91Ó°ÊÓ

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.

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 \(^{4}\mathrm{He}\): $$m(^4\mathrm{He}) = m(^{15}\mathrm{N}) + m(\mathrm{p}) - m(^{12}\mathrm{C})-\Delta m$$

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: $$d = R_\mathrm{p} + R_\mathrm{N} = r_0 (1 + 15^{1/3})$$

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": $$K_\text{min} = U(d) = k\frac{e^2}{r_0 (1 + 15^{1/3})}$$