Need help Finding binding energy

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SUMMARY

The discussion focuses on calculating the binding energy of lithium-7 (3Li^7) using the mass defect method. The initial calculation yielded a binding energy of approximately 0.126788 MeV, which was identified as incorrect. Participants highlighted the need to account for the mass defect accurately, emphasizing that 1 atomic mass unit (amu) is equivalent to 931.5 MeV. The correct approach involves ensuring the speed of light (c) is squared in the energy calculation, which should yield a result closer to 6 x 10^-12 J.

PREREQUISITES
  • Understanding of atomic mass units (amu) and their conversion to energy (MeV).
  • Familiarity with the mass defect concept in nuclear physics.
  • Proficiency in applying Einstein's mass-energy equivalence formula (E=mc^2).
  • Basic knowledge of lithium isotopes, specifically lithium-7 (3Li^7).
NEXT STEPS
  • Review the calculation of binding energy using mass defect for different isotopes.
  • Learn about the significance of mass defect in nuclear stability and reactions.
  • Explore the conversion factors between atomic mass units and MeV in detail.
  • Investigate common pitfalls in energy calculations involving nuclear physics.
USEFUL FOR

Students studying nuclear physics, educators teaching atomic structure, and anyone interested in understanding binding energy calculations for isotopes.

Couture
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I need help with this problem: Find Binding energy (in MEV) for lithium 3Li^7. (atomic mass =7.016003u).

3Li^7 (seven is the nucleon number and three is the atomic number= total number of neutons and protons)


This is my attempt:

3 protons ; (7-3)= 4 Neutrons

3(1.007276u) + 4(1.008665u)=7.056488

7.056488 -7.016003u= 0.040485

0.040485u (1.66053888e-27kg/u * (3.00e8)^2=2.0314e-20J
(using equation E= delta mc^2) E=mc^2

2.0314e-20J 1ev/1.6022e-19J = 0.126788 meV


I entered this answer into my online homework, but its says its wrong. And my teacher says for this problem you don't have to account for the electrons when solving for the mass defect (delta m).

Help asap would be greatly appreciated my online homework will be do soon.
thankyou
 
Last edited:
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Couture said:
7.056488 -7.016003u= 0.040485

Looks OK to here.

0.040485u (1.66053888e-27kg/u * (3.00e8)^2=2.0314e-20J
(using equation E= delta mc^2) E=mc^2

2.0314e-20J 1ev/1.6022e-19J = 0.126788 meV

There is plainly something wrong with this because 1 amu is equivalent to 931.5 MeV, so your result should be about 1/25 of that. Check your calculation here again:

0.040485u (1.66053888e-27kg/u * (3.00e8)^2=2.0314e-20J

You should be getting something on the order of 6 x 10^-12 J. (Did you remember to square the c ?)
 

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