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I came about the following doubts (extremely comprehensive in nature) in a logical sequence, so I have made it a point to explain either in full detail and then proceed to the interrelation between the two. I have also shared my hypothesis for the for the observed events.

In the following reaction for radioactive dating:

$$\ce{C^{14}_6 \longrightarrow N^{14}_7 + e_{-1}^0 + \overline{\nu_e}}$$

The neutral carbon-14 atom initially had $6e^-$, $6p$ and $8n$ in which a neutron decayed into a proton and an electron leaving a nitrogen atom with $7p$, $6e^-$and $7n$ and the high energy electron from the neutron decay escaped from the atom :

$$n^1_0\longrightarrow p^{1}_1 + e_{-1}^0 + \overline{\nu_e}$$

$\hskip2in$ Bound neutron decay

Fig 1: Bound neutron decay
Now it is evident that the nitrogen atom has a positive charge and in totality, it is a positively charged gas (which could be a matter of great concern since a lot of charge at one place could ionize the surroundings and could even lead to the dielectric breakdown of the atmosphere, provided there is a lot of charge.)

The above predicted results are not normally observed partly because it has a half life of about 5700 years (corresponding to extremely low concentrations) and that the nitrogen atom "might" take an electron from the surroundings, or the electron might just not escape from the gas... (This is unclear.)

Moreover, in the following reaction : (with a half life of about $11s$)

$$\ce{Mg^{23}_{12}(g) \longrightarrow {Na^{23}_{11}}^-(g) + e^{0}_{+1} + \nu}$$

The magnesium-23 has $12p$, $12e^-$ and $11n$ to start with and after the $\beta^+$ decay, sodium-23 atom with $12n$, $11p$ and $12e^-$ is left behind, which is a negatively charged ion. The $\beta^+$ radiation proceeds according to the reaction:

$$p^1_1\longrightarrow n^{1}_0 + e_{+1}^0 + \nu_e$$

With a high concentration of Mg-23 atoms we could(or should) achieve a negatively charged sodium gas in less time. Now the same doubt arises that whether the ions will remain so, or if lose an electron to the surrounding (which would still not solve the problem), or capture the positron (which is also unlikely due to the high speed).

Now, to neutralize the charge on the atom, it should lose/gain an $e^-$. Here comes the k capture reaction in which an electron and a proton combine to form a neutron according to the following reaction:

$$p^1_1+ e_{-1}^0\longrightarrow n^{1}_0 + \nu_e$$

Accordingly when an electron is "captured" by a charged species to neutralize the charge, that electron orbits the nucleus rather than go through all the orbitals and collide with the nucleus (which is made clear by the following observed reactions).

$$\ce{{Na^{22}_{11}}^+(g) + e_{-1}^0 \longrightarrow Na^{22}_{11}(g)}$$

where the positively charges sodium atom takes an electron (which revolves around the nucleus) to neutralize its net charge and the reaction isn't known to follow as:

$$\ce{{Na^{22}_{11}}^+(g) + e_{-1}^0 \longrightarrow Na^{22}_{10}(g)}$$

wherein we assume it to be a k capture reaction and algebraically add the subscripts and the superscripts. Since it isn't a nuclear reaction, the electron wouldn't really go through towards the nucleus.

This way I don't really see why an $e^-$ should ever come on the left hand side in a nuclear reaction (while it does in the k capture reaction).

But the k-capture or electron capture is found to occur in nature, so how can an electron really collide with the nucleus? Even if the innermost electron is allowed to be "sucked in" by the nuclear force, the angular momentum would still be conserved and the electron would revolve with faster and faster tangential velocities and much lower radial velocity (relativistic effects might have to be considered). This way, wouldn't it take infinite time for it to happen, with the radial speeds becoming infinitesimal?

Also, when the electron and a proton are really close to each other, the binding energy is really high, and the point when the electron and the proton "collide", the whole system would have an undefined potential energy (most probably $-\infty$ eV, since in the preceding case, the potential energy was decreasing). Wouldn't it result in the emission of huge amount of energy (after obviously subtracting the kinetic energy)?

P.S: It is slightly (rather wildly) off topic, but since we are considering the case when the electron is revolving at speeds near $c$ while coming towards the nucleus, do we take the increased mass into the consideration of the mass of the atom altogether, or is it frame dependent (even galaxies move at speeds beyond $c$ with respect to us)?

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    $\begingroup$ I wish more people were like you, in the sense of making their formatting right on their first question. Welcome to Chem.SE. $\endgroup$ – Ivan Neretin Sep 2 '16 at 19:29
  • $\begingroup$ I'm still not sure what the question is after reading this twice through, other than you have confusion on nuclear physics. As for you P.S., you have an incorrect view of electrons in orbitals - they do not revolve around the nucleus, they occupy (are) orbital wave functions. $\endgroup$ – Jon Custer Sep 2 '16 at 19:47
  • $\begingroup$ Yeah, I understand your point @JonCuster, I am aware of the orbitals (as also mentioned somewhere in the question, and was used just for the ease of asking). The question is in two parts - 1. What happens to the positively charged nitrogen ion (to support the doubt, I have also cited a faster reaction in the question). 2. Assuming it takes an electron from the surroundings(by still not really solving the issue), I have a doubt in the k capture reaction. With knowing this, reading the question once more might explain my question exactly. If not, please let me know. :) $\endgroup$ – Pranshu Malik Sep 3 '16 at 12:38
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It seems that the gist of the question is about the distance in the force equation for Coulombic attraction. $$ F = k*\frac{q_1 * q_2}{r^2}$$ Naively, as the distance between the charges goes to zero, the force should go to infinity. That obviously doesn't happen. The same problem exists for black holes; the gravity should be at a singularity which doesn't really exist. The point is that the traditional formulas are "good enough" under most circumstances, but fail under extreme conditions. For instance the theory of relativity predicts that time slows for a body which is motion. GPS satellites actually need such a correction.

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  • $\begingroup$ I'd rather underline that Coulomb's law still works at these distances; it is just that an electron is not a dot. $\endgroup$ – Ivan Neretin Sep 3 '16 at 9:33
  • $\begingroup$ @MaxW I understand, but when the electron is really near the nucleus and not far, we would have a really low potential energy, but the speed might exceed in such configuration which is why we don't really have anything below the 1s orbital. So when you lower your potential energy, you gain kinetic energy, but when they "collide" would they not realease some energy, understanding that the kinetic energy will approach infinity (due to relativistic corrections), so the mass would also increase, so isn't this really going weird? $\endgroup$ – Pranshu Malik Sep 3 '16 at 12:43
  • $\begingroup$ The excess energy gets emitted as an electron neutrino. See en.wikipedia.org/wiki/Electron_capture // Reread my post. As the distance approaches zero the energy doesn't go to infinity. The proton and the electron have real volumes as particles. Also at the distance that they merge Coulomb's Law doesn't apply. Rather Coulomb's law works in the other direction. As the electron from a beta particle gets 100 atomic diameters away from the nucleus it is at essentially an infinite distance since we can measure the energy loss to 4 significant figures (I'd guess). $\endgroup$ – MaxW Sep 3 '16 at 15:02
  • $\begingroup$ I should have pointed out that the simple Coulomb's law also breaks down for the inner orbitals of heavy elements. Ab inito calculation for the inner orbitals (ie 1s) have to be corrected for relativity. So some of the gained energy appears to be converted to mass. $\endgroup$ – MaxW Sep 3 '16 at 15:37

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