Problem 23 - Chemistry The Molecular Science

Question

Boron has two stable isotopes, \({ }^{10} \mathrm{~B}\) (abundance \(=\) \(19.78 \%\) ) and \({ }^{11} \mathrm{~B}\) (abundance \(=80.22 \%\) ). Calculate the binding energies per nucleon of these two nuclei and compare their stabilities. The required masses (in \(\mathrm{g} / \mathrm{mol}\) ) are \({ }_{1}^{1} \mathrm{H}=1.00783\); \({ }_{0}^{1} \mathrm{n}=1.00867 ;{ }_{5}^{10} \mathrm{~B}=10.01294 ;\) and \({ }_{5}^{11} \mathrm{~B}=11.00931 .\)

Step-by-Step Solution

Answer

^{11}B is more stable than ^{10}B with a higher binding energy per nucleon of approximately 6.931 MeV compared to 6.475 MeV for ^{10}B.

Step 1: Mass Defect Calculation for 10B

Calculate the mass defect for ^{10}B by finding the difference between its total nuclear mass (sum of protons and neutrons) and its actual mass. The mass of the nucleus should include 5 protons and 5 neutrons.Total mass of free particles = 5 \times 1.00783 \, \text{g/mol} + 5 \times 1.00867 \, \text{g/mol} = 10.0825 \, \text{g/mol}.Mass defect (\Delta m) for ^{10}B = \text{total mass of free particles} - \text{actual mass of \(^{10}\)B} = 10.0825 - 10.01294 = 0.06956 \, \text{g/mol}.

Step 2: Binding Energy Calculation for 10B

Convert the mass defect to energy using Einstein's equation, E=mc², where c is the speed of light (\(3.00 \times 10^8 \ \text{m/s}\)). Remember to convert the mass defect from \(\text{g/mol}\) to \(\text{kg/mol}\) as 1 g = \(10^{-3}\) kg.\[\Delta E = \Delta m \times c^2 = 0.06956 \times 10^{-3} \times (3.00 \times 10^8)^2 \, \text{J/mol}.\]Convert energy from J/mol to MeV/nucleon by factoring in Avogadro's number (\(6.022 \times 10^{23}\) mol⁻¹) and 1 J = \(6.242 \times 10^{12}\) MeV.Binding energy per nucleon for ^{10}B \(\approx \dfrac{0.06956 \times 10^{-3} \times (3.00 \times 10^8)^2 \times 6.242 \times 10^{12}}{10 \times 6.022 \times 10^{23}} \approx 6.475 \, \text{MeV/nucleon}\) (approximate).

Step 3: Mass Defect Calculation for 11B

Calculate the mass defect for ^{11}B similarly, by considering 5 protons and 6 neutrons.Total mass of free particles = 5 \times 1.00783 \, \text{g/mol} + 6 \times 1.00867 \, \text{g/mol} = 11.09051 \, \text{g/mol}.Mass defect (\Delta m) for ^{11}B = \text{total mass of free particles} - \text{actual mass of \(^{11}\)B} = 11.09051 - 11.00931 = 0.08120 \, \text{g/mol}.

Step 4: Binding Energy Calculation for 11B

Similarly, compute the binding energy for ^{11}B.\[\Delta E = \Delta m \times c^2 = 0.08120 \times 10^{-3} \times (3.00 \times 10^8)^2 \, \text{J/mol}.\]Convert energy from J/mol to MeV/nucleon:Binding energy per nucleon for ^{11}B \(\approx \dfrac{0.08120 \times 10^{-3} \times (3.00 \times 10^8)^2 \times 6.242 \times 10^{12}}{11 \times 6.022 \times 10^{23}} \approx 6.931 \, \text{MeV/nucleon}\) (approximate).

Step 5: Comparison of Stabilities

Compare the binding energies per nucleon to evaluate the nuclear stability. A higher binding energy per nucleon generally indicates a more stable nucleus. For ^{10}B, the binding energy per nucleon is approximately 6.475 MeV/nucleon, while for ^{11}B, it is approximately 6.931 MeV/nucleon. Hence, ^{11}B is more stable than ^{10}B.

Key Concepts

Stable Isotopes

Boron is a fascinating element primarily due to its two stable isotopes, \(^{10}B\) and \(^{11}B\). Stable isotopes do not undergo radioactive decay, which means they maintain a constant presence over time without changing into other elements.
These isotopes can coexist naturally owing to their balanced nuclear forces.

• \(^{10}B\), with its lower abundance of approximately 19.78%, has a slightly less stable nucleus.
• \(^{11}B\) is the more common isotope, making up about 80.22% of boron's natural abundance.

The stability of these isotopes affects the element's chemical properties and plays a crucial role in various applications, such as its use in nuclear reactors and certain medical therapies.
The balance of protons and neutrons in the nuclei provides insights into why some isotopes are more stable than others.

Mass Defect

Mass defect is a pivotal concept in nuclear physics, illustrating the mass conservation principle in nuclear reactions. When the nuclei of atomic particles come together to form an atom, not all the mass is retained.
This lost mass, known as mass defect, converts into binding energy that holds the nucleus together.
For instance, in the case of \(^{10}B\):

• The expected total mass of the nucleus, summing the masses of free protons and neutrons, is 10.0825 g/mol.
• Its actual mass is measured as 10.01294 g/mol, showing a mass defect of 0.06956 g/mol.

Similarly, for \(^{11}B\):

• Calculated total mass stands at 11.09051 g/mol, while the actual mass is 11.00931 g/mol.
• This results in a mass defect of 0.08120 g/mol.

This mass defect is crucial in calculating the binding energy and understanding an isotope's nuclear stability.

Nuclear Stability

Nuclear stability is determined predominantly by the binding energy per nucleon. A nucleus is considered more stable when it possesses a higher binding energy per nucleon.
This concept helps in assessing the energy required to separate a nucleus into individual protons and neutrons.
For \(^{10}B\) and \(^{11}B\):

• \(^{10}B\) has a binding energy of approximately 6.475 MeV/nucleon.
• \(^{11}B\), on the other hand, exhibits a higher binding energy of around 6.931 MeV/nucleon.

As a rule of thumb, nuclear stability increases with higher binding energy values. Thus, \(^{11}B\) is significantly more stable than \(^{10}B\).
This superior nuclear stability is why \(^{11}B\) is found in greater abundance naturally.

Boron Isotopes

The element Boron, bearing the atomic number 5, naturally exists primarily as two isotopes: \(^{10}B\) and \(^{11}B\). These isotopes differ in their neutron count while maintaining the same number of protons.
Here are some key characteristics:

• \(^{10}B\) contains 5 protons and 5 neutrons in its nucleus.
• \(^{11}B\) contains 5 protons and 6 neutrons.

The difference in neutron numbers accounts for the disparity in their nuclear properties such as mass and stability.
The variation also influences boron's role in technological and scientific fields, as each isotope has specific applications based on its properties.
For example:

• \(^{11}B\) is often used in nuclear reactors for neutron absorption due to its higher stability.
• \(^{10}B\) finds application in medical therapies, like cancer treatment, by leveraging its neutron capture capabilities.

Thus, understanding these isotopes helps unlock boron's full potential in various industries.