Post a New Question


posted by .

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.

  • Physics -

    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

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question