Calculate the binding energy e of the boron nucleus 11 5b (1ev=1.602×10−19j). express your answer in millions of electron volts to four significant figures.

Depends on the precision you're working to. proton mass ~ 1.00728 amu neutron mass ~ 1.00866 amu electron mass ~ electron mass = 0.000549 amu Binding mass is: mass of constituents - mass of atom Eg for nitrogen: (7*1.00728)-(7*1.00866)-(7*0.000549) -14.003074 = 0.11235amu Binding energy is: E=mc^2 where c is the speed of light. Nuclear physics is usually done in MeV[1] where 1 amu is about 931.5MeV/c^2. So: 0.11235 * 931.5 = 104.6MeV Binding energy per nucleon is total energy divided by number of nucleons. 104.6/14 = 7.47MeV This is probably about right; it sounds like the right size! Do the same thing for D/E/F and recheck using your numbers & you shouldn't go far wrong :) 1 - have you done this? MeV is Mega electron Volts, where one electronVolt (or eV) is the change in potential energy by moving one electron up a 1 volt potential. ie energy = charge * potential, so 1eV is about 1.6x10^-19J (the same number as the charge of an electron but in Joules). It's a measure of energy, but by E=mc^2 you can swap between energy and mass using the c^2 factor. Most nuclear physicists report mass in units of MeV/c^2 - so you know that its rest mass energy is that number in MeV.