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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

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

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