convert the nuclear binding energy from Joules to mega electron volts (MeV). One electron volt (eV) is equal to 1.602 × 10^{-19} Joules, so 1 MeV = 1.602 × 10^{-13} Joules. Nuclear binding energy (MeV) = Nuclear binding energy (Joules) / 1.602 × 10^{-13} Joules/MeV

Key Concepts

Mass defect

In the context of nuclear physics, the term **mass defect** refers to the difference between the mass of a completely assembled nucleus and the sum of the masses of its individual protons and neutrons. This difference in mass is crucial because it represents the mass that has been converted into binding energy, holding the nucleus together. Consider a nucleus made up of several protons and neutrons. By calculating the "total expected mass" of these particles when they are separate, and comparing it to the mass of the actual nucleus, we observe a deficit. This missing mass is what we call the mass defect, and it is directly proportional to the binding energy of the nucleus:

• A larger mass defect indicates a greater binding energy, signifying a more stable nucleus.
• Mass defect provides insight into nuclear reactions and energy output.

Understanding mass defect helps us grasp concepts such as fission and fusion, as these processes involve alterations in nuclear structure and the release of significant amounts of energy.

Mass-energy equivalence

Albert Einstein's mass-energy equivalence equation, represented as \(E=mc^2\), forms the fundamental basis for understanding how mass can be converted into energy and vice versa. This equation tells us that a small amount of mass can be transformed into a large amount of energy, given the constant speed of light \(c\) is a large number.

Key Points to Understand

• "\(m\)" refers to the mass defect that gets converted into energy \(E\).
• \(c\) squared magnifies even a tiny mass into considerable amounts of energy, thanks to the magnitude of the speed of light \(\approx 2.9979 \times 10^8\, \text{m/s}\).

This principle is the backbone of nuclear technology, explaining why nuclear reactors and atomic bombs release such enormous energies. When a nucleus changes its configuration, like during nuclear fusion or fission, some mass is transformed into energy, powering the Sun and other stars through fusion processes.The understanding of mass-energy equivalence has revolutionized science by linking mass directly to energy, providing insights not just in nuclear physics but also in cosmic phenomena and energy technologies.

Atomic mass units

To describe the mass of atoms and subatomic particles, we use the atomic mass unit (abbreviated as u). This scale offers a convenient way to express very tiny masses encountered at the atomic level, such as those of protons, neutrons, and electrons.

Understanding Atomic Mass Units

• The atomic mass unit is defined based on the carbon-12 isotope, where one unit is one-twelfth of the mass of a carbon-12 atom.
• The value of 1 atomic mass unit is approximately \(1.660539 \times 10^{-27}\) kg.

This unit simplifies calculations and comparisons when dealing with atomic and nuclear scales, as using kilograms or grams directly would involve impractical decimal places.When performing nuclear calculations, like determining the binding energy of a nucleus, converting atomic mass units to kilograms allows us to apply the mass-energy equivalence principle to obtain energy values in standard units.

Protons and neutrons

Protons and neutrons are the primary building blocks of the atomic nucleus, collectively known as nucleons. They play crucial roles in defining the properties and identity of elements.

Characteristics of Protons and Neutrons

• **Protons** have a positive charge and determine the atomic number, which identifies an element.
• **Neutrons** are neutral, having no charge, and contribute to the atomic mass but not the atomic number.

Both protons and neutrons have similar masses, approximately 1 atomic mass unit each. Neutrons add stability to the nucleus, helping to offset protons' repulsive electrostatic forces due to their positive charges. Understanding the composition of the nucleus is essential for calculating its mass, predicting stability, and performing nuclear reactions. By examining the ratio of protons to neutrons, scientists can predict the stability of an isotope, important in nuclear physics and chemistry for identifying isotopes that may exist in nature or decay over time.