For lead 82Pb207 (atomic mass = 206.975880 u) obtain (a) the mass defect in atomic mass units, (b) the binding energy (in MeV), and (c) the binding energy per nucleon (in MeV).

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## Answers (1)

proton mass = 1.00727638 amu

neutron mass = 1.0086654 amu

Mass of 82 protons + 125 neutrons =

207 + 82*(0.00727638) + 125*(0.0086654) = whatever

Subtract 206.975880 from the "whatever",

and you will have the "mass defect".

(b) E = mc^2

Plug in m = "mass defect" from part (a),

and c = speed of light in a vacuum,

and note that 1 MeV = 1.602 x 10^(-13) joules.

(c) Divide the answer from part (b) by 207