Mass Defect: Pb-210 Nuclear Binding Energy

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SUMMARY

The discussion focuses on calculating the mass defect and binding energy for lead-210 (Pb-210) with atomic mass 209.984163 u. The mass defect was initially calculated as 1.72 u, but the user realized the error was due to neglecting the mass of the 82 electrons. The correct approach requires subtracting the mass of the electrons from the total atomic mass to obtain accurate results for the binding energy and binding energy per nucleon.

PREREQUISITES
  • Understanding of atomic mass units (u)
  • Knowledge of nuclear binding energy concepts
  • Familiarity with basic nuclear physics equations
  • Ability to perform unit conversions in physics
NEXT STEPS
  • Calculate the binding energy of Pb-210 using the corrected mass defect
  • Learn about the significance of binding energy per nucleon in nuclear stability
  • Explore the concept of mass defect in other isotopes
  • Study the role of electrons in atomic mass calculations
USEFUL FOR

Students in nuclear physics, educators teaching atomic structure, and anyone interested in the calculations related to nuclear binding energy and mass defect.

MrDMD83
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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.
 
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Nevermind. I forgot to initially subtract the mass of the 82 electrons from the mass of entire atom.
 

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