Weak contribution to nuclear binding Does the weak nuclear force play a role (positive or negative) in nuclear binding?
Normally you only see discussions about weak decay and flavour changing physics, but is there a contribution to nuclear binding when a proton and neutron exchange a $W^\pm$ and thus exchange places? Or do $Z$ exchanges / neutral currents contribute?
Is the strength of this interaction so small that it's completely ignorable when compared to the nuclear binding due to residual strong interactions?
 A: A back of the envelope calculation suffices, meaning all factors of $2,3, \pi$ etc.
have been ignored.
The residual strong force is mainly due to pion exchange and can be modeled
by this plus a short range repulsion due to exchange of $\omega$ mesons. The Yukawa potential from pion exchange is $e^{-m_\pi r}/r$. The weak interactions arise from $W^\pm$ and $Z$ exchange and will give rise to a similar potential with
$m_\pi$ replaced by $m_Z$ (in the spirit of the calculation I take $m_W \sim m_Z$). At typical nuclear densities of $.16/(\rm fermi)^3$ nuclei are separated by distances which are within a factor of two of $1/m_\pi$. Thus the ratio of the Yukawa potential due to weak exchange to the Yukawa potential due to pion exchange at these densities is roughly $e^{-m_W/m_\pi} \sim e^{-640}$. I've ignored the fact that there are different coupling constants in front of the two potentials, but this difference is irrelevant compared to the factor of $e^{-640}$. So yeah, you can ignore the weak contribution to the binding energy.
A: This question is experimentally accessible, despite the feebleness of the weak interaction, because the strong and electromagnetic interactions are symmetric under parity transformations and the weak interaction is not.  
The contribution to the binding energy is small enough that it's not a good way to think of things.  Better is to continue the process of trying to describe nuclear energy eigenstates as linear combinations of different spin-orbit states.  For instance, the deuteron ground state has isospin zero and spin, parity $J^P=1^+$, and so must be a linear combination of the even-$L$ spin triplets 
$\left|{}^3S_1{}^{T=0}\right>$ 
and 
$\left|{}^3D_1{}^{T=0}\right>$; 
the d-wave component famously contributes about 4% of the wavefunction and was the first evidence for the tensor nature of the nuclear force.  But because the weak interaction contributes to the nuclear interaction, the ground state isn't an exact eigenstate of the parity operator (or, for that matter, of isospin) and there's a little bit of p-wave mixed in:
$$
\left|\text{deuteron}\right>
=
\sqrt{0.96}\left|{}^3S_1{}^{T=0}\right>
+
\sqrt{0.04}\left|{}^3D_1{}^{T=0}\right>
+
\epsilon_0\left|{}^3P_1{}^{T=0}\right>
+
\epsilon_1\left|{}^1P_1{}^{T=1}\right>
$$
In the formation of deuterium by neutron capture on hydrogen, you get interference between parity-allowed capture to the $S$- and $D$-wave states and parity-forbidden capture to the $P$-wave states.  These interferences manifest as asymmetries or spontaneous polarizations in the photons emitted during capture which are more or less linear in the amount of $P$-wave mixing; typical asymmetries are a few parts per billion.
In heavier nuclei (e.g. helium & beyond) you lose the luxury of a ground-state wavefunction which can be described in a paragraph, or even at all.  However, a perturbation-theory way of describing the influence of the weak interaction is to say that a particular physical eigenstate with, say, positive parity $\left|\psi^+_\text{physical}\right>$ will be mostly given by a strong-force eigenstate with definite parity, but contain contributions from nearby opposite-parity states due to the weak interaction:
$$
\left| \psi^+_\text{physical} \right>
=
\left| \psi^+ \right>
+
\sum_i
\left| \psi^-_i \right>
\frac{
\left< \psi^-_i \middle|
H_\text{weak}
\middle| \psi^+ \right>
}{
E_i - E_+
}
$$
In heavy nuclei with a dense forest of excited states, you sometimes find same-spin, opposite-parity states which have very different lifetimes and very similar energies; these states are prime candidates to exhibit parity mixing due to the weak interaction.
There's a famous excitation in lanthanum which decays by emitting photons with a 10% parity-forbidden directional asymmetry.
Microscopically, your other answers are correct that the nucleus is too large and the overlap between nucleons too small for appreciable exchange of $W$ and $Z$ bosons.  But you can of course say the same thing about nucleons and exchange of gluons.  The effective theory of the weak interaction between nucleons models the nuclear force as an exchange of strong mesons (the $\pi,\rho,\omega$) where each nucleon-nucleon-meson vertex with a given set of quantum numbers has a particular parity-nonconserving amplitude.  (There was some effort a few years ago to move into the twenty-first century and come up with an "effective field theory" which described the nucleon-nucleon weak interaction without mesons; a big pile of work seems to have produced a one-to-one relationship between the coupling constants in the modern effective field theory and the coupling constants in the old meson theory.)
This has been a pretty long-winded preparation for my answer to your question: the contribution of the weak interaction to the energy of any particular nuclear state is pretty small, for the same reason that the Coulomb-force contribution to the energies of light nuclei can generally be neglected.  What's more interesting is to try an use the short-range nature of the weak interaction to peek at high-energy physics hiding inside of stable nuclei.
A: From this blog post of mine, one should be inclined to think that for large nuclei it can at least contribute to the spin-orbit force, and then to the correction to N=50 and N=82 nuclear shells. 
As QGR noted above, the exponential suppression of the potential does not need to be all the history. Note that usually the total potential in a nucleus is Woods-Saxon, where the exponential does not factor out.
Independently of particle mass, a low-momentum exchange is able to see all the nucleus; when a beta decay happens, the electron carries a momentum of less than 100 MeVs and then it is delocalized across all the nucleus. This is the same that when a electronic transition happens in an atom: the photon carries a momentum of only some electron-Volts, and then it is delocalised across all the orbital.
Consider that a typical contribution to nuclear shells must be about 2 or 3 MeVs. Most (well, a lot) of the nuclear shell correction comes from exchange of $\omega$ and other mesons, which have a mass around the 750 MeV range. Quotient by W mass, and we could expect linear corrections of the order of a 1%. 
So, if some collective effect can do a linear correction to appear, it could be more or less in the range to be noticeable: (750 MeV)^2/81GeV=6.9 MeV.   
To look for collective effects was the idea of the post I have referred in the first line: check the favored masses when the "liquid drop" nucleus splits in two pieces. The small fragment of fission channel S3 happens to have an average mass of 79.21 $\pm$ 1.14 GeV across a range of nuclei from 233Pa to 245Bk; the small frament of channel S2 happens to have an average mass of 92.34 $\pm$ 2.91 GeV.   
A: The answer is, there is a very small affect. If you take a look at the Wikipedia article, you will note that the weak force is about 10^-13 weaker then the strong force. Therefore it's just not going to have much of an affect. 
