Consequences of a neutron decay (beta radiation) and reasoning for an electron capture

Question

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 :

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

$\hskip2in$

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)?

Answer 1

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.