Consider the induced nuclear reaction 1H2 + 7N14 ---->6C12 + 2He4. The atomic masses are 1H2 (2.014102 u), 7N14 (14.003074 u), 6C12 (12.000000 u), 2He4 (4.0026030 u). Determine the energy (in MeV) released when the 6C12 and 2He4 nuclei are formed in this manner.
Reactant mass = 2.014102+14.003074 = 16.017176 amu
Product mass = 16.002603 amu
Mass loss = 0.014573 amu
= 2.419*10^-29 kg
Multiply by c^2, in m^2/s^2, for the energy release in Joules. Then convert that to MeV
To determine the energy released in this nuclear reaction, we can use Einstein's mass-energy equivalence equation: E = Δm * c^2, where E is the energy released, Δm is the change in mass, and c is the speed of light.
To find the change in mass, we need to subtract the total mass of the reactants from the total mass of the products.
Total mass of reactants = mass of 1H2 + mass of 7N14 = 2.014102 u + 14.003074 u = 16.017176 u.
Total mass of products = mass of 6C12 + mass of 2He4 = 12.000000 u + 4.0026030 u = 16.002603 u.
Δm = Total mass of products - Total mass of reactants = 16.002603 u - 16.017176 u = -0.014573 u.
Now, we need to convert the change in mass to kilograms since the speed of light is specified in meters per second (m/s). 1 atomic mass unit (u) is equal to 1.66053906660 x 10^-27 kg.
Δm in kilograms = -0.014573 u * (1.66053906660 x 10^-27 kg/u) = -2.41867 x 10^-29 kg.
Substituting the values into the equation E = Δm * c^2, we can calculate the energy released.
E = (-2.41867 x 10^-29 kg) * (3.0 x 10^8 m/s)^2.
E = -2.41867 x 10^-29 kg * (9.0 x 10^16 m^2/s^2).
E = -2.17685 x 10^-12 J.
To convert this energy to MeV, we can use the conversion factor: 1 MeV = 1.60218 x 10^-13 J.
Energy in MeV = (-2.17685 x 10^-12 J) / (1.60218 x 10^-13 J/MeV).
Energy in MeV = -13.57 MeV.
Therefore, the energy released when the 6C12 and 2He4 nuclei are formed in this reaction is approximately 13.57 MeV.