# How can I explain why the weak nuclear interaction between individual nucleons is 'weak'?

By considering the energy-time uncertainty principle, estimate the range of the weak nuclear interaction at low energies. Compare this range to the size of a typical nucleon (for example, a proton) and therefore explain why the weak nuclear interaction between individual nucleons is 'weak'.

I'm a bit stuck with this question: to estimate the range of weak nuclear interactions do I need to put the energy levels of a quark into the energy-time principle? Beyond doing this I am a bit lost!

Any comments would be much appreciated; this is an assignment so the work has to be original.

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Again, you should be thinking about the weak boson exchange. Those bosons have mass and the energy to produce one is not "really" available...so where does it come from and how does that limit the range of the interaction? Another question for the student: why is this not a problem for the electromagnetic interaction? – dmckee Sep 13 '11 at 16:32

You are supposed to take 1/M and turn it into a length using dimensional analysis with hbars and c's． This is the compton wavelength of the force carrier, which gives the range of the interaction.

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There's a relationship between the mass of a force boson and the range of the force that should have been covered in class (or is easy to figure out). It seems that you need the size of the proton. Not sure how you're supposed to get that; maybe looking it up is enough.

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Defining the "size" of a proton turns out to be harder than one would imagine, but depending on the metric you choose you generally get an answer between $0.5$ and $1.5\text{ fm}$ (these limits are the distance scale of the nucleon--nucleon hard core potential and the RMS charge distribution of the proton respectively). For BOTE work $1\text{ fm}$ is both easy and reasonably correct. – dmckee Sep 14 '11 at 22:44
@dmckee was it not defined as fermi in honor of Fermi who contributed so much to understanding nuclear interactions, because it is of the order of the effective range of nuclear interactions? – anna v Oct 10 '11 at 7:22
@annav $fm$ stands for both fermi and femtometer (they are the same unit). Yes, it's in honor of Enrico Fermi, though I don't know exactly why. Most probably due to his contribution to nuclear physics. – Manishearth Feb 8 '12 at 17:25
@annav From Wikipedia: "The fermi is named after the Italian physicist Enrico Fermi (1901–1954), one of the founders of nuclear physics. The term was coined by Robert Hofstadter in a 1956 paper published in the Reviews of Modern Physics journal, "Electron Scattering and Nuclear Structure"." enwp.org/Femtometer – Manishearth Feb 8 '12 at 17:26

The range of a force (at the level of this question) is proportional to the de Broglie wavelength associated with the mass of the exchange boson. Try googling the formula for it and compute the corresponding range for the mass of the $W^+$, $W^-$ and $Z^0$ and compare what you get with the range of the massless $\gamma$

:-)

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