Problem 43 - Chemistry: Principles and Reactions

Question

Consider the fusion of boron- 10 with an alpha particle. The products of the fusion are carbon-13 and a proton. (a) Write a nuclear reaction for this process. (b) How much energy is released when \(1.00 \mathrm{~g}\) of \(\mathrm{B}-10\) is fused with an \(\alpha\) -particle?

Step-by-Step Solution

Answer

Answer: The nuclear reaction for the fusion process of boron-10 with an alpha particle is: $$^{10}\text{B} + ^{4}\text{He} \rightarrow ^{13}\text{C} + ^{1}\text{H}$$ The energy released when 1 gram of B-10 is fused with an alpha particle is approximately \(1.82 \times 10^{25} \text{ MeV}\).

Step 1: (a) Writing the nuclear reaction

For this fusion process, we have boron-10 and an alpha particle on the reactant side. Boron-10 is represented by \(^{10}\text{B}\) and an alpha particle is represented by \(^{4}\text{He}\). The products of the reaction are carbon-13 and a proton, represented by \(^{13}\text{C}\) and \(^{1}\text{H}\), respectively. The nuclear reaction can be written as: $$^{10}\text{B} + ^{4}\text{He} \rightarrow ^{13}\text{C} + ^{1}\text{H}$$

Step 2: (b) Calculating the energy released

We will now use mass-energy equivalence to find the energy released per fusion reaction. The mass-energy equivalence formula is: $$E = (m_{i} - m_{f})c^2$$ where \(E\) is the energy released, \(m_{i}\) is the initial mass, \(m_{f}\) is the final mass, and \(c\) is the speed of light. To find the energy released per fusion reaction, we need the masses of the reactants and products: - Mass of boron-10: \(m_{B-10} = 10.01294 \text{ u}\) - Mass of alpha particle: \(m_{He-4} = 4.001506 \text{ u}\) - Mass of carbon-13: \(m_{C-13} = 13.003355 \text{ u}\) - Mass of proton: \(m_{H-1} = 1.007825 \text{ u}\) Calculating the initial mass \(m_{i}\) and the final mass \(m_{f}\): $$ m_{i} = m_{B-10} + m_{He-4} = 10.01294 \text{ u} + 4.001506 \text{ u} = 14.014446 \text{ u} $$ $$ m_{f} = m_{C-13} + m_{H-1} = 13.003355 \text{ u} + 1.007825 \text{ u} = 14.011180 \text{ u} $$ Now, we can calculate the energy released per fusion reaction using mass-energy equivalence formula: $$ E = (m_{i} - m_{f})c^2 = (14.014446 \text{ u} -14.011180 \text{ u}) (931.5 \frac{\text{MeV}}{\text{u}}) = 3.02 \text{ MeV} $$ Next, we need to find the number of fusion reactions when 1 gram of B-10 is fused with an alpha particle. Since 1 mole of B-10 is approximately 10 grams, the number of B-10 atoms in 1 gram is: $$ N = \frac{1 \text{ g}}{10 \text{ g/mol}} \times \text{Avogadro's number} $$ $$ N = 0.1 \times 6.022 \times 10^{23} \text{ atoms} = 6.022 \times 10^{22} \text{ atoms} $$ Finally, we can calculate the total energy released when 1 gram of B-10 undergoes fusion by multiplying the energy released per reaction by the number of fusion reactions: $$ E_{total} = E \times N = 3.02 \text{ MeV} \times 6.022 \times 10^{22} \text{ atoms} = 1.82 \times 10^{25} \text{ MeV} $$ So, the energy released when 1 gram of B-10 is fused with an alpha particle is approximately \(1.82 \times 10^{25} \text{ MeV}\).

Key Concepts

Mass-Energy Equivalence

The concept of mass-energy equivalence is a cornerstone of modern physics, derived from Einstein's famous equation, \[ E = mc^2 \]. This equation tells us that mass and energy are interchangeable; a small amount of mass can be converted into a large amount of energy because the speed of light, \( c \), is a very large number.
This principle is particularly important in nuclear reactions, where mass differences between reactants and products translate into energy release or absorption. In our example, the slight mass difference between the boron-10 and alpha particle compared to their fusion products, carbon-13 and a proton, is converted into energy.
Understanding this concept helps us appreciate how nuclear fusion processes can release massive amounts of energy, which is the principle behind both the power of the Sun and potential nuclear energy sources on Earth.

Nuclear Reactions

Nuclear reactions involve changes in an atom's nucleus, leading to the transformation of elements. Unlike chemical reactions that deal with electrons and bonds, nuclear reactions require a deeper delve into protons and neutrons.
In our exercise, boron-10 fuses with an alpha particle (a helium nucleus) to form carbon-13 and a proton. The nuclear reaction equation is written as:

• \(^{10}\text{B} + ^{4}\text{He} \rightarrow ^{13}\text{C} + ^{1}\text{H}\)

This equation must show conservation of both mass number and charge—essential to ensuring the laws of physics are satisfied.
Nuclear reactions like this fusion process are the key behind the energy produced in stars, including our Sun. Understanding these principles gives us insight into both natural and human-made processes that harness or demonstrate nuclear energy.

Energy Release Calculation

Calculating the energy release in a nuclear reaction involves assessing the mass difference between reactants and products and converting that difference into energy.
In our example, the initial mass is the sum of the boron-10 and alpha particle:

• Initial mass \( m_i = 14.014446 \text{ u} \)

The final mass is the sum of the carbon-13 and proton:

• Final mass \( m_f = 14.011180 \text{ u} \)

By using the mass-energy equivalence formula \( E = (m_i - m_f)c^2 \), the energy released per reaction is calculated to be 3.02 MeV.
To find the total energy released when 1 gram of boron-10 undergoes fusion, we convert the mass to moles and then calculate the number of reactions. As 1 mole of boron-10 is approximately 10 grams, 1 gram represents \(6.022 \times 10^{22}\) atoms.
Finally, multiplying the energy per reaction by the number of atoms, we find the total energy release is approximately \(1.82 \times 10^{25}\) MeV. This calculation emphasizes how even small reactions can result in significant energy output, highlighting the power of nuclear processes.