Need help Finding binding energy

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  • #1
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
 
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Answers and Replies

  • #2
dynamicsolo
Homework Helper
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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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