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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) enter image description here

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.

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  • $\begingroup$ Your 2nd paragraph ("The Basic cause of radioactivity is dominant electrostatic repulsion...") can't be quite right, otherwise tritium wouldn't decay to helium-3; you'd expect helium-3 to decay to tritium instead. $\endgroup$
    – PM 2Ring
    Apr 22, 2020 at 17:26
  • $\begingroup$ @PM2Ring I know that is only happening due to $β^-$ decay only.I know β^- decay occurs in nature but i did not understand β^- decay theoritically.I think bill's is approaching towards my question.I think you misunderstood my question. $\endgroup$
    – sarthak
    Apr 22, 2020 at 17:47
  • $\begingroup$ @PM2Ring The statement is given in my book.If you need it ,I may upload it, $\endgroup$
    – sarthak
    Apr 22, 2020 at 17:48

4 Answers 4

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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.


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.

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  • $\begingroup$ I did not meant that i know every physical meaning abot n/p rather i wanted that no one should explain just by simply dividing numbers. $\endgroup$
    – sarthak
    Apr 22, 2020 at 18:38
  • $\begingroup$ @sarthak Okay. Did you understand the reasoning I gave in my answer? $\endgroup$
    – J. Murray
    Apr 22, 2020 at 21:30
  • $\begingroup$ I understood from your answer that we have to consider arrangement in energy levels and spin too.can please apply it on alpha or β^+ decay $\endgroup$
    – sarthak
    Apr 23, 2020 at 2:17
  • $\begingroup$ @sarthak The case of $\beta^+$ decay is basically the same - simply swap the protons in my answer for neutrons. The nucleus with 7 protons and 5 neutrons is Nitrogen-12, which decays via $\beta^+$ to Carbon-12 for the same reason that the nucleus with 5 protons and 7 neutrons (Boron-12) decays to Carbon-12 via $\beta^-$. $\endgroup$
    – J. Murray
    Apr 23, 2020 at 4:55
  • $\begingroup$ @sarthak $\alpha$ decay is a slightly different beast. Your question specifically refers to $\beta$ decay, and so my answer is primarily aimed in that direction. If you want to know about $\alpha$ decay, you should ask a separate and well-formulated question about it. $\endgroup$
    – J. Murray
    Apr 23, 2020 at 4:57
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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.

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  • $\begingroup$ But i did not understood how adding neutron will result in high energy than adding proton.Please consider explaining in detail. $\endgroup$
    – sarthak
    Apr 22, 2020 at 17:40
  • $\begingroup$ @sarthak Edited $\endgroup$
    – Bill N
    Apr 22, 2020 at 17:43
  • $\begingroup$ I am just high school student,I would definitely try to understand the theory given in link in my free time.Is there any simpler and intiutive explaination for high school level $\endgroup$
    – sarthak
    Apr 22, 2020 at 17:53
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    $\begingroup$ @sarthak Only that nuclear behavior cannot be totally understood from electrostatic potential. The nuclear potential and the way protons and neutrons interact causes behavior that seems counter-intuitive. Adding neutrons, as you say, can help counter electrostatic repulsion, but that also adds energy. Sometimes the "nuclear" energy added is greater than the reduction in electrostatic energy. Plus, there is that thing called spin, which complicates the system even further. $\endgroup$
    – Bill N
    Apr 22, 2020 at 18:03
  • $\begingroup$ While this is not directly related to beta decay, the semi-empirical mass formula may help you understand the different competing effects in the nucleus. @bill-n is right, electrosatic repulsion is not the only affect. en.wikipedia.org/wiki/Semi-empirical_mass_formula $\endgroup$ Apr 23, 2020 at 16:54
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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.

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  • $\begingroup$ Neutrons have an average bigger mass than protons , not each neutron has bigger mass than each proton. $\endgroup$
    – user256968
    Apr 22, 2020 at 16:55
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    $\begingroup$ @JellyStrawberry I would appreciate seeing a reference on that statement. $\endgroup$
    – Bill N
    Apr 22, 2020 at 17:12
  • $\begingroup$ BTW, the mean lifetime of the free neutron is about 14m 40s, its halflife is therefore about 10m 11s. $\endgroup$
    – PM 2Ring
    Apr 22, 2020 at 17:16
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    $\begingroup$ Wkiedia: Masss of neutron 939.5 MeV . Mass of proton 938.3 MeV . Mean lifetime of unbound neutron 881 sec. I misrememeber half life versus mean lifetime. All free neutrons have the same mass! $\endgroup$
    – mike stone
    Apr 22, 2020 at 17:19
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    $\begingroup$ @ Bill N. Thanks fior the edit. Those pesky neutrinos just keep going missing. $\endgroup$
    – mike stone
    Apr 22, 2020 at 17:26
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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.

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  • $\begingroup$ I tried to summarise what i understood from other's answer. $\endgroup$
    – sarthak
    Apr 23, 2020 at 4:32

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