Problem 35 - Chemistry and Chemical Reactivity

Question

Calculate the binding energy per mole of nucleons for calcium- \(40,\) and compare your result with the value in Figure \(25.4 .\) Masses needed for this calculation are (in g/mol) \(_{1}^{1} \mathrm{H}=1.00783,_{0}^{1} \mathrm{n}=1.00867\) and \(_{20}^{40} \mathrm{Ca}=39.96259\)

Step-by-Step Solution

Answer

The binding energy per nucleon for calcium-40 is approximately 8.25 × 10¹¹ J/nucleon.

Step 1: Determine number of nucleons

Calcium-40 ( _{20}^{40}Ca) contains 40 nucleons total, consisting of 20 protons and 20 neutrons.

Step 2: Calculate mass of nucleons if unbound

To find the mass of the nucleons if they are unbound, we calculate: Mass of protons: 20 protons × mass of H = 20 × 1.00783 g/mol = 20.1566 g/mol Mass of neutrons: 20 neutrons × mass of n = 20 × 1.00867 g/mol = 20.1734 g/mol Total mass of unbound nucleons = 20.1566 + 20.1734 = 40.33 g/mol

Step 3: Determine mass defect

The mass defect is the difference between the mass of the nucleons when unbound and the mass of the calcium-40 nucleus: Mass defect = mass of unbound nucleons - mass of _{20}^{40}Ca = 40.33 g/mol - 39.96259 g/mol = 0.36741 g/mol

Step 4: Convert mass defect to energy (binding energy)

Using Einstein’s equation, E = mc², convert the mass defect to energy. First, we convert the mass defect from grams to kilograms: Mass defect = 0.36741 g/mol × (1 kg/1000 g) = 0.00036741 kg/mol. Then, using the speed of light c = 2.998 × 10⁸ m/s: Binding energy = mass defect × c² = 0.00036741 kg/mol × (2.998 × 10⁸ m/s)² = 3.299 × 10¹³ joules/mol.

Step 5: Calculate binding energy per nucleon

Now, we divide the binding energy by the total number of nucleons: Binding energy per nucleon = 3.299 × 10¹³ joules/mol ÷ 40 nucleons = 8.2475 × 10¹¹ joules per nucleon.

Key Concepts

Mass Defect

The mass defect refers to the difference in mass between a nucleus and its constituent nucleons when they are unbound. In simple terms, when protons and neutrons come together to form a nucleus, the total mass of the nucleus is slightly less than the sum of the individual masses of these nucleons. This missing mass is called the mass defect.
Understanding why mass is lost when nucleons bind requires considering nuclear forces. These forces, which bind the nucleons together, are incredibly strong but only act over a very short range. As a result of these forces, some mass is converted to energy, which is the essence of binding energy. The mass defect is a direct measure of how much energy binds the nucleons together within the nucleus.

• Mass of unbound protons and neutrons is their total weight as if they were separate particles.
• The mass defect indicates that mass has been transformed into energy within the nucleus.
• Greater mass defect signifies a more stable nucleus with higher binding energy.

Einstein's Mass-Energy Equivalence

Einstein's Mass-Energy Equivalence is encapsulated in the famous equation \(E = mc^2\). This groundbreaking formula states that energy (E) is equal to mass (m) times the speed of light (c) squared. In the context of nuclear physics, this relationship explains how the mass defect is related to the binding energy that holds a nucleus together.
In a nuclear reaction, the small loss in mass (mass defect) results in a significant release of energy due to the large value of the speed of light squared. This energy is observed as binding energy, which keeps the nucleus stable and bound.

• Binding energy plays a critical role in nuclear reactions, such as fusion and fission.
• It explains why relatively small changes in mass lead to large energy releases.
• The energy released can be calculated by converting the mass defect into energy using \(E = mc^2\).

Nucleons in Calcium-40

Calcium-40 is an isotope with a total of 40 nucleons, split evenly between protons and neutrons. These nucleons are bound together in the nucleus through strong nuclear forces, contributing to the stability and characteristics of the isotope.
Nucleons are the building blocks of an atomic nucleus and their binding is fundamental to the element's properties. In the case of Calcium-40:

• There are 20 protons and 20 neutrons.
• The binding energy per nucleon is a measure of the energy required to remove a single nucleon from the nucleus.
• The more energy required, the more stable the nucleus.

Analyzing the nucleons allows for the evaluation of the nucleus' energy, stability, and potential uses, particularly in scientific and medical applications.