I am struggling to understand the idea of holes being able to move across the junction, however.
Holes move in just the same way electrons do.
One minor difference is that because the curvature of the E-k curve for the valence band is different from the curvature for the conduction band, holes and electrons have different effective masses.
Other than that, they move the same way. If electrons can move across the junction, holes can do the same thing.
In comments you added,
As I understand it, the driving force behind electron diffusion is the repulsion of like charges. The holes are all in the valence band and so unlike electrons cannot move freely unaided, and thus I wouldn't expect them to naturally repel and move into the n-type region.
This is incorrect on a couple of points.
First, just like when uncharged molecules diffuse in a gas, the driving force behind diffusion is nothing more than that the particles are moving randomly, so if there are more of them at point A and fewer at a nearby point B, there will be a net flow of particles from A towards B.
To the extent that the mutual repulsion of the electrons causes a current, we would call that part of the drift current, not part of the diffusion current. (Also remember that, for example, in the n-region there are vast numbers of electrons and they repel each other, there are also fixed positive charges at the ionized donor sites that attract the mobile electrons, and the associated fields can cancel each other out)
Second, electrons can't move freely in the valence band because nearly all electron states are full. But this doesn't prevent holes from moving in the valence band. Holes in the valence band move readily, because the valence band isn't jam-packed with holes, it's jam-packed with electrons. Holes in the valence band move nearly as readily as electrons in the conduction band (see the point about different effective masses above).
In your comment reply to boyfarrel, you said,
Would it be correct to say that the diffusion of the free electrons from the n-type region into the p-type region causes electrons from thermally generated pairs to do the same, thus causing holes to gradually progress further away from the p-type region (and so into the n-type)?
This is again not quite right.
There are (relatively) many electrons in the n-region. These not only got excited from the donor sites into the conduction band, but thermodynamics says they'll take a certain distribution of energies above the conduction band edge. Because of this, a small fraction of them will have enough energy to overcome the potential barrier of the junction, should they randomly happen to move in that direction.
Similarly, on the p-side there are relatively many holes, and a small fraction of these have sufficient energy to overcome the junction barrier should they randomly happen to move in that direction.
The small fraction of the holes from the p-side that happen to cross the barrier over to the n-side form an excess population relative to the equilibrium hole population on the n-side. So, through random motion they have net motion away from the junction.
At the same time, as they randomly move around on the n-side, they have a probability to recombine with the vastly more numerous electrons present on the n-side. So we see (if the n-side is big enough) an exponential drop in this excess hole population as we move away from the junction. (the same way there's an excess electron population on the p-side, dropping exponentially as we move further into the p-side)
The p-side excess electrons aren't there because of anything the holes did, and the n-side excess holes aren't there because of anything the electrons did. They're there because there's a big honking crowd of electrons or holes on the other side of the junction, and a small fraction of those have managed to "jump the fence" and end up on the "wrong" side of the junction.
Best Answer
What happens when you touch an object with a positively charged object?
Ans: It gets positively charged.
Now, you have connected a semiconductor to a positive end of battery. What do you expect?
Ans: Yes, it gets positively charged.
Yes, it will.
But the question is to what extent?
You figured it out correctly that undoped silicon is unwilling to give away electrons, or accept holes, because that would rip gaps in the stable crystalline bonding
Think of a comb getting charged when rubbed with hair. What do you observe?
Ans: You see that charge on comb is localised and do not spread out. The charge don't pass to your hand.
Similarly, the p semiconductor gets locally charged . The point in direct contact with battery gets charged and remaining remains uncharged.
Extent of positive charge and effective distance depends on strength and voltage of battery.
Now, when you connect it to negative potential
This will give you similar result but with little increased magnitude of negative charge on it as it has a lot of holes to be occupied.