Reason for a thin base of transistor I have been recently asked the following question:

Why is the base of a transistor kept thin?

MY ASSUMPTION
Base is kept thin so that not many charge carriers could be neutralized in the process of biasing ...
AM I CORRECT?
 A: Your assumption is superficially OK, but some key specifics are not there. For example, which charge carriers are you thinking about - those injected from the emitter, or those injected through the base contact? Next, what do you mean by 'neutralized' - are you thinking of recombination and detailed balance of the majority and minority carriers or not?
So: with no base current, forward biasing the emitter-base junction will inject majority carriers from the emitter into the base, where they are minority carriers. The base width is chosen such that those excess minority carriers will recombine before reaching the collector. Here, the wider the base the better to some extent.
Now add in base current - what happens now? There are additional minority carriers and these can dramatically shift, through detailed balance describing carrier recombination, the steady state minority carrier concentration. Thus, some of the minority carriers will survive the base to get to the collector. Small changes in base injection will cause large shifts in the steady state minority carrier population, accounting for the gain of the device. Under these conditions, a narrower base is advantageous - a greater fraction of the emitter-base current will make it to the collector at a given steady-state minority carrier population. 
So, the base has to be wide enough to not have complete recombination of minority carriers injected from the emitter when there is no base current. And, it has to be thin enough so that minority carriers will transit the base under appropriate base currents. But, it is all about carrier recombination and the difference between equilibrium (no injection) and steady-state (with injection) minority carrier concentrations.
A: The base of bipolar transistors is kept thin mainly to improve their speed of operation, which is related to the transit time $t_T$ of the electrons from the emitter to the collector which decreases with decreasing thickness of the base. In modern transistors, recombination in the base is practically negligible. The so-called transit frequency $$f_T=\frac {1}{2\pi t_T}$$ is an important figure of merit of a bipolar transistor which is related to the upper frequency of operation of a bipolar transistor. 
