Question

The nuclear masses of \({ }^{7} \mathrm{Be},{ }^{9} \mathrm{Be},\) and \({ }^{10} \mathrm{Be}\) are \(7.0147,9.0100,\) and 10.0113 amu, respectively. Which of these nuclei has the largest binding energy per nucleon?

Short answer

The isotope with the largest binding energy per nucleon is \({ }^{9}\)Be, with a binding energy of approximately 4.809 MeV/nucleon.

Step-by-step solution

Recall the formula for binding energy of a nucleus

The binding energy of a nucleus is given by the mass defect multiplied by the speed of light squared: \[ E = \Delta m \times c^2 \] where \(\Delta m\) is the mass defect and \(c\) is the speed of light.

Calculate the mass defect for each isotope

To calculate the mass defect, we will subtract the nuclear mass from the combined mass of all the protons and neutrons present in the nucleus. For beryllium, we know that it has 4 protons and the number of neutrons varies depending on the isotope. The mass of a single proton is 1.00728 amu and the mass of a single neutron is 1.00867 amu. For \({ }^{7}\)Be: \[ \Delta m = (4 \times 1.00728) + (3 \times 1.00867) - 7.0147 = 0.02919 \text{ amu} \] For \({ }^{9}\)Be: \[ \Delta m = (4 \times 1.00728) + (5 \times 1.00867) - 9.0100 = 0.04647 \text{ amu} \] For \({ }^{10}\)Be: \[ \Delta m = (4 \times 1.00728) + (6 \times 1.00867) - 10.0113 = 0.03939 \text{ amu} \]

Calculate the binding energy for each isotope

Now that we have the mass defects, we can use the formula for binding energy to find the binding energy for each isotope. We will use the conversion factor \(1 \text{ amu} = 931.5 \text{ MeV/c}^2\) to convert the mass defects into MeV. For \({ }^{7}\)Be: \[ E = 0.02919 \times 931.5 = 27.179 \text{ MeV} \] For \({ }^{9}\)Be: \[ E = 0.04647 \times 931.5 = 43.281 \text{ MeV} \] For \({ }^{10}\)Be: \[ E = 0.03939 \times 931.5 = 36.696 \text{ MeV} \]

Calculate the binding energy per nucleon for each isotope

Finally, we will divide the binding energy by the number of nucleons in each isotope. For \({ }^{7}\)Be: \[ \frac{E}{A} = \frac{27.179}{7} = 3.883 \text{ MeV/nucleon} \] For \({ }^{9}\)Be: \[ \frac{E}{A} = \frac{43.281}{9} = 4.809 \text{ MeV/nucleon} \] For \({ }^{10}\)Be: \[ \frac{E}{A} = \frac{36.696}{10} = 3.670 \text{ MeV/nucleon} \]

Identify the isotope with the largest binding energy per nucleon

Comparing the binding energy per nucleon for each isotope: \[ { }^{7} \mathrm{Be}: 3.883 \text{ MeV/nucleon}\\ { }^{9} \mathrm{Be}: 4.809 \text{ MeV/nucleon}\\ { }^{10} \mathrm{Be}: 3.670 \text{ MeV/nucleon} \] We can see that \({ }^{9}\)Be has the largest binding energy per nucleon, which is approximately 4.809 MeV/nucleon.

Key concepts

Nuclear Mass

The concept of nuclear mass is pivotal in understanding atomic nuclei. It refers to the mass of an entire nucleus, including both protons and neutrons, but excluding the electrons. It is often measured in atomic mass units (amu). Each nucleus displays a unique nuclear mass, influenced by the number and type of nucleons (protons and neutrons) it contains. For example, the nuclear masses of beryllium isotopes like

• d{}^{7} Be,
• d{}^{9} Be, and
• d{}^{10} Be, are given in the problem as 7.0147 amu, 9.0100 amu, and 10.0113 amu, respectively. The nuclear mass is less than the sum of separate masses of protons and neutrons due to the binding energy, which holds the nucleus together.

Mass Defect

Mass defect is a key idea that helps explain why nuclei are stable. It represents the difference between the sum of individual masses of protons and neutrons and the actual nuclear mass. This difference arises because some energy is released when nucleons bind together to form a nucleus. According to Einstein's famous equation, \(E = \Delta m \times c^2\), this released energy corresponds to the mass defect. Calculating the mass defect for isotopes of beryllium involves subtracting the nuclear mass from the combined mass of its protons and neutrons. For instance:

• For \({ }^{7}\)Be, the mass defect is calculated as 0.02919 amu.
• For \({ }^{9}\)Be, it is 0.04647 amu.
• For \({ }^{10}\)Be, it is 0.03939 amu.This mass defect corresponds to the binding energy required to hold the nucleus together.

Beryllium Isotopes

Beryllium isotopes are variants of the beryllium element that differ in neutron number. The most common isotopes included in discussions are

• \({ }^{7}\)Be,
• \({ }^{9}\)Be, and
• \({ }^{10}\)Be. All these isotopes have four protons but differ in the number of neutrons.
  • For example, \({ }^{7}\)Be has three neutrons, while \({ }^{9}\)Be has five, and \({ }^{10}\)Be has six. These variations contribute to the differences in nuclear mass and binding energy per nucleon. Understanding isotopes is crucial for interpreting nuclear reactions, stability, and decay processes. Each isotope has a different binding energy per nucleon, affecting its stability.

Proton and Neutron Masses

The mass of a proton is approximately 1.00728 amu, while a neutron is slightly heavier at 1.00867 amu. These masses are fundamental to calculating the nuclear mass and mass defect of an isotope. In any nucleus, protons provide positive charge and are balanced by electrons in the atom, while neutrons add mass without charge, contributing to isotopic variations. When calculating nuclear properties, the total mass is often slightly less than the sum of separate protons and neutrons due to binding energy. This energy reflects the stability of the nucleus and is a crucial factor in nuclear physics. Understanding these masses helps in determining the binding energy and stability of the nucleus, as seen in calculations for beryllium isotopes.