Radioactive Decay - Beta Particles

pfalk
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One thing that is really confusing me is Beta Particles.

This site states that Beta Particles are electrons that are shot out of the nucleus (atomic core) of an atom.
http://home.clara.net/darvill/nucrad/morebeta.htm

To rationalize this claim it states that a neutron will decay to a proton and an electron. And that the electron gets shot out of the nucleus.

This doesn't make much sense to me. Protons are 2 up quarks and a down quark and Neutrons are 2 down quarks and an up quark. When a neutron decays to a proton one of the down quarks changes to an up quark.
Does this process release an electron?
Or is the site wrong?

Thanks!
 
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I only know that neutrons do decay into and electron and a neutrino, giving off radiation. and the free electron is called the beta particle. As for the quark equations I do not know, but the way quarks interact with each other at that level, are much different than what other particles usualy do.

A lot of the math is explained here:
http://en.wikipedia.org/wiki/Neutron

My thoughts are that there are massive particles composed of multiple fundamental particles and free, fast fundamental particles. Our current univese likes to have particles that are in the middle. With the amount of gravity pressent, neutrons are much to massive to be floatting around alone, and then break apart, losing some mass, and then releasing energy. Thus giving off some radiation and some smaller and faster moving particles.

But in nuetron starts, were gravity just has enogh striength to over-come the electromagnetic force..so the electrons collide with the protons, and then become nuetrons again. It seems that the stablity of particles really realize on the amount gravity and pressure on the object.
 
pfalk said:
When a neutron decays to a proton one of the down quarks changes to an up quark.
Does this process release an electron?

Yes. It also releases an anti-electron-neutrino. The complete process is

d \rightarrow u + e^{-} + {\overline \nu}_e

This is understood as a two-step process involving a virtual W^{-}:

d \rightarrow u + W^{-}

followed by

W^{-} \rightarrow e^{-} + {\overline \nu}_e
 
Thanks for the answers.
But I'm alittle bit confused now.

Initially I thought that when a neutron decayed to a proton that it also released (as stated in the replies) an electron and an anti-neutrino.
I was talking with a condensed matter physicist and got chewed apart for saying that.
She told me that electrons are leptons and that leptons are fundamentally different from quarks. And that an electron would never be produced by the decay of a neutron to a proton.
She stated that the notion of a neutron = a proton + an electron + an anti-neutrino is wrong. That this formula is only used because it accounts for the differing mass between a neutron and a proton.

So, is she incorrect?
Thanks!
 
jtbell said:
Yes. It also releases an anti-electron-neutrino. The complete process is

d \rightarrow u + e^{-} + {\overline \nu}_e

This is understood as a two-step process involving a virtual W^{-}:

d \rightarrow u + W^{-}

followed by

W^{-} \rightarrow e^{-} + {\overline \nu}_e

Hi jtbell,
I don't mean to be a pest.
But could you explain/define the terms being used?
I'm a novice's novice when it comes to this.
Thank you very much.
 
She is incorrect pfalk.

The nomenclature by jtbell is as follows:

d \rightarrow u + e^{-} + {\overline \nu}_e

d is a d-quark
u is a u-quark
e^- is an electron
{\overline \nu}_e is an anti-electron-neutrino

The W^- is the negative charged W-boson, the force carrier of the weak nuclear force.

The arrow means that what's on the left hand side is converted into what is on the right hand side.

That one puts an electron in the equation where the neutron goes to a proton just for the sake of mass conservation is not true. You should give her a copy of this conversation + an introductory textbook on subatomic physics.
 
Thanks again for the responses!
Alot of things are getting cleared up for me.
 
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