If the base-emitter junction of an NPN transistor is forward biased, electrons flow from emitter to base. If the collector-base junction is reverse biased, how can they cross it?
If I connect the emitter to negative and collector to positive terminal of a battery and don't connect any battery to base-emitter junction, what would happen? Will electrons cross the emitter-base junction? What will stop them?
1) The essence of an NPN transistor is that it is not just 2 PN diodes in anti series (like NP - PN ) but an NPN structure where the P-layer is quite thin.
So what happens is that the electrons coming from the emitter (at some speed) while traveling through that very thin base, have a large chance of going too far and ending up in the collector. This happens despite the base-collector junction being reverse biased.
The thing to remember: thin base !
Also, the doping in the base is also such that the base cannot "catch" (recombine) all the electrons so many electrons make it "safely" through the base without recombining (with the holes in the base) and end up in the collector. I hope you realize that having a thin base region helps as well.
2) You mean, leave the base floating.
I only discus the majority carrier currents here. There is also a much smaller minority carrier current, see this question and the answer Since the effect of this is much smaller, I ignore it here.
When the base is floating, the Emitter-base junction cannot come into forward mode. There would then be no electric field to accelerate the electrons from the emitter so they cannot make it into the collector.
So the NPN will not conduct (between collector and emitter) in this case as the electrons from the emitter cannot make it to the collector as their energy is not sufficient. They need the base to be at a higher potential (the base-emitter junction must be in forward mode) in order to accelerate/gain energy to make it into the collector.