Which one is larger: an atomic mass unit based on the current standard or one based on the mass of a Be-9 atom?

My attempt:

Current standard is $1/12$th the mass of a C-12 atom. New standard is $1/9$th the mass of a Be-9 atom.

Now I understand that the only the basic unit of measurement of masses is changing. The absolute masses of the different atoms in the universe remains the same. That means the absolute mass of an He-atom in the old system and in the new system must be the same.

Now, absolute mass (of an atom in any system) = no. of "amu"s of that atom * mass of 1 amu (in that system)

We are supposed to determine the relation between the mass of 1 amu in the old and in the new system. Obviously, the one having more "amu"s would have a smaller mass of 1 amu.

And now I am stuck. I can't proceed further than this much explanation. I've looked on Wikipedia and some other questions but can't find similarities. Hints are appreciated!

UPDATE: although not explicitly mentioned, the book does NOT expect readers to look up data tables to find the answer. It expects a possible reason as to why is the Be-9/C-12 scale larger.

  • $\begingroup$ It might be worth looking into the original definition of amu, which was based on the mass of $\ce{O^{16}}$ rather than carbon. chemistry.oregonstate.edu/courses/ch121-3s/ch121/… $\endgroup$
    – Tyberius
    Mar 29, 2017 at 22:12
  • $\begingroup$ Not oxygen-16. The earlier standard was developed before it was realized there were isotopes, so just plain oxygen (which, we know now, is mostly oxygen-16 in nature) was 16. $\endgroup$ Mar 30, 2017 at 13:45
  • $\begingroup$ @OscarLanzi Physicists used the atomic mass of the nuclide $\ce{^16O}$ whereas chemists used the average atomic mass of natural oxygen. Anyway, the amu became obsolete when IUPAP (1960), IUPAC (1961), ISO, CIPM (1967) and CGPM (1971) agreed to assign the value 12 to the relative atomic mass of the nuclide $\ce{^12C}$. $\endgroup$
    – user7951
    Jul 15, 2018 at 12:59

2 Answers 2



Which one is larger: an atomic mass unit based on the current standard or one based on the mass of a Be-9 atom?


It is easy enough to look up the weights of $\ce{^{12}C}$ (12.00000000...) and $\ce{^9Be}$ (9.0121822) based on $\ce{^{12}C}$ scale and make a decision from that data. Thus a mass scale based on $\ce{^9Be} =9.00...$ would be heavier than an mass scale based on $\ce{^{12}C} = 12.00...$.


The OP has asserted that the problem is not supposed to be just a "lookup problem" but that some reasoning should be offered.

I content that the question simply asks which is larger. It doesn't ask why. Furthermore I'd assert that there isn't any "simple" reasoning by which the problem can be analyzed other than lookup.

Neutron to proton ratio reasoning

In his answer QuarkyLittleThing's reasoned that a neutron is heavier than a proton and since $\ce{^9Be}$ has a higher neutron to proton ratio (5:4) than $\ce{^{12}C}$ (6:6), it ought to be heavier.

Using the neutron to proton ratio reasoning, $\ce{^9B}$ only has 4 neutrons but 5 protons, so a $\ce{^9B}$ scale should be lighter than the $\ce{^{12}C}$ scale. But the mass of $\ce{^9B}$ is 9.0133288 u, so the neuron to proton ratio isn't entirely sound, and a mass scale based on $\ce{^9B} =9.00...$ would in fact be heavier than a mass scale based on $\ce{^{12}C} = 12.00...$.

Lower nuclear binding energy reasoning

Mithoron's made a comment on QuarkyLittleThing's answer asserting that the reason was "lower nuclear binding energy."

To properly analyze Mithoron's comment one must note that the nuclear binding energy is based on the energy differences of the actual fission/fusion of nuclei. Thus as NicolauSakerNeto pointed out on an earlier version of this answer, this is really not a "reason", but a tautology due to $\text{E}=mc^2$. So a decrease in the mass of the nucleus is proportional to an increase in the nuclear binding energy for that nucleus.


The current system, takes 1 amu as $1/12th$ that of a $\ce{C^12 atom}$. The $\ce{C^12 atom}$ consists of 6 protons and 6 neutrons, so the average mass would be the average of a proton and a neutron. Now if a $\ce{Be^9 atom}$ is considered, it has 4 protons and 5 neutrons, making the average tilted toward the neutron side, therby giving the value of 1 amu as slightly more than that given by the $\ce{C^12 atom}$

Hence the one based on the $\ce{Be^9 atom}$ is larger.

Hope this helps

  • 1
    $\begingroup$ It would be bigger, but because of lower nuclear binding energy $\endgroup$
    – Mithoron
    Mar 19, 2017 at 17:44
  • 3
    $\begingroup$ It is easy enough to look up the weights of C-12 (12.00000000...) and Be-9 (9.0121822) based on C-12 scale and make a decision from that data. Trying to rationalize why is apt to get you into trouble. $\endgroup$
    – MaxW
    Mar 19, 2017 at 18:23

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