$β^-$ decay should be impossible

Question

I have gone through texts about Radioactivity.What I understood is that

The Basic cause of radioactivity is dominant electrostatic repulsion over nuclear force of attraction between nucleons due to this the nucleons either try to reduce their size so nuclear force again gets strengthen (this is done by $α$ Decay) or They decrease the protons either by $β^+$ decay or by electron capturing.

But $β^-$ decay confuses me.Why a nucleus would try to reduce number of neutrons.Because how much more neutrons are there that much lower the net potential energy of system(as attrative nuclear would cause a more negative potential)Hence system's energy is lower then why nucleus should undergo $β^+$ decay and increase system's energy.(As it is some kind of law of nature that system always tends to go to lower energy)

I don't want any explaination using n/p ratio as I myself understand that simple concept.I want that if i had made some mistake in understanding then please help me figure that out or else is it not true that system does not tends to decrease the energy (If you go with this then you have to explain why electron jumps to lower shell once excited.)

Thank you.

Answer 1

The key lies in the Pauli exclusion principle. As a simple toy model, imagine a 1D box of length $L$, into which we will place some number of neutrons and some number of protons. For the moment, pretend that none of the particles interact with each other, and let $m_p \approx m_N \approx m$ be the mass of each particle.

The single-particle energy levels of this system are given by

$$\mathcal E_n = \frac{n^2 \pi^2\hbar^2}{2mL^2}= n^2 \epsilon$$

Consider the case of 5 protons and 7 neutrons, which we might consider as a very rough model of the Boron-12 nucleus. What is the ground state energy of the system? Recall that due to the Pauli exclusion principle, we can have a maximum of two protons and two neutrons in each energy level. Therefore, we would have two protons in the first energy level, two in the second, and one in the third; for the neutrons, we'd have two in the first, two in the second, two in the third, and one in the fourth.

As a result, the ground state energy of the system would be

$$E_{5,7} = \underbrace{(2\cdot 1^2 + 2\cdot 2^2 + 3^2)\epsilon}_{\text{protons}} + \underbrace{(2 \cdot 1^2 + 2\cdot 2^2 + 2\cdot 3^2 + 4^2)\epsilon}_{\text{neutrons}} = 19\epsilon + 44\epsilon = 63\epsilon$$

What about the ground state energy with six protons and six neutrons, which would correspond to the Carbon-12 nucleus?

$$E_{6,6}= \underbrace{(2\cdot 1^2 + 2\cdot 2^2 + 2\cdot 3^2)\epsilon}_{\text{protons}} + \underbrace{(2 \cdot 1^2 + 2\cdot 2^2 + 2\cdot 3^2 )\epsilon}_{\text{neutrons}} = 28\epsilon + 28\epsilon = 56\epsilon$$

The conclusion we draw is that due to Pauli exclusion, a nucleus with a significant imbalance between protons and neutrons has a higher energy than one with the same number of nucleons but a more balanced proton-to-neutron ratio.

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Nuclear stability is a balancing act. All nucleons feel a short-range attractive force due to each other due to the residual strong force. Protons contribute a long range repulsive force due to their charge. Pauli exclusion doesn't contribute a force per se, but it acts to effectively increase the energy of the nuclei with an imbalance between protons and neutrons. The spins of the various nucleons can also contribute. These interactions are all summarized in the crude, empirical, but remarkably accurate liquid drop model of the nucleus.

So to directly answer your question, $\beta^-$ decay happens at least in part due to an excess of neutrons in the nucleus, which causes the ground state energy of the nucleus to be higher than it would be if the ratio were more balanced. You say

I don't want any explanation using n/p ratio as I myself understand that simple concept

but I'm not sure that you do, as it is a major contributing factor to nuclear stability.

Answer 2

Electrostatic repulsion is not the only consideration! The nuclear potential results in a structure of energy levels which, in most models, are split between neutrons and protons (see Nuclear shell model and Nuclear structure). The energy levels for neutrons versus protons can be (and usually are) different.

Because of this difference, the addition of a neutron to a stable nucleus could result in a higher total potential energy than the addition of a proton to the original stable nucleus. If that's true, then, along with other factors, $\beta^-$ decay of the nucleus might be favorable.

This is a simple explanation, and those "other factors" would include large total mass vs small total mass, shell closure, angular momentum, etc.

Answer 3

As it is more massive than a proton, an isolated neutron is unstable, having a half life of about 15 minutes and decaying according to $$ n\to p+e^-+\bar{\nu_e}. $$ Neutrons bound to a nucleus have effectively lower mass because of the nuclear binding energy, but too many of them can cause the nucleus with one neutron replaced by a proton to have a lower mass than the original, and so the beta decay occurs.

Answer 4

The basic reason for radioactivity is possibility of reducing energy of system of nucleons with very low or no energy input from surrounding

(very low input is required for $α$ decay to provide binding energy and in other types of decay no energy is required) If we just think $β^-$ decay with the nuclear and electrostatic energies (or forces) then this type of decay seems impossible but that's not true we jave to consider energy levels and spin of nucleus then this seems possible as in that case energy after $β^-$ decay will decrease.Moreover it occurs because mass defect per nucleon is small(as binding energy per nucleon is small) hence neutron has higher mass than proton although after forming nucleus hence it decays into proton.