Problem 26 - Chemistry and Chemical Reactivity

Question

Calculate the binding energy per nucleon for calcium-40, and compare your result with the value for calcium-40 in Figure \(23.4 .\) Masses needed for this calculation are $$_{1}^{1} \mathrm{H}=1.00783,_{0}^{1} \mathrm{n}=1.00867, \text { and }_{20}^{40} \mathrm{Ca}=39.96259.$$

Step-by-Step Solution

Answer

The binding energy per nucleon for calcium-40 is 8.56 MeV.

Step 1: Identify the components

To calculate the binding energy per nucleon for calcium-40, we need to identify the components of the nucleus. Calcium-40 ( _{20}^{40}Ca) contains 20 protons and 20 neutrons.

Step 2: Calculate the total mass of separated nucleons

The total mass of separated nucleons is calculated by adding the masses of 20 protons and 20 neutrons.\[\text{Total mass} = 20 imes 1.00783 + 20 imes 1.00867\]\[= 20.1566 + 20.1734 = 40.3300\,\text{u (unified atomic mass units)}\]

Step 3: Calculate the mass defect

The mass defect is the difference between the total mass of the separated nucleons and the actual mass of the calcium nucleus.\[\text{Mass defect} = 40.3300 - 39.96259 = 0.36741\,\text{u}\]

Step 4: Convert the mass defect into energy

Using Einstein's equation \(E = mc^2\), convert the mass defect into energy. The conversion factor for atomic mass units to MeV is approximately \(931.5\,\text{MeV/u}\).\[\text{Binding energy} = 0.36741 \times 931.5 = 342.24\,\text{MeV}\]

Step 5: Calculate the binding energy per nucleon

To find the binding energy per nucleon, divide the total binding energy by the number of nucleons in the calcium nucleus, which is 40.\[\text{Binding energy per nucleon} = \frac{342.24}{40} = 8.56\,\text{MeV}\]

Key Concepts

Mass Defect

Understanding mass defect is essential in nuclear physics. When we calculate the total mass of protons and neutrons (collectively known as "nucleons") within a nucleus, it is typically more than the actual mass of the nucleus itself. This difference is known as the "mass defect."

The mass defect can be understood as a measure of the mass "lost" when the nucleons bind together. This "missing" mass is not really lost but instead converted into binding energy, which holds the nucleus together.

• The mass defect for Calcium-40 ( Ca) was calculated as 0.36741 u (unified atomic mass units).
• This calculation is crucial for determining the energy dynamics within an atomic nucleus.

Nucleons

Nucleons are the building blocks of atomic nuclei, consisting of protons and neutrons. In any given atom, its characteristic stability and identity depend heavily on the number of these particles.

Calcium-40 is an isotope of calcium where the nucleus contains a total of 40 nucleons. Divided equally, it has 20 protons and 20 neutrons.

• A proton is a positively charged particle, while a neutron is neutral.
• The overall mass and energy characteristics of the nucleus greatly depend on the individual masses of these nucleons.

Energy Conversion

Energy conversion in the context of mass defect is explained by Einstein's famous equation, \(E = mc^2\). This equation tells us that mass can be converted into energy and vice-versa.

In nuclear physics, the mass defect is converted into binding energy. This energy represents how strongly the nucleons are held together in the nucleus.

• The conversion factor used is approximately 931.5 MeV/u, which helps in translating mass in atomic mass units to energy in megaelectronvolts (MeV).
• This conversion yielded 342.24 MeV of total binding energy for the calcium nucleus.

Calcium-40

Calcium-40, denoted as \(^{40}_{20}{Ca}\), is a stable isotope of the chemical element calcium. It contains 20 protons and 20 neutrons within its nucleus.

This isotope is often used in calculations of nuclear properties due to its balanced nucleon count and resultant stability.

• In the context of binding energy calculations, Calcium-40's mass is crucial for determining the mass defect and subsequent energy.
• The exercise showed that the binding energy per nucleon for Calcium-40 was 8.56 MeV, showcasing its nuclear stability.