Consider the fusion reaction of a proton and a deuteron: $^{1} \mathrm{H}+_{1}^{2} \mathrm{H} \rightarrow \mathrm{X} .$ (a) Identify the reaction product \(\mathrm{X}\). (b) The binding energy of the deuteron is about \(1.1 \mathrm{MeV}\) per nucleon and the binding energy of "X" is about 2.6 MeV per nucleon. Approximately how much energy (in MeV) is released in this fusion reaction? (c) Why is this reaction unlikely to occur in a room temperature setting?

In a fusion reaction, two lighter nuclei combine to form a heavier nucleus. In this case, we have a proton (\(^1\)H) and a deuteron (\(^2\)H) that are undergoing fusion. To find the product nucleus, we just need to add their nucleon numbers (protons+neutrons) and their atomic numbers (protons). Protons: \(1_1+1_1=1+1=2\) Neutrons: \(\frac{2}{1}+\frac{1}{1}=2+1=3\) So, the reaction product X is a nucleus with 2 protons and 3 nucleons. This is a nucleus of helium with mass number 3, denoted as \(^3\)He. The reaction can be written as: \(^1\mathrm{H}+_1^2\mathrm{H} \rightarrow _2^3\mathrm{He}\)

Given that the binding energy of \(^2\)H is about \(1.1\;\text{MeV}\) per nucleon, and "X" (\(^3\)He) has a binding energy of about \(2.6\;\text{MeV}\) per nucleon, we can calculate the total binding energies for each nucleus: Binding energy of \(^1\)H: 0 (protons have no neutrons bound to them, so no binding energy) Binding energy of \(^2\)H: \(1.1\;\text{MeV}\times2\;\text{nucleons}=2.2\;\text{MeV}\) Binding energy of \(^3\)He: \(2.6\;\text{MeV}\times3\;\text{nucleons}=7.8\;\text{MeV}\)

To calculate the energy released in the fusion reaction, we need to find the difference in total binding energy before and after the reaction: Energy released: \(E_\text{released} = \text{(Binding energy of products)} - \text{(Binding energy of reactants)}\) \(E_\text{released} = 7.8\;\text{MeV} - (0 + 2.2\;\text{MeV}) = 5.6\;\text{MeV}\) So, approximately \(5.6\;\text{MeV}\) of energy is released in this fusion reaction.

At room temperature, the average kinetic energy of particles is very low compared to the energies involved in nuclear reactions (in this case, MeV). As a result, the probability of two nuclei approaching close enough for the strong nuclear force to overcome the electrostatic repulsion and cause a fusion reaction is very small. This is due to Coulomb's barrier, which is the energy barrier that must be overcome by the nuclei to get close enough for the interaction to occur. To increase the likelihood of these reactions, the kinetic energies of the nuclei must be increased, which can be achieved by raising the temperature to very high values, such as in the sun's core or in experimental fusion reactors. In conclusion, this fusion reaction between a proton and a deuteron is unlikely to occur in a room temperature setting because the kinetic energy of the particles at that temperature is too low to overcome the electrostatic repulsion between their positive charges.