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?

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: $$^{10}\text{B} + ^{4}\text{He} \rightarrow ^{13}\text{C} + ^{1}\text{H}$$

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 \(1.82 \times 10^{25} \text{ MeV}\).