PN Junction paradox or something else?
It does work as you say, with holes and electrons moving away from the junction.  But only for a short moment.  Then it stops.  It stops because the holes in the p-type side moved away from the junction, and the electrons in the n-type side also moved away from the junction.  This creates a wide insulating layer, located at the junction.   (The word &quot;insulating&quot; actually means &quot;lacks mobile charges.&quot;)
In other words, during reverse-bias of any diode, the voltage of the power supply is sweeping out the charge-carriers from the junction.  The &quot;depletion layer&quot; widens.
But, because any doped semiconductor material is always full of immobile charged atoms (the original dopant ions,) this means that the semiconductor depletion-layer has pos&amp;neg &quot;plates,&quot; and it behaves as a static-charged capacitor.  The n-type will be strongly positive-charged whenever its electrons are swept out.  The p-type side is opposite, and the two together form an insulating dielectric with a built-in voltage.   The wider the depletion-layer, the higher the voltage across the &quot;capacitor plates.&quot;
If we apply 30V reverse bias, then the holes and electrons are pulled away, and the depletion layer widens.   But it only becomes exactly wide enough, until the &quot;newly-exposed&quot; ions in the insulating semiconductor have 30V on their &quot;charged capacitor&quot; plates.  This voltage exactly opposes the applied RB reverse-biase voltage from an external power supply.
Doped semiconductors are &quot;electrets,&quot; they are insulating materials with strong charges locked into their matrix.   But also, semiconductors are filled with an equal amount of opposite, movable charges.  These opposite charges cause the semiconductor to become neutral-charged, and also conductive.   However, if we can sweep the charges out of the semiconductor, then it becomes a charged insulator, an &quot;electret.&quot;
(An &quot;electret&quot; is the electrostatic analogy to the permanent magnet.  Rather than being a permanent-magnetized ferromagnet, an &quot;electret&quot; is a permanent-charged dielectric.)
ADDED:
During reverse bias, immense numbers of hole/electron pairs AREN'T created in the depletion layer.  Yes, during forward-bias the hole/electron pairs are canceled out as they fall together, emitting IR photons as they do.   But the opposite doesn't happen (no thermodynamic backwardds-time events, where photons magically come in for no reason, to break covalent bonds and thus create any large population of hole-electron pairs in the depletion layer.)

HOWEVER, this hole/electron &quot;pair production&quot; process DOES occur elsewhere in the diode.  It occurs at the spot where the external metal terminal touches the p-doped side of the diode.
If we touch a metal terminal against p-type silicon, we should be creating anothee diode in series with the man pn junction (creating a Schottky rectifier, a metal/semiconductor junction.)  This produces an unwanted pn junction, since this new junction faces backwards when compared to the main pn junction (and therefore, the diode as a whole would always be turned off!  Opposite junctions connected in series will block diode current.)
To eliminate this unwanted effect, we must form a nonrectifying contact, a metal/semiconductor bond that acts more like a copper-iron thermocouple than like a Schottky diode.  Look up &quot;nonrectifying contact&quot; or &quot;nonrectifying junctions&quot; chapter in semiconductor theory-textbooks.
To intentionally produce non-rectifying contacts, first we heavily dope the semiconductor (called ++p doping, or --n doping.)  Because of some mysterious[note1] quantum-mechanical effects, the heavy doping causes the semiconductor to act like a metal, with a dense charge-cloud inside, and only insignificant depletion layers will be formed by contact with other materials.  (In other words, iron and copper in contact will not form insulating depletion zones, and don't act like diodes.  Same with metal against p++ or n-- doped layers.)
So, first apply ++p doping to the p-type end of your diode, then attach the metal wire there.  When the main pn diode junction is forward-biased, this non-rectifying contact is reverse biased, and new holes are being created at the metal/semiconductor contact.  Covalent bonds of the semiconductor are spontaneously breaking via QM tunneling or thermal vibrations, and a cloud of holes flies off into the ++p type semiconductor, and an equal cloud of electrons flies off into the metal wire.  (Side effect: the temperature of this junction starts dropping.)
(Another classic nonrectifying junction is found in &quot;TE modules&quot; or &quot;Peltier coolers,&quot; where numerous ++p-type and --n-type semiconductor blocks are attached to thin copper connecting straps, yet none of these junctions create any diodes.  They only create thermocouples, where the built-in contact-potential is much larger than is expected for metal-metal thermocouples.)
[1] Not entirely mysterious: any depletion zones formed by metal-against-++p semiconductor will be incredibly thin:  thinner than the wavelength of electron matter-waves.  When electron particles behave as waves, they cannot &quot;see&quot; the insignificant thinness of electrostatic energy-barriers in depletion-zones produced by metal-metal contacts.  Or in other words, if a depletion zone is too thin, electrons will tunnel freely across, even if the depletion zone offers a high energy-barrier, and where Classical Mechanics would predict that it should be an insulator.  The same is true both for metal-against-(++p), and metal-against-(--n) semiconductor contacts.   Then, when this too-thin depletion zone is reverse-biased by an external power supply, the &quot;wave-like&quot; covalent-bond electrons &quot;tunnel out&quot; of the ++p semiconductor (also they absorb thermal vibrations as they jump to the higher energy-band of the metal wire, so the metal-to-(++p) contact becomes colder.  It's just the Peltier Effect of thermocouple junctions.)  What's the size of the small voltage on this junction?  DOESN'T MATTER, because it's exactly opposed by the equal-opposite small voltage appearing at the other metal contact of the diode (as well as opposed by the tiny voltage appearing at the places where +p touches ++p, etc.)  As a whole, it behaves like a copper-iron-copper thermocouple, which creates zero net voltage on its terminals, because the two identical thermocouples in series are wired in opposite directions.