Calculate the difference between the binding energy of a nucleus of carbon 12 and the sum of the binding energies of three alpha particles. Assuming the carbon 12 is composed of three alpha particles in a triangular structure, with three effective "alpha bonds" joining them, what is the binding energy per alpha bond?
I have calculated the binding energy for the alpha particles using the liquid drop model and using BE = (A-Z)mn + Zmp - ma (where mn is the mass of the neutron, mp that of the proton and ma that of the atom in question). When I do it the latter way, I get around 23 MeV, and when I do it the former way I get around 28 MeV which according to my googling is the corrent answer. I'm wondering why this would be so?
The Attempt at a Solution
Also, assuming we use the first equation for BE, then the different between the two binding energies is 0 (since we would have 3 times the BE for an alpha particle equals the binding energy of carbon 12). This doesn't seem right to me.
Any comments or advice would be appreciated.