the nuclear binding energy, EB, can be found from the mass defect using Einstein's equation for mass-energy equivalence: EB=delta mc^2, where c is the speed or light (3x10^8m/s). if mass is measured in kilograms and energy in joules, then 1kg <--> 9x10^16J. But in the nuclear domain, masses are often expressed in atomic mass units(1amu=1.66x10^(-27)kg), and energy is expressed in electronvolts (1eV=1.6x10^(-19)J). In terms of these units, the equation for the nuclear binding energy, EB=deltamc^2, can be written as EB(in eV)= delta m (in amu)x 931 MeV
the above paragraph was VERY confusing and I tried for 2 hours to make sense out of it but failed, can you Please explain that paragraph and how does it relate to the below problem, and why do they ignore the speed of light from the EB equation?:
1) the mass defect of helium nucleus is 5x10^(-29)Kg what is the nuclear binding energy?
my answer:
since EB=delta mc^2
then EB= 5x10^(-29)Kg ( 3x10^8m/s)= 1.5x10^(-20)
the correct answer:
the equation EB=delta mc^2 implies that 1kg <--> 9x10^16J, so a mass defect of 5x10^(-29)Kg is equivalent to an energy of (5x10^(-29)Kg) ( 9x10^16J)=4.5x10^(-12)J
thank you very much
ChemBuddy | 
Chem Reddit | 
Topic: what is the nuclear binding energy of Helium nucleus? (Read 19105 times)