# Binding energy per nucleon of the nucleus

• leminn
leminn
Homework Statement
I'm trying to calculate the binding energy in MeV per nucleon of the nucleus for Sulphur-34. However, I am told that my calculation is wrong and that the mass of the electrons do not have to be taken into account.
Relevant Equations
Binding Energy per nucleon of the nucleus in MeV
How exactly would it then be calculated? Here's what I have tried:
proton mass: 1.007276 amu
Neutron mass: 1.008665 amu
Sulphur-34 mass: 33.9678668 amu

Calculation:

1.##(16\cdot1.007276)+(18\cdot1.008665)+(16 \cdot0.000549)=34.28117##
2.##34.28117-33.9678668=0.3133032##
3. 1 atomic mass unit = ##\frac{931.5 \mathrm{MeV}}{c^2}##
4. ##[E=mc^2]=0.3133032 ×\frac{931.5 MeV}{c^2} =291.840051 (\text{total})##
##\frac{291.840051}{34}=8.583530912\mathrm{MeV}## (per nucleon)

I would really appreciate the communities support on finding my error.

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What's supposed to be wrong with this?

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Well, actually I think your computation is quite OK, indeed I get the result of 0.313301997u for the binding energy.
I don't know who told you that you shouldn't take the masses of the electrons into account, but as far as I understand I think you must use them. At least if the mass you give for sulphur-34 is the atomic mass (and the value for the atomic mass that I have is 33.967867004u, so I think it is).
Of course, if you are using atomic masses, these computations usually neglect the binding energy of the electrons, but those values are usually in the eV range, so I think that can safely be neglected vs the 300MeV you get.

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I get the same: ##8.58 MeV## per nucleon (subtracting the mass of the electrons from the atomic mass).

And, I get ##8.34 MeV## per nucleon (ignoring the mass of the electrons). The mass of the electrons is significant compared the the defective mass and ought to be taken into account.

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I guess the question is, where did you get this: Sulphur-34 mass: 33.9678668 amu? Does it include the electrons or not?

Homework Helper
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I guess the question is, where did you get this: Sulphur-34 mass: 33.9678668 amu? Does it include the electrons or not?
That's the atomic mass; not the mass of the nucleus.

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That's the atomic mass; not the mass of the nucleus.
How do you know that? What is the source? Normally atomic masses are for a mix of all naturally occurring isotopes. Sulfur has 4 - S-32, S-33, S-34, and S-36. So is this the measured atomic mass of isotopically pure S-34? Unless we know how that mass was derived, how do we know whether it includes the electrons or not?

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How do you know that? What is the source? Normally atomic masses are for a mix of all naturally occurring isotopes. Sulfur has 4 - S-32, S-33, S-34, and S-36. So is this the measured atomic mass of isotopically pure S-34? Unless we know how that mass was derived, how do we know whether it includes the electrons or not?
That's the (relative) atomic mass of Sulphur-34; the predominant isotope is Sulphur-32 (95%), so the (average) atomic mass of Sulphur is nearer ##32## amu.

It's here for example:

https://en.wikipedia.org/wiki/Isotopes_of_sulfur

Homework Helper
As I said in my previous post, the atomic mass of S-34 is 33.967867004 u (from Nuclear Data Center at KAERI) this is really similar to the value leminn uses, so I would assume it is the atomic mass. Also there they say that the Binding energy per nucleon is 8.583498 MeV

• PeroK
OK, I'll shut up. I just wanted to know the source of the data and how we knew whether it contained the electrons or not.

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Homework Statement:: I'm trying to calculate the binding energy in MeV per nucleon of the nucleus for Sulphur-34. However, I am told that my calculation is wrong and that the mass of the electrons do not have to be taken into account.
Relevant Equations:: Binding Energy per nucleon of the nucleus in MeV

How exactly would it then be calculated? Here's what I have tried:
proton mass: 1.007276 amu
Neutron mass: 1.008665 amu
Sulphur-34 mass: 33.9678668 amu

Calculation:

1.##(16\cdot1.007276)+(18\cdot1.008665)+(16 \cdot0.000549)=34.28117##
2.##34.28117-33.9678668=0.3133032##
3. 1 atomic mass unit = ##\frac{931.5 \mathrm{MeV}}{c^2}##
4. ##[E=mc^2]=0.3133032 ×\frac{931.5 MeV}{c^2} =291.840051 (\text{total})##
##\frac{291.840051}{34}=8.583530912\mathrm{MeV}## (per nucleon)

I would really appreciate the communities support on finding my error.
I found this excellent summary of binding energy. One of the points made is that this is something of a special case where we need particles masses to a high level of precision and cannot neglect the electrons: