Calculating the weight of 1 atomic mass unit

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  • #1
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I'm trying to deduce the weight of 1 atomic mass unit (##1u##) in ##kilograms## from the following scenario:

One atomic mass unit is 1/12 of the weight of a ##^{12}C## atom in its ground state. A ##^{12}C## atom consists of 6 protons, 6 neutrons and 6 electrons. This means that
$$1u = \frac{1}{12} \cdot mass(6 protons + 6 neutrons + 6 electrons) = \frac{1}{2} \cdot (m_p + m_n + m_e)$$
The masses of the proton, neutron and electron in kg are:

Proton: ##1.67262189821 \cdot 10^{-27} kg##
Neutron: ##1.67492747121 \cdot 10^{-27} kg##
Electron: ##9.1093835611 \cdot 10^{-31} kg##

Filling these weights in the above formula shows that ##1u = 1.67423015389 \cdot 10^{-27} kg##. However, Wiki shows that ##1u = 1.66053904020 \cdot 10^{−27} kg## which already differs from the second decimal. What is the cause of this difference?
 

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  • #2
Drakkith
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The sum of the masses of 6 free particles of each type is more than the mass of the Carbon-12 atom. This is because the binding together of these particles releases energy from the system (atom), requiring that it have less mass than the total mass of the free particles. This is known as binding energy. I haven't crunched the numbers, but I'd bet that's the reason for the discrepancy.
 
  • #3
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The sum of the masses of 6 free particles of each type is more than the mass of the Carbon-12 atom. This is because the binding together of these particles releases energy from the system (atom), requiring that it have less mass than the total mass of the free particles. This is known as binding energy. I haven't crunched the numbers, but I'd bet that's the reason for the discrepancy.
So this would be an example of mass->energy conversion through
E2=p2c2 + m2c4
?
 
  • #4
Borek
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[itex]E=mc^2[/itex] is perfectly enough.
 
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  • #5
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The sum of the masses of 6 free particles of each type is more than the mass of the Carbon-12 atom. This is because the binding together of these particles releases energy from the system (atom), requiring that it have less mass than the total mass of the free particles. This is known as binding energy. I haven't crunched the numbers, but I'd bet that's the reason for the discrepancy.
Thanks for the clarification. Is there a way to prove that my concluded difference in mass is exactly the binding energy release? For example, by looking at how much energy is needed per binding?
 
  • #6
Borek
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Doable, but not straightforward, as it would require finding energies released in several reactions yielding the C12 nucleus (it is not a single step 6p + 6n → C12, google for nuclear synthesis of C12). Typically we do it the other way around - we calculate the binding energy from the mass deficit, as this is much easier.
 
  • #7
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Doable, but not straightforward, as it would require finding energies released in several reactions yielding the C12 nucleus (it is not a single step 6p + 6n → C12, google for nuclear synthesis of C12). Typically we do it the other way around - we calculate the binding energy from the mass deficit, as this is much easier.
Thanks, I'll try and see what I can get. One other question, does the weight of a proton, neutron or electron differ even further depending on how many bindings there are in an atom? For example, a proton has a lower mass than its rest mass when it's bounded in an atom with 6 protons, 6 neutrons and 6 electrons. Does a proton have an even lower mass when it's bounded in an atom with 12 protons, 12 neutrons and 12 electrons?
 
  • #8
Borek
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Thanks, I'll try and see what I can get. One other question, does the weight of a proton, neutron or electron differ even further depending on how many bindings there are in an atom? For example, a proton has a lower mass than its rest mass when it's bounded in an atom with 6 protons, 6 neutrons and 6 electrons. Does a proton have an even lower mass when it's bounded in an atom with 12 protons, 12 neutrons and 12 electrons?
Number of electrons doesn't matter for the nucleus mass.

Yes, the mass deficit is a function of the nucleus size, it has maximum value around nuclei containing 56 nucleons. For larger nuclei the binding energy per nucleon becomes lower again (which is why large nuclei can undergo fission).

500px-Binding_energy_curve_-_common_isotopes.svg.png

(image from wikipedia)
 
  • #9
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Number of electrons doesn't matter for the nucleus mass.

Yes, the mass deficit is a function of the nucleus size, it has maximum value around nuclei containing 56 nucleons. For larger nuclei the binding energy per nucleon becomes lower again (which is why large nuclei can undergo fission).

500px-Binding_energy_curve_-_common_isotopes.svg.png

(image from wikipedia)
Very interesting to read its Wiki page! So I've found the binding energy of a Carbon-12 atom on Wiki which is 92161.753 ± 0.014 keV. I'd assume that this is the energy-mass equivalent to my calculated mass difference.

My calculated mass difference in my OP is:

$$1.67423015389 \cdot 10^{−27} kg - 1.66053904020 \cdot 10^{-27} kg = 1.369111369 \cdot 10^{-29} kg$$

The released bindingenergy has a mass-equivalent of:

$$\frac{92161753 eV \cdot (1.60217656535 \cdot 10^{-19} J/eV)}{c^2} = 1.640660010 \cdot 10^{-28} kg$$

This is already a factor of 10 difference. What am I missing? Should I instead use the net energy release during the triple alpha process?

EDIT: When using the net energy release during the triple alpha process instead, which is 7.273 MeV, I'd get a mass difference of ##1.296529960 \cdot 10^{-27} kg##, which is even a factor of 100 difference. Am I doing something wrong?
 
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  • #10
mjc123
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Your calculated difference is what you worked out for 1u. 12C is 12u. So multiply your value by 12...
 
  • #11
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Your calculated difference is what you worked out for 1u. 12C is 12u. So multiply your value by 12...
Of course! Didn't pay attention to that. Dividing the mass-equivalent of the binding energy 92161753 eV by 12 would give ##1.367216675 \cdot 10^{-29} kg## while my calculated difference is ##1.369111369 \cdot 10^{-29} kg##.
I might be very nitpicky here but is there an explanation other than inacurracy why there's a difference of ##0.001894694 \cdot 10^{-29}##? I can see that the measured binding energy has an inacurracy range of +/- 0.014 keV. Perhaps this is the cause..
 
  • #12
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Number of electrons doesn't matter for the nucleus mass
I have one question regarding if we're talking about the total mass of an atom (not just the nucleus);

Doesn't the mass of an electron change depending on how many electrons there are in an atom and how massive the nucleus is? The cause being that these 2 factors would result in different velocities for an electron and thus a different (relativistic) mass?
 
  • #13
DrClaude
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Doesn't the mass of an electron change depending on how many electrons there are in an atom and how massive the nucleus is? The cause being that these 2 factors would result in different velocities for an electron and thus a different (relativistic) mass?
The mass of an electron doesn't change ("relativistic mass" is an outdated concept). But yes, different electrons in the atom have different momenta, so they contribute unequally to the mass of the atom.
 
  • #14
DrClaude
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I can see that the measured binding energy has an inacurracy range of +/- 0.014 keV. Perhaps this is the cause..
Try and calculate that. It's a good exercise.
 
  • #15
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different electrons in the atom have different momenta, so they contribute unequally to the mass of the atom.
Aren't these different momenta caused by the individual velocities, though?
 
  • #16
Borek
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Aren't these different momenta caused by the individual velocities, though?
Velocity is a classic concept, which doesn't apply here easily. It made sense in a planetary model of an atom, but it doesn't work for an orbital (being better described as a standing wave).
 
  • #17
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Velocity is a classic concept, which doesn't apply here easily. It made sense in a planetary model of an atom, but it doesn't work for an orbital (being better described as a standing wave).
Was about to ask the same question as Comeback City. So does a planetary model of an atom, wherein the mass of an electron is calculated relativistically, not give an accurate total mass of the atom?

Another question that I'm curious about: Does the released binding energy quantity include the compensation for the increase of total atom mass because of the orbiting electrons having momenta?
 
  • #18
Borek
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Try to estimate what fraction of the atom mass is given by its electrons, and try to estimate by how much would the mass of an atom change when the electron gets excited (you can use ionization energies for that - first ionization is in a way equivalent to the maximum amount of energy that the atom can absorb while getting excited). Compare these numbers with the atom mass.
 
  • #19
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Try to estimate what fraction of the atom mass is given by its electrons, and try to estimate by how much would the mass of an atom change when the electron gets excited (you can use ionization energies for that - first ionization is in a way equivalent to the maximum amount of energy that the atom can absorb while getting excited). Compare these numbers with the atom mass.
Before I attempt this, do different isotopes of elements have different ionization energies?
 
  • #20
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Try to estimate what fraction of the atom mass is given by its electrons, and try to estimate by how much would the mass of an atom change when the electron gets excited (you can use ionization energies for that - first ionization is in a way equivalent to the maximum amount of energy that the atom can absorb while getting excited). Compare these numbers with the atom mass.
I take it you mean that the increase in atom mass because of the electrons developing momenta is negligible compared to the mass loss from the protons and neutrons when they bind?
 
  • #21
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Try and calculate that. It's a good exercise.
Just did. When using the two extremes (92161753 eV plus 14 eV and minus 14 eV) as binding energies I get a range of
$$1u = 1.66053904842 \cdot 10^{-27} kg - 1.66053905258 \cdot 10^{-27} kg$$
This range is still above the given value of ## 1u = 1.66053904020 \cdot 10^{-27} kg## according to Wiki. The minimum difference being ##8.22156574774 \cdot 10^{-36}##, a difference of 0.000000495%. I have a feeling that such a small difference is probably caused by the number of significant digits (speed of light and binding energy at 8 digits accuracy while the weight of 1u at 11 digits accuracy).
 
  • #22
Borek
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I take it you mean that the increase in atom mass because of the electrons developing momenta is negligible compared to the mass loss from the protons and neutrons when they bind?
Not only negligible, but orders of magnitude beyond accuracy of our mass measurements.
 
  • #23
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Not only negligible, but orders of magnitude beyond accuracy of our mass measurements.
This link gives a pretty good description regarding this. Seems that the energy needed to make an electron have such a large momentum to shoot out of an atom (ionization) is on the order of a million times smaller than the released binding energy when protons and neutrons bind, let alone a momentum of an electron that still keeps it orbiting a nucleus.

Is there a formula that gives the relationship between the increase in mass of an atom and the momentum of an electron? Because I only know a formula that shows that an increase in momentum means an increase in velocity (## m \cdot v = √(2 \cdot E \cdot m)##). I can use the Lorentz transformation to calculate the increase in mass but as it is said; calculating the relativistic mass of an electron is an outdated concept. Furthermore, I read that the velocity in that formula is not the classical momentum of a particle but a phase velocity of its wave.
 
  • #24
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500px-Binding_energy_curve_-_common_isotopes.svg.png
Guys, I've got a few questions regarding this diagram to make sure I understand it well.

1. Do fusion of atoms that are heavier than Fe require energy as well?

2. Is it correct to say that fusing atoms that are heavier than Fe will not yield any energy? And, if the answer to question 1 is yes, that the given energy to fuse these heavy atoms actually turns into mass and make the nucleons heavier?

3. Can I say that if atoms that are heavier than Fe would fall apart, it will yield energy while if atoms lighter than Fe would fall apart, they would actually need energy? In the latter case (atoms lighter than Fe), the needed energy would be turned into mass and make the nucleons heavier?

4. How can it be that the total released binding energy of a ##^{12}C## atom is 92161.753 keV while the triple alpha process to make a ##^{12}C## atom actually releases 7.367 MeV?
I understand that the triple alpha process goes through more steps but shouldn't the end net released energy be equal to the released binding energy of a ##^{12}C## atom from single nucleons?
 
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  • #25
Borek
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make the nucleons heavier
I feel like you are in general on the right track, but I don't like this statement. I don't think we can assign mass to nucleons in nuclei, at best we can calculate their average mass.

I understand that the triple alpha process goes through more steps but shouldn't the end net released energy be equal to the released binding energy of a [itex]^{12}C[/itex] atom from single nucleons?
What is the starting point of the triple alpha process?
 

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