Finding binding energy and converting into MeV

  • #1
Question and formulas
The nucleus of a Sulphur atom has an atomic mass of 32 and an atomic number of 16. If the mass of this atomic nucleus is 31.972072 amu (atomic mass unit), find its binding energy in MeV.
Table of conversions/constants
mass of proton 1.007826
mass of neutron 1.008665
Speed of light 2.99792458 x 10^8
1 amu = 1.6606x 10^-27 kg
1 Mev= 1.6022x 10^-13 J
E=mc^2, Be= (#n)(mass of n) + (#p)(mass of p) - (nucleus's mass)

Attempt at the solution
We have its amount of neutrons and protons, so we multiply 16p with 1.007826 and add it with 16n times 1.008665. We get 32.263856 and subtract it by 31.972072 amu to get its mass defect (0.291784). Afterwards we multiply it to get its binding energy using the formula E=mc^2 . So E= (0.291784) (1.6606x10^-27) (2.99792458 x 10^8) ^2 to get it into joules. Then we divide it by 1.6022x 10^-13 J to turn it into MeV. However i end up with the result 271.8010848 MeV but my online assignment says its wrong help?
 

Answers and Replies

  • #2
gneill
Mentor
20,912
2,860
Your proton mass in amu looks a bit high. Can you check that value?
 
  • #3
Your proton mass in amu looks a bit high. Can you check that value?
Thats how it was given in the table of conversion, i cant change it myself, i double checked its still the same :/
 
  • #4
gneill
Mentor
20,912
2,860
Thats how it was given in the table of conversion, i cant change it myself, i double checked its still the same :/
I understand. These things happen.

If you Google "proton mass amu" you'll see a different value :)

Of course, it could be that the authors of the problem set decided to finagle the constant a bit to make a "new" answer set, but that's pretty unlikely.

Your approach seems to be fine to me, so I went looking for discrepancies.
 
  • #5
I understand. These things happen.

If you Google "proton mass amu" you'll see a different value :)

Of course, it could be that the authors of the problem set decided to finagle the constant a bit to make a "new" answer set, but that's pretty unlikely.

Your approach seems to be fine to me, so I went looking for discrepancies.
Oh okay thanks good to know then ill ask my physics teacher tomorrow then thanks!! :)
 
  • #6
gneill
Mentor
20,912
2,860
Oh okay thanks good to know then ill ask my physics teacher tomorrow then thanks!! :)
Good plan. You might, if you have any attempts left, try the established value of 1.00726 amu for the proton mass in your calculations to confirm or refute the suspicion of "typographical incident" :)
 

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