Question on Baryons and Specific Decays

In summary, beta decay occurs when the daughter nuclei has a mass that is lower than the parent nuclei. The reason for beta decay is to reduce the mass of the constituent particles.
  • #1
olv
1
0
1) We know in case of beta decay if there is excess of protons or neutrons in the nucleus beta decay take place.
Force involved - weak interaction
Reason involved - to stabilize, to obtain proton neuton 1:1 ratio in the nucleus.
E. g., Beta minus - Carbon 14 transforms into Nitrogen 14

2) Case of Lambda & Sigma baryons -

Force involved - weak
Reason involved - ?

Does anyone know the reason why Lambda Λ0 (uds - quark composition) decays into,
p+ + π−
or
n0 + π0

Also why Sigma Σ+(uss - quark composition) decays into,
p+ + π0
or
n0 + π+

What mechanism make them go decay into proton or neutron( ignore the π0 & π+) ? What is the reason involved ?
 
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  • #2
Beta decay of nuclei occur when the daughter nuclei has lower mass than the parent nuclei.
You have to specify what "stabilize" means. Here it means minimizing the mass.

Decays occur whenever it is possible. If something is possible in quantum physics, there is a non zero probability for it to happen.
 
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  • #3
olv said:
1) We know in case of beta decay if there is excess of protons or neutrons in the nucleus beta decay take place.
Force involved - weak interaction
Reason involved - to stabilize, to obtain proton neuton 1:1 ratio in the nucleus.
E. g., Beta minus - Carbon 14 transforms into Nitrogen 14

2) Case of Lambda & Sigma baryons -

Force involved - weak
Reason involved - ?

Does anyone know the reason why Lambda Λ0 (uds - quark composition) decays into,
p+ + π−
or
n0 + π0

Also why Sigma Σ+(uss - quark composition) decays into,
p+ + π0
or
n0 + π+

What mechanism make them go decay into proton or neutron( ignore the π0 & π+) ? What is the reason involved ?
The short answer is the answer to the question "Why does any particle decay?" It's always the same answer: because the decay products are more stable than the particle that is decaying.

The reason ##C^{14}## decays into ##N^{14}## is because ##N^{14}## has a more stable nucleus than ##C^{14}## does, not that we need the number of protons to equal the number of neutrons. (That reasoning only works for the smaller atomic numbers.)

Also, note that the ##\Lambda ^0## has two decay modes:
##\Lambda ^0 \rightarrow p^+ + \pi ^-##

and
##\Lambda ^0 \rightarrow p^+ + e^- + \overline{ \nu }_e##

When it comes to subatomic particles it is almost always true that the higher the mass, the quicker it decays. So (generally, not always) the reason for the decay is to reduce the mass of the constituent particles.

-Dan
 
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  • #4
"More stable" has nothing to do with why particles decay.
Particles decay if the sum of the masses of the decay products is less than the mass of the decaying particle.
 
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  • #5
Meir Achuz said:
"More stable" has nothing to do with why particles decay.
Particles decay if the sum of the masses of the decay products is less than the mass of the decaying particle.
That's more or less the content of the lifetime amplitude. But wouldn't that define the measure of the stability of a particle? That's the way I've always looked at it, anyway. Is there a more exact definition of stability?

-Dan
 
  • #6
I usually consider the lifetime to be a measure of stability, with the shorter the lifetime, the less stable.
 
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  • #7
Meir Achuz said:
Particles decay if the sum of the masses of the decay products is less than the mass of the decaying particle.
And if the conservation laws are fulfilled, like electric charge etc
 
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  • #8
malawi_glenn said:
And if the conservation laws are fulfilled, like electric charge etc
Of course.
 
  • #9
topsquark said:
That's more or less the content of the lifetime amplitude. But wouldn't that define the measure of the stability of a particle? That's the way I've always looked at it, anyway. Is there a more exact definition of stability?
Uranium-238 with a half life of 4.5 billion years decays to thorium-234 with a half life of 24 days (which then decays to 234mPa with a half life of just 1 minute). I see no way how the thorium isotope could be called "more stable" than its parent.

##\Lambda## with a lifetime of 2.6*10-10 seconds has a 1/3 chance to decay to a neutron plus ##\pi^0##, the latter has a lifetime of just 8*10-17 seconds.
 
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  • #10
mfb said:
Uranium-238 with a half life of 4.5 billion years decays to thorium-234 with a half life of 24 days (which then decays to 234mPa with a half life of just 1 minute). I see no way how the thorium isotope could be called "more stable" than its parent.

##\Lambda## with a lifetime of 2.6*10-10 seconds has a 1/3 chance to decay to a neutron plus ##\pi^0##, the latter has a lifetime of just 8*10-17 seconds.
A very good point. Even though the problem started with a nucleus, I hadn't really thought that one through. (I usually ignore anything bigger than a proton. :) )

Thanks for the catch!

-Dan
 
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1. What are baryons and how are they different from other subatomic particles?

Baryons are a type of subatomic particle that are composed of three quarks. They are different from other particles, such as leptons, because they are affected by the strong nuclear force and can be found in the nucleus of an atom.

2. How do baryons decay and what is the significance of specific decays?

Baryons can decay through the weak nuclear force, which is responsible for radioactive decay. Specific decays refer to the specific combinations of quarks that make up a baryon and how they decay into other particles. Studying these specific decays can provide insight into the fundamental forces and interactions of the subatomic world.

3. What is the role of baryons in the Standard Model of particle physics?

Baryons are an essential component of the Standard Model, which is a theory that describes the fundamental particles and forces in the universe. Baryons, along with other particles, are classified and organized in the Standard Model based on their properties and interactions.

4. Can baryons be created or destroyed?

Baryons can be created or destroyed through high-energy collisions, such as those that occur in particle accelerators. However, in everyday life, baryons are conserved and cannot be created or destroyed, only transformed into other particles through decay or interactions.

5. How are baryons studied and what are some current research topics in this field?

Baryons are studied through experiments in particle accelerators, as well as through theoretical models and simulations. Current research topics in this field include the search for new and exotic baryon states, the study of baryon asymmetry in the universe, and the investigation of baryon properties and interactions at high energies.

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