Question

The last step in the carbon cycle that takes place inside stars is \(\mathrm{p}+^{15} \mathrm{N} \rightarrow^{12} \mathrm{C}+(?) .\) This step releases \(5.00 \mathrm{MeV}\) of energy. (a) Show that the reaction product "(?)" must be an \(\alpha\) particle. (b) Calculate the atomic mass of helium-4 from the information given. (c) In order for this reaction to occur, the proton must come into contact with the nitrogen nucleus. Calculate the distance \(d\) between their centers when they just "touch." (d) If the proton and nitrogen nucleus are initially far apart, what is the minimum value of their total kinetic energy necessary to bring the two into contact?

Short answer

Answer: The minimum total kinetic energy required is given by the formula: $$K_\text{min} = k\frac{e^2}{r_0 (1 + 15^{1/3})}$$ where \(k \approx 8.988 \times 10^9 \,\mathrm{N m^2 C^{-2}}\), \(e \approx 1.602 \times 10^{-19} \,\mathrm{C}\), \(r_0 \approx 1.2 \times 10^{-15} \,\text{m}\), and \(d = r_0 (1 + 15^{1/3})\) is the distance between the centers when they just "touch."

Step-by-step solution

(a) Identify the unknown particle in the nuclear reaction

For the given nuclear reaction, we need to identify the unknown product, and conserve the number of protons, and neutrons: Number of protons before reaction = Number of protons after reaction Number of neutrons before reaction = Number of neutrons after reaction Before the reaction, there is \(\mathrm{p}\) with 1 proton and no neutrons, and \(^{15} \mathrm{N}\) with 7 protons and 8 neutrons. Hence, protons before = 1 + 7 = 8 neutrons before = 0 + 8 = 8 After the reaction, there is \(^{12} \mathrm{C}\) with 6 protons and 6 neutrons, and the unknown product. Therefore, unknown protons = 8 - 6 = 2 unknown neutrons = 8 - 6 = 2 Since the unknown particle has 2 protons and 2 neutrons, it must be an \(\alpha\) particle or helium-4 nucleus.

(b) Calculate the atomic mass of helium-4

The energy released in the reaction is \(5.00 \,\mathrm{MeV}\) or \(5.00 \times 10^{6} \, \mathrm{eV}\). Converting the energy released to mass using the mass-energy equivalence relation: $$\Delta m = \frac{\Delta E}{c^2}$$ where \(c \approx 3 \times 10^8 \,\mathrm{m/s}\) (speed of light) and \(\Delta E = 5.00 \times 10^6 \, eV\). Here, remembering that \(1 \,\mathrm{eV} \approx 1.6 \times 10^{-19} \, \mathrm{J}\), we have: $$\Delta m = \frac{(5.00 \times 10^6 \, \mathrm{eV})(1.6 \times 10^{-19} \, \mathrm{J/eV})}{(3 \times 10^8 \, \mathrm{m/s})^2}$$ Now we know that the mass of \(^{15}\mathrm{N}\) and a proton is \(m(^{15}\mathrm{N}) + m(\mathrm{p})\). Also, the mass of \(^{12}\mathrm{C}\) after the reaction is \(m(^{12}\mathrm{C})\). Using the mass-energy equivalence, we can find the mass of \(^{4}\mathrm{He}\): $$m(^4\mathrm{He}) = m(^{15}\mathrm{N}) + m(\mathrm{p}) - m(^{12}\mathrm{C})-\Delta m$$

(c) Calculate the distance between the centers of the proton and nitrogen nucleus when they just "touch"

The distance between the centers of the proton and nitrogen nucleus is the sum of their radii when they just "touch." Using the nuclear radius formula: $$R = r_0 A^{1/3}$$ where \(r_0 \approx 1.2 \times 10^{-15} \,\text{m}\) and \(A\) is the mass number. So for the proton and nitrogen nucleus, we have: $$R_\mathrm{p} = r_0 A_\mathrm{p}^{1/3} = r_0 \times 1^{1/3} = r_0$$ $$R_\mathrm{N} = r_0 A_\mathrm{N}^{1/3} = r_0 \times 15^{1/3}$$ Now, the distance between their centers when they just "touch" is: $$d = R_\mathrm{p} + R_\mathrm{N} = r_0 (1 + 15^{1/3})$$

(d) Calculate the minimum value of the total kinetic energy to bring the proton and nitrogen nucleus into contact

To determine the minimum total kinetic energy, we consider the Coulomb potential energy between the proton and the nitrogen nucleus when they just "touch": $$U(r) = k\frac{e^2}{d}$$ where \(k \approx 8.988 \times 10^9 \,\mathrm{N m^2 C^{-2}}\), \(e \approx 1.602 \times 10^{-19} \,\mathrm{C}\), and \(d\) is the distance between the centers when they just "touch." Therefore, the minimum total kinetic energy required is equal to the potential energy when they just "touch": $$K_\text{min} = U(d) = k\frac{e^2}{r_0 (1 + 15^{1/3})}$$