# Largest Atomic Weight

1. Jul 11, 2008

### nuby

What is the maximum atomic weight of a single atom (in theory)? Could there be an element with an atomic number greater than 150, somewhere in the universe? How is it possible?

2. Jul 11, 2008

### nuby

How is it known or calculated?

3. Jul 11, 2008

### DaveC426913

The problem with heavier and heavier atoms is that they
a] require increasingly larger amounts of energy to create them
b] decay increasingly more rapidly.

eg. element 118 decays after a mere 10-100 microseconds.

Can one say one has created a stable atom if it only lasts billionths of a second?

4. Jul 11, 2008

### melior

There is speculation that continuing the noble gas series will give relatively stable atoms. Counter-acting this is the problem starting around uranium where the electro-magnetic force in the nucleus first equals and then surpasses the strong force with increasing atomic number.
We have a naming scheme that can handle an infinite number of elements, but anything higher than 111 and you have to answer DaveC426913's question.

5. Jul 11, 2008

### nuby

I read an article somewhere (can't remember where) that the largest theoretical element was 135 (?) ... And I believe it mentioned something about the orbital size. (?)

So if the conditions are right then could you say there is no limit on the mass/size of an atom?

6. Jul 11, 2008

### melior

Your right, I forgot about the island of stability. The nucleus has shells that act a lot like the stacked orbits of electrons. If you get a filled shell, the total energy is lower than if you had a partially filled shell. If you can get enough of an energy bonus out of filling a shell than the increase in EM interactions, it should have higher stability.
Attempts to use this idea involve filling in the next shell of neutrons and the next shell of protons to get element 114 with 184 neutrons. The problem is that it is really hard to get enough neutrons and protons to stick to a nucleus long enough to get a full shell. Anything short of a full shell will very rapidly by fission or alpha decay.
I don't know about element 135, but the chances of finding higher transuranic elements in detectable quantities are basically 0 in nature.

The article is a bit old, but it gives a good over view:
http://discovermagazine.com/1994/aug/anislandofstabil417

7. Jul 12, 2008

### nuby

Is there a formula for calculating the most stable "proton : neutron" ratio? It seems like 2/pi seems to be ideal.

Any idea why? i.e. uranium (Z/N) is 92/146 .. is .630

8. Jul 12, 2008

### Mr Noblet

The nucleus is more stable when it has a certain number of nucleons. The numbers are refered to as magic numbers. If both nuleon numbers are a magic number, then the nucleus is at its most stable. The known numbers are 2, 8, 20, 28, 50, 82, 126

http://en.wikipedia.org/wiki/Magic_number_(physics [Broken])

Last edited by a moderator: May 3, 2017
9. Jul 12, 2008

### malawi_glenn

nuby, there is quite hard to determine the most stable proton-neutron ratio from theory, since we dont know all ingrediences of the nucleon-nucleon force plus that we can't solve systems of particles with numerical methods when the number of particles are above approx 30..

So it depends on what nuclear model you use and what approximations etc you use, then it is up to experiments to verify and justify the models.

The magic numbers that Mr_Noblet presented to you was the first start of trying to theoretically solve the nuclear many body 'problem' -> 'The nuclear shell model'

And if such an element would exists, then it must be created via the r-process in stars (stellar nuclear synthesis), so even if it exists in theory an 'island' of stability above the valley of instability in the nuclear shart, it is extremely unprobable that the r-process can go that far PLUS that we also have the neutron drip line which occurs in all the nuclear structure theories.

10. Jul 12, 2008

### humanino

If you exclude the concept of neutron stars as "giant nuclei", there is a theoretical limit on the mass of an atom, but it is comparatively little known.

Let us assume the following : the EM coupling constant is exactly 1/137, and the nuclei are point-like (we remove those after, it is only for the sake of explanation). The coupling to the EM field does not only involve the coupling constant, but also the charge Z. At Z=137, the perturbative series can not converge anymore. Physically, the EM vacuum becomes unstable : an electron from a virtual pair is absorbed by the nucleus, and the positron flies away. This imposes a maximum Z. Together with the neutron drip line, which imposes a maximum N for each Z, this makes an absolute maximum A possible for an atom.

When you remove the simplifying conditions, you make model-dependent evaluations all pointing towards a maximum Z around 185. For one thing, this is very high compared to what we can do today, and much higher than the anticipated next island of stability. For another think, this instability is due to EM and not to the strong force, so technically, people would still call nuclei with very high Z (larger than this limit) unstable "bound states under QCD".

Note that the neutron star limit occurs much much much higher in A, and is somehow irrelevant here.

This mechanism is known as supercritical binding, was first discussed by the russian school following Pomeranchuk, and if you are interested in it (or want to know why it is interesting) you can read Gribov conception of quark confinement, which stems from the (possible) same phenomenon in QCD with light (physical) quarks.

edit
"Anomalous Positron Peaks from Supercritical Collision Systems", Phys. Rev. Lett. 54, 1761 - 1764 (1985)

Last edited: Jul 12, 2008
11. Jul 12, 2008

### nuby

Would (2pi/c^3)/proton_mass work?

Last edited: Jul 12, 2008
12. Jul 12, 2008

### humanino

:uhh: why ?

13. Jul 12, 2008

### nuby

Well, while trying to figure the maximum mass of an atom, I came up with a few ideas (for fun). It involved the smallest possible 'orbital' radius of hydrogen, and c. I wasn't sure it was real/used equation, or if I just did something odd, but it produced a result close to the Em coupling constant.

14. Jul 12, 2008

### humanino

Can you explain that concept ? I'm not sure what you are talking about.

Can you provide enough information for us to understand your claims ?

15. Jul 12, 2008

### nuby

humanino, I'll send it to you in a pm.

16. Jul 13, 2008

### nuby

delete,dup

Last edited: Jul 13, 2008
17. Jul 13, 2008

### nuby

This isn't a theory, and I'm not making any claims. I'm just curious why I get the following results (sorry about the format, I was having a problem posting)
If atomic_radius_total = r = E/(2*pi)
Then apply to Hydrogen:
r = 2.4e-11
1.) r*c = .00720 (close to fine structure constant 'a')
2.) r*c^2 = 2.16e6 (close to calculated electron velocity of H '2.19e6')
3.) (m*c^4)/(2*pi) = 2.16e6 (same as #2)
4.) (2*r)/c = 1.60e-19 (close to elementary charge constant 'e')
5.) If maximum result for #3 is 'c', then:
c = (m*c^4)/(2*pi)
m = maximum_mass = ((2*pi)/c^3) = 2.32e-25 (protons only)
maximum_mass/proton_mass = 139
or: 1/(r*c)

18. Jul 14, 2008

### malawi_glenn

Nuby, it seems to me that you only are making wild guesses :P and using totaly akward forumulas. I mean, what does:

"Then apply to Hydrogen:
r = 2.4e-11"

resemble? What kind of radius? And where comes the forumula E = 2r*pi from? 1) You can't use classical mechanics dealing with quantum entities like atoms and nuclei, 2) you must use UNITS in your calculations.

Also take notice that there is not linear relationship between Z/N of the stable nuclei, http://www.nndc.bnl.gov/chart/ so talking about a stable configuration at '2/pi ' and similar doesent work.

At last, take notice that neutrons stars are held toghether also by gravity, while in ordinary nuclei gravity doesent play a role.

19. Jul 22, 2008

### nuby

r=2.4e-11 Is supposed to represent an electron orbital radius within a hydrogen atom.

r=E/(2*pi) Was a random idea I had, which relates the energy/mass of an atom to my made up physical dimensions, i.e. (4/3)*pi*(E/(2*pi))^3 could be Mass , and 2*pi*r could be Energy .
Works for atoms with a standard atomic weight around 178.

And the rest were just observations I made while screwing around. Nothing to take seriously.

20. Jul 22, 2008

### malawi_glenn

nuby, you have to use units when you are doing physics.

(4/3)*pi*(E/(2*pi))^3 could be Mass (No this has unit energy^3)

2*pi*r could be Energy (No, it has unit length)

Nuby, largets atomiv weight you get by doing advanced nuclear structure physics calculations, nothing else.

21. Jul 22, 2008

### nuby

22. Jul 22, 2008

### kanato

What about them? Ratios of quantities that have the same units end up being dimensionless, but if the quantities don't have the same units then their ratio will have units.

You haven't answered the question of what units you're using for r. Angstroms? Centimeters? Meters? Ok, I'll assume you're using SI units, so r = 2.4e-11 m. Then if r is in units of m:

1) r*c has units $$(m^2/s)$$. The fine structure constant is dimensionless.
2) r*c^2 has units of $$(m^3/s^2)$$ which is not a velocity.
3) mc^4/(2*pi) has units of $$(J m^2/s^2)$$ which, like #2, is not a velocity. This result is not even comparable to the result of #2 because the units are different.
4) 2r/c has units of seconds. This is totally unrelated to the elementary charge.
5) If you set c = (m*c^4)/(2*pi), then m must have units of $$s^3/m^3$$ which has nothing to do with mass.

and lastly, 1/(r*c) is not 139, it's one hundred thirty nine seconds per meter squared. You're getting numbers which look interesting because you've just randomly combined numbers without regard to units until you get one number that looks close to another physical number, yet with the units ignored. You have to pay attention to the units, otherwise you could say things like "5 inches is larger than 1 foot, because 5 > 1," which is just absurd.

I hope that clarifies why units are important for this sort of thing.

23. Jul 24, 2008

### nuby

Makes sense. Thank you.

But, can you explain where the units come from in the electric constant, or magnetic constant?

The magnetic constant is 4*pi * 10^-7 .. But where does (H/m) come from?

And the electric constant 1/(c^2 * 4 * pi * 10^-7) .. which has units: A^2 * s^4 / kg * m^3

Where do these units come from? Or is it different with fundamental constants?

Thanks.

24. Jul 25, 2008

### malawi_glenn

Yes they are different because they are DEFINED.

The same is for the speed of light in vacuum, the gravity constant, plancks constant etc.

As an example, you measure the force between two charges as a function of distance between them, and you'll find that the force is proportional to r^(-2)

$$F = \text{Const.} \frac{q_1q_1}{r^2}$$

Since units must be the same on Left hand side as right hand side, the unit of the constant must be: [ Nm2C−2 ]