Beta Decay: Unstable Nuclei & Why It Occurs

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Beta decay occurs when a nucleus has an excess of neutrons compared to protons, leading to instability. An unstable nucleus has a probability of spontaneously splitting into smaller parts to achieve a lower energy state. Nature favors lower energy configurations, which explains why larger nuclei tend to decay. The stability of a nucleus varies with atomic number, requiring a specific neutron-to-proton ratio for stability. Understanding the energy dynamics and configurations of nucleons is essential to grasping the principles of beta decay.
Fabian901
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I've started studying physics at a basic level and I've ready that beta decay takes place when the number of neutrons exceed the number of protons in a nucleus. Why does this excess number of neutrons compared to protons make the nucleus unstable?
Also, what does an unstable nucleus mean?
 
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Unstable means there is a probability the nucleus will spontaneously split into smaller parts.

Nature prefers lower energy states. The energy of larger nucleus is higher than smaller ones so they will try to spilt to go to a lower energy state.

Exactly why it does this theoretically I don't know - nuclear physics is not really something I am up on.

But I suspect it's related to why excited electrons spontaneously emits photons to go to lower energy states - its tied up with Quantum Field Theory and the vacuum. Vacuum fluctuations are responsible - my suspicion its its the same with spontaneous nuclear splitting.

Thanks
Bill
 
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Thanks a lot for the answer! Where does that energy in the nucleus actually come from?
Many thanks!
 
Fabian901 said:
I've started studying physics at a basic level and I've ready that beta decay takes place when the number of neutrons exceed the number of protons in a nucleus. Why does this excess number of neutrons compared to protons make the nucleus unstable?
Also, what does an unstable nucleus mean?
Not quite true. For increasing atomic numbers stability requires more neutrons than protons. If the ratio is too large or too small, there will be decay.

Heaviest stable nuclides are those of lead, 82 protons and 124 to 126 neutrons.
 
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Imagine that the protons and neutrons fill up two "parallel" sets of energy levels. For lower proton numbers the energy levels are nearly the same, so for a given total number of nucleons you get the lowest energy when you have half neutrons and half protons. See the picture associated with the asymmetry term in the Wiki article on the semi-empirical mass formula:

http://en.wikipedia.org/wiki/Semi-empirical_mass_formula#Asymmetry_term

For larger proton numbers, the electrostatic repulsion "spreads" the proton energy levels upwards and further apart, so the lowest-energy configuration has fewer protons than neutrons.
 
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I see, thanks a lot for the answers!
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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