Need help Finding binding energy

[Couture]

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

 

[dynamicsolo]

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 ?)

 

[Moetasim]

Hi there,

I can see that you have made a good attempt at solving this problem. However, there are a few things that could be improved upon in your approach.

Firstly, when calculating the mass defect (delta m), you should only consider the mass of the nucleons (protons and neutrons) in the nucleus. Therefore, the correct calculation would be 3(1.007276u) + 4(1.008665u) - 7.016003u = 0.040485u.

Next, to convert this mass defect into energy, you should use the equation E = (delta m)c^2. In your calculation, you have used the total mass (including the mass of the electrons) and then converted it to energy using the equation E = mc^2. This could be why your answer was incorrect.

Finally, to convert the energy into milli-electron volts (meV), you should use the conversion factor of 1.6022e-19J/1meV. This would give you a final answer of 0.1268 meV, which is the correct binding energy for lithium-7.

I hope this helps and good luck with your homework! Remember to always double check your calculations and units to ensure accuracy.