I think not. Look at http://upload.wikimedia.org/wikiped...inding_energy_curve_-_common_isotopes.svg.png. It shows e.g. that removing one nucleon from C12 will reduce the binding energy by a bit more than one twelfth.desmond iking said:Same?
this is homeworkharuspex said:I think not. Look at http://upload.wikimedia.org/wikiped...inding_energy_curve_-_common_isotopes.svg.png. It shows e.g. that removing one nucleon from C12 will reduce the binding energy by a bit more than one twelfth.
I don't think there's a simple answer to your question. Is it homework/coursework?
That'll certainly work, but the casual wording of the question suggests the author thinks there is some generic route to an answer. Desmond, is this the exact wording, or have you reworded it to make it sound more generic?TSny said:If you remove a proton, what type of nucleus is left? Can you find the binding energy of this nucleus?
The "Energy Req'd to Remove Nucleon from 12_6 C Atom" refers to the amount of energy needed to remove a nucleon (proton or neutron) from a 12_6 C atom, which is a carbon atom with 12 nucleons (6 protons and 6 neutrons).
This energy is important because it determines the stability and binding energy of the nucleus. Higher energy requirements mean that the nucleus is more tightly bound, while lower energy requirements indicate a less stable nucleus.
The energy required is calculated using the mass defect of the nucleus and the binding energy per nucleon. The mass defect is the difference between the mass of the nucleus and the sum of the masses of its individual nucleons. The binding energy per nucleon is the amount of energy needed to hold each nucleon in the nucleus.
The unit of measurement for this energy is electron volts (eV) or mega electron volts (MeV).
The energy required varies for different atoms and isotopes. Generally, larger atoms and heavier nuclei require more energy to remove a nucleon, while smaller atoms and lighter nuclei require less energy.