The minimum energy required to remove one proton from the nucleus of 14N is calculated using the mass defect method and is found to be 7.545 MeV.
Step-by-step explanation:
To determine the minimum energy needed to remove one proton from the nucleus of 14N, we need to use the concept of nuclear binding energy which is related to the mass defect and energy-mass equivalence (E=mc2). To remove a proton from 14N, the mass of the resulting 13C nucleus and the separated proton should be compared with the original 14N nucleus.
Firstly, we calculate the mass defect:
Mass of 14N: 14.0031 u
Mass of 13C: 13.0034 u
Mass of a proton: 1.0078 u
The total mass of 13C and a proton is 13.0034 u + 1.0078 u = 14.0112 u. The mass defect is then the difference between the total mass of 13C and a proton, and the mass of 14N:
Mass defect = 14.0112 u - 14.0031 u = 0.0081 u
Converting this mass defect into energy:
Energy = Mass defect × 931.5 MeV/c2
Energy = 0.0081 u × 931.5 MeV/c2
Energy = 7.545 MeV (rounded to three decimal places)
This is the minimum energy required to remove a proton from the nucleus of 14N.
The probable question can be: Determine the minimum energy (in MeV) needed to remove just one proton from the nucleus 14N. (Let the mass of a proton be 1.0078 u, the mass of 14N be 14.0031 u, and the mass of 13C be 13.0034 u.
MeV