Mass Defect: Pb-210 Nuclear Binding Energy

[MrDMD83]

Homework Statement

For lead N=210 Z=82 Pb (atomic mass = 209.984163 u) obtain each of the following:
(a) The mass defect in atomic mass units
u
(b) the binding energy (in MeV)
MeV
(c) the binding energy per nucleon (in MeV)
MeV

Homework Equations

The Attempt at a Solution

a) 82 x 1.6726e-27 kg + 128 x 1.6749e-27 kg= 3.515e-25 kg

3.515e-25kg x 1u/1.6605e-27= 211.708

211.708-209.984163= 1.72u but it says my answer is wrong.

 

[MrDMD83]

Nevermind. I forgot to initially subtract the mass of the 82 electrons from the mass of entire atom.

 

[m2003]

Can you explain why?

b) E=mc^2

= 3.515e-25 kg x (3.00e8 m/s)^2

= 3.16e-8 J

1 J= 6.24e18 MeV

3.16e-8 J x 6.24e18 MeV/1 J= 1.97e11 MeV

c) 1.97e11 MeV/210 nucleons= 9.38e8 MeV per nucleon

The mass defect is correct, but the binding energy and binding energy per nucleon seem to be incorrect. The binding energy should be the difference between the mass of the nucleus and the sum of the masses of its individual protons and neutrons. In this case, it would be the difference between 209.984163u and 211.708u, which is -1.724u. To convert this to MeV, we use the conversion factor 931.5 MeV/u, giving us a binding energy of -1.603 MeV.

For the binding energy per nucleon, we divide the binding energy by the number of nucleons, which in this case is 210. This gives us a value of -0.0076 MeV per nucleon. Note that the negative sign indicates that energy is released when the nucleus is formed, as the mass of the nucleus is less than the sum of its individual particles.