# Beta+ decay of a proton

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2020 Award
The A level book that AS students have for the AQA course has a Feynman diagram of a proton decaying into a neutron - via a W boson, producing a beta+ and a neutrino.

afaiaa, protons have a lifetime of 1033 years, so this doesn't seem a likely event. Is this just a 'fictional' event, put in the book to show symmetry with beta- decay of a neutron or is there something else involved? If some energy is added to make it happen (i.e. whilst the proton is in a nucleus), how come it isn't shown in the Feynman diagram, in some way?

All this stuff was found out twenty years after I started at Uni and general reading doesn't always help a chap with questions like this.
I must say, subjecting kids to this sort of thing, only months after they staggered through a skimpy GCSE Science course, seems pointless. They get presented with this even before the Photoelectric effect, the definition of the Electron Volt and Equations of motion!!!!
Am I just a dinosaur or what? (None of them seem to understand it either)

I'm guessing that the book is referring to a proton in the nucleus.AFAIK it is free protons which are reputed to have such long lifetimes.

Astronuc
Staff Emeritus
afaiaa, protons have a lifetime of 1033 years, so this doesn't seem a likely event.
That would be a free proton, e.g., nucleus of a hydrogen atom. In a nucleus with an excess of protons or deficiency of neutrons, positron emission is one way for the nucleus to drop to a lower energy state, and the nucleus and atom assume a different identity. Electron capture is another method of reducing Z. The bottom line is the transformation of a proton to a neutron.

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html#c4

http://hyperphysics.phy-astr.gsu.edu/hbase/particles/lepton.html#c2

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2020 Award
OK, I thought it would have to be something like that but how come the Feynman diagram doesn't include the effect of another particle - or is it the W boson that comes along from elsewhere to make things happen?
It's a bit confusing.

fzero
Homework Helper
Gold Member
OK, I thought it would have to be something like that but how come the Feynman diagram doesn't include the effect of another particle - or is it the W boson that comes along from elsewhere to make things happen?
It's a bit confusing.

Nucleons are bound states of up and down quarks. The weak interaction allows for a down quark to decay to an up quark via

$$d \rightarrow W^- + u$$

and the $$W^-$$ will decay as

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

However a free proton is the minimum energy configuration of uud. The d quark is not free to decay because the resulting end products would have a higher energy than the initial state. But a sufficiently excited state of uud would have enough energy for a u quark to change to a d, making a neutron configuration udd. The base process would be

$$u + \text{energy} \rightarrow W^+ + d \rightarrow e^+ + \nu_e + d$$.

Since the nucleons in a nucleus are constantly exchanging energy, it is possible for a proton in a nucleus to enter an excited state allowing this process to happen.

Astronuc
Staff Emeritus
W-boson is an intermediate state

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html#c2

http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html#c5

http://hyperphysics.phy-astr.gsu.edu/hbase/particles/expar.html#c3
http://www2.slac.stanford.edu/vvc/theory/weakbosons.html

http://www2.slac.stanford.edu/vvc/theory/weakinteract.html

If one looks at the chart of nuclides, one sees that electron capture (K-captured) is actually preferred to positron emission, although positron emission is more like for those radionulides further away from the 'line of stability'.
http://www.nndc.bnl.gov/chart/reCenter.jsp?z=20&n=18 (move along the diagonal - lower-left to upper-right)

What happens is that the proton changes into a neutron by means of the weak force(in terms of quarks uud to udd) with the release of the W plus exchange particle which then transforms into the positron and neutrino.
(For reasons similar to yours I also find it confusing and I just opened the book that I guess you are referring to.CGP AS Physics page 107 )

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2020 Award
Not the CGP book but the Nelson Thornes book, endorsed by AQA (Jim Breithaupt).

So I am right in saying that the Feinman Diagram doesn't tell the whole story??? 'Something' must allow the formation of a W but the simple 'one and three out' diagram doesn't show it.

fzero
Homework Helper
Gold Member
Not the CGP book but the Nelson Thornes book, endorsed by AQA (Jim Breithaupt).

So I am right in saying that the Feinman Diagram doesn't tell the whole story??? 'Something' must allow the formation of a W but the simple 'one and three out' diagram doesn't show it.

The Feynman diagram provides a path for a process to happen, but the probability that the process will happen still depends on the energy of the input particles (as well as physical constants and masses, etc). The process must ultimately satisfy energy and other conservation rules as well.

A worthy analogy might be that a road over a hill gives a path for a car to take from point A to point B, but if nothing is propelling the car, it's not going to make it.

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2020 Award
That makes excellent sense. Thanks very much everyone.

phyzguy
afaiaa, protons have a lifetime of 1033 years, so this doesn't seem a likely event.

Just a comment on this. The 10^33 years is a lower bound on the free proton lifetime. No confirmed proton decay has been observed, and it is possible that the proton is completely stable.

Astronuc
Staff Emeritus