The proton separation energy is the minimum energy required to remove a single proton from a nucleus. For the nucleus N 14 7 , this separation energy is given as 7 . 452 MeV . If the atomic mass of C 13 6 is 13 . 00335 u and the mass of a hydrogen atom H 1 1 is 1 . 00783 u , what is the atomic mass of N 14 7 ? (Take 1 u = 931 . 5 MeV / c 2 )

Step-by-step solution

The nuclear reaction for the removal of a proton is: N 14 7 + E p → C 13 6 + H 1 1 The proton separation energy E p is the energy equivalent of the mass defect Δ m : Δ m = E p 931 . 5 = 7 . 452 931 . 5 = 0 . 00800   u The mass defect is also given by the difference between the sum of the product masses and the mass of the parent nucleus: Δ m = M ( C 13 6 ) + M ( H 1 1 ) - M ( N 14 7 ) Rearranging to solve for the mass of N 14 7 : M ( N 14 7 ) = M ( C 13 6 ) + M ( H 1 1 ) - Δ m M