Calculate the minimum kinetic energy (in megaelectronvolts) that \(\alpha\) particles must possess to produce the nuclear reaction $$_{2}^{4} \mathrm{He}+^{14}_{7} \mathrm{N} \longrightarrow^{17}_{8} \mathrm{O}+_{1}^{1} \mathrm{H}.$$ The nuclidic masses are \(_{2}^{4} \mathrm{He}=4.00260 \mathrm{u}\); \(_{7}^{14} \mathrm{He}=14.00307\mathrm{u}\);\(_{1}^{1} \mathrm{H}=1.00783 \mathrm{u}\);\(_{8}^{17} \mathrm{H}=16.99913 \mathrm{u}\);

First, we need to calculate the mass energy of each particle, because energy is conserved in nuclear reactions. The mass energy can be calculated by using the relation E=m*c², where m is the mass of particle and c is the speed of light. The mass of particles must be converted into kg from atomic mass unit (u). The energy obtained will be in joules. As speed of light c = \(3 \times 10^8\) m/s, and 1 u = \(1.66 \times 10^{-27}\) kg, we find the energy of each particle: \(E_{^{4}He} = 4.00260 u \times 1.66 \times 10^{-27} kg/u \times (3 \times 10^{8} m/s)^2 = 5.582 \times 10^{-10}\) J, \(E_{^{14}N} = 14.00307 u \times 1.66 \times 10^{-27} kg/u \times (3 \times 10^{8} m/s)^2 = 1.5602 \times 10^{-9}\) J, \(E_{^{1}H} = 1.00783 u \times 1.66 \times 10^{-27} kg/u \times (3 \times 10^{8} m/s)^2 = 1.411 \times 10^{-10}\) J, \(E_{^{17}O} = 16.99913 u \times 1.66 \times 10^{-27} kg/u \times (3 \times 10^{8} m/s)^2 = 1.8911 \times 10^{-9}\) J

The mass defect is the difference between the initial mass and final mass. It can be calculated as ∆m = (mass of reactants – mass of products) = (total energy of reactants - total energy of products)= ((5.582 \times 10^{-10}) J + (1.5602 \times 10^{-9}) J) - ((1.411 \times 10^{-10}) J + (1.8911 \times 10^{-10}) J) = 4.64 \times 10^{-10} J

The energy calculated is in joules, but in nuclear reactions, the energy is usually represented in electronvolts (eV), specifically in megaelectronvolts (MeV). To convert J to MeV, 1 J = \(6.242 \times 10^{12}\) MeV. The energy in MeV is thus given by 4.64 \times 10^{-10} J * \(6.242 \times 10^{12}\) MeV/J = 289.61 MeV