16
$\begingroup$

If protons have a positive charge and electrons have a negative charge, can we add up several protons and electrons together to create a new element, without adding neutrons to hold the nuclei together?

$\endgroup$
  • 2
    $\begingroup$ Helium-2 has been detected--but the half life is so short that it's not even known. Likewise Lithium-3. Beryllium-4 hasn't even been detected. I see no point in trying to walk further up the table. $\endgroup$ – Loren Pechtel Sep 19 '15 at 3:49
21
$\begingroup$

Yes and no.

Elements are defined by the number of protons only. It does not matter if (say) a carbon nucleus has six or seven (or eight) neutrons, they will all react the same.*

With that, to create new elements, you would need to get up to some 115 or so protons fused together. But there is a reason for neutrons: all the positively charged protons in the nucleus repel each other electrostatically so neutrons are there to stabilise the nucleus — you can think of it as cushioning protons apart. For each element there is a set number of neutrons that will create a stable nucleus. E.g. for hydrogen (one proton) it is zero or one, for oxygen it is eight, for carbon six or seven, for tin it can be $62, 64, 65, 66, 67, 68, 69, 70, 72$ or $74$.

Nuclei with other neutron numbers decay radioactively on half-life scales from femtoseconds to millions of years. I would expect no-neutron nuclei (non-hydrogen) to be on the very short edge of that spectrum (although I’m no radiochemist).


*: There are reactions in certain systems that depend a lot on hydrogen being $\ce{^1H}$ rather than $\ce{^2H}$ or $\ce{D}$ because they rely on the tunneling effect whose probability decreases with mass (I think by the factor $m^2$). That’s also why one shouldn’t drink $\ce{D2O}$

$\endgroup$
  • 1
    $\begingroup$ The comments section is not for extended discussion, therefore I have moved them into a seperate chatroom. I also imported the conversation from the periodic table to there. $\endgroup$ – Martin - マーチン Sep 18 '15 at 5:04
  • $\begingroup$ Just for fun, is there any way to extrapolate what the half-life of, say, ununennium-119 would be? (I know it would be based on ridiculous and unrealistic assumptions, but we might get a hilariously small number.) $\endgroup$ – Nate Eldredge Sep 18 '15 at 5:59
  • $\begingroup$ The Belt of Stability chart: chemwiki.ucdavis.edu/Physical_Chemistry/Nuclear_Chemistry/… sum it very well. $\endgroup$ – Orace Sep 18 '15 at 8:36
  • $\begingroup$ @NateEldredge It wouldn't form in the first place so its half life is undefined. $\endgroup$ – bon Sep 18 '15 at 9:14
  • $\begingroup$ Aren't all reaction rates dependent on the particular isotope (thus dependent on the number of neutrons)? - with the relative difference being highest for deuterium? $\endgroup$ – Peter Mortensen Sep 19 '15 at 14:35
19
$\begingroup$

Jan's answer is correct. I will try to fill in a few details about why neutrons are essential to creating stable nuclei.

All stable isotopes excepting Hydrogen-1 have neutrons in their nuclei. Hydrogen, for example has two stable isotopes: The first simply has a proton with no neutrons in the nucleus, while the second, often called Deuterium has a proton and one neutron.

Oxygen has three stable isotopes (O-16, O-17 and O-18), and by stable, I mean that they NEVER decay. The reason they never decay is that they cannot transmute (decay) to any other configuration of protons and neutrons without existing in a higher energy state. Since they don't have the energy, the protons and neutrons that make up those nuclei are stuck in that configuration (until something gives them the necessary energy to go to a higher energy state).

Now, as to your question: Neutrons play an important role in stabilizing the nucleus. Protons and neutrons respond to the Strong Nuclear Force (SNF) and are attracted to each other by that force. Protons, with charge, ALSO respond to the electrostatic force. Since all protons are positively charged, the more of them present in a nucleus, the larger the force that is trying to rip the nucleus apart.

Neutrons, in a nucleus bring added SNF to the nucleus without adding any repulsive electrostatic force. Thus they play an important role in creating stable nuclei. They increase the tendency of a nucleus to stick together. Thus all elements beyond Hydrogen have some neutrons in their nuclei.

And that explains why we don't find atoms with large numbers of protons and very few neutrons. By the time we get to Uranium, we find that nature needed more than one neutron for each new proton in order to create stable nuclei. U-238 is the most abundant form of Uranium and it has 92 protons and 146 neutrons.

But that leaves us with one other important dilemma. If neutrons attract each other, and there is NO electrostatic repulsive force between them, one could reason that we should find atoms with large numbers of neutrons and very few protons. Why not Hydrogen 99? 1 proton and 98 neutrons would seem to be extremely stable, yet it is not found in nature. In fact, if you add even one more neutron to Deuterium, you find that it (Tritium) decays with about a 12.32 year half-life.

The answer, is that there exists yet another force besides the SNF and the electrostatic force. It is called the Weak Nuclear Force because it is weaker than the SNF.

For an introduction to the Weak Nuclear Force I would go here:

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

So, we need neutrons to make atoms stable. Electrons cannot exist inside the nucleus due to the Uncertainty principle, but that's another issue entirely.

$\endgroup$
  • 8
    $\begingroup$ One additional note: Neutrons on their own are not stable, they decay into a proton, an electron and a neutrino. This is the WNF in action. The same things happen if you have a nucleus with too many neutrons. Having the right number of protons and neutrons together stops this somehow. (Your Wikipedia link explains this, but I think it should be said directly) $\endgroup$ – Stig Hemmer Sep 18 '15 at 7:25
8
$\begingroup$

I'll focus on the nucleus only. As far as I'm aware, there is little to no variation in radioactive decay lifetimes as a function of ion charge state; that is to say that the number of electrons surrounding a nucleus has little to no impact on whether the nucleus will decay. I'm open to corrections from measurements at, say, the Hollifield facility at Oak Ridge.

The 1-hydrogen nucleus is stable - one proton, no neutrons.

The 4-lithium nucleus is listed as having a decay time of 8E-23 seconds per the Chart of the Nucleides. So, three protons and one neutron is really, really unstable. A 3-lithium nucleus has never been observed. So, three protons with no neutrons is not stable.

The 2-He nucleus has been put forth as the mechanism for the observed 'exotic 2-proton decay' of even-Z nuclei. However, a recent paper on 2-proton 19-Mg decay says 'modeling the decay as the tunneling of a diproton ($^{2}$He nucleus) typically yielded lifetimes several orders of magnitude too short.' Thus, it would appear that the 2-He nucleus, two protons and no neutrons, which has never been observed alone, is not the object leading to observed 2-proton decays. Thus, while it has been postulated, it does not seem to occur.

A no-neutron universe would then consist of hydrogen, and hydrogen alone.

$\endgroup$
  • $\begingroup$ Thank you for elaborating , so the 2-He is actually a good example for what i was asking, even if it has a short lifespan , does it contain 2 electrons too ? $\endgroup$ – soundslikefiziks Sep 17 '15 at 23:16
  • $\begingroup$ Even if it would be created in 2-proton decay those protons would get unbound in time similar to that for 4-lithium no electron would "even notice that" :) $\endgroup$ – Mithoron Sep 17 '15 at 23:38
  • $\begingroup$ @Mithoron Ah, i see your point there :) so the question is how did the researchers ever "notice that" . $\endgroup$ – soundslikefiziks Sep 18 '15 at 0:19
  • $\begingroup$ Ever notice what? That 2-He was postulated was to try and account for the observed 2-proton decay via one particle leaving the nucleus (solar to an alpha), and then decaying into the two observed protons. You seem very keen still to link electrons with nuclei more closely than chemistry allows or nuclear physics requires. $\endgroup$ – Jon Custer Sep 18 '15 at 1:14
  • 2
    $\begingroup$ While in almost all cases the electronic structure/surroundings of an atom has negligible influence on its radioactivity, there are reported cases showing an effect. Here's an interesting read on the subject. $\endgroup$ – Nicolau Saker Neto Sep 18 '15 at 9:15
4
$\begingroup$

Without getting into the specifics of nuclear forces, there is a simple way to think about this.

What matter for chemistry is the number of protons in the nucleus of the atom as this determines the nature of the electron orbitals the atom will have which is what chemistry is about.

So, if you could persuade electrons to joint the nucleus and that led to a proton turning into an neutron then you would indeed have a new element. Whether that would be a stable element depends on all sorts of complicated stuff involving nuclear stability (and the strong and weak nuclear forces I'm trying to avoid dealing with). Besides, we already mostly know what nuclei are stable or not as nuclear physics has explored the possibilities extensively.

Moreover, this process actually happens, though it isn't common. There is a rare form of radioactive decay for proton rich nuclei where the nucleus "captures" and electron transforming a proton to a neutron and thereby creating a different element. An example would be rubidium-83 (37 protons, 46 neutrons) decaying to krypton-83 (36 protons, 47 neutrons). Wikipedia lists more examples.

So the answer is: yes you can add electrons to a nucleus to create a new element. But the fact that it isn't common suggests it is hard. And this reflects both the instability of most of the products you could get (even if it could happen the products would be too unstable to see in most cases) and the difficulty of the reaction mechanism (there is a big energy barrier preventing it happening for most nuclei).

$\endgroup$
  • $\begingroup$ Re "What matter for chemistry is the number of protons": So we would be OK drinking heavy water instead of regular water? $\endgroup$ – Peter Mortensen Sep 19 '15 at 14:38
  • $\begingroup$ @PeterMortensen I wouldn't advise drinking heavy water, though it can reverse some of the effects of drunkenness. Hydrogen is the one major exception to the "only protons matter" rule as doubling the nuclear mass makes a significant difference to hydrogen chemistry. Otherwise it is a good rule of, thumb. $\endgroup$ – matt_black Sep 20 '15 at 0:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.