# Radioactive Decay - Beta Particles

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.

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!

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

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

My thoughts are that there are massive particals composed of multiple fundamental particals and free, fast fundamental particals. Our current univese likes to have particals 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 particals.

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 particals really realise on the amount gravity and pressure on the object.

jtbell
Mentor
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

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

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!!

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.

malawi_glenn
Homework Helper
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 whats 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.