Question: Why is it that the produced clock frequency is 6.144 MHz, but internally it (8085 processor) uses only 3.072 MHz. Also what leads to the specific value of 6.144 in a clock. 

I found an answer at yahoo.... <http://answers.yahoo.com/question/index?qid=20080810090119AAurr2i>

but I must admit I didn't still get it it well. Could any of you guys throw out a few lines on this, please?

Answer: 
There are lots of reasons to split the instruction cycle into multiple clock cycles. A good example is accessing the main memory bus.

Most modern processors are Von-Neumann architectures; that is, their code and data both exist in the same memory chip. Well, if you want to read an instruction, and that instruction is going to load a variable from memory...that's two memory accesses. But most memory is only single-port (that is, it can only do one read or write per cycle). So how do you read the instruction *and* read your variable?

The solution is to use a two-stage instruction cycle. The first stage will fetch the instruction from memory, and the second stage can then read (or write!) the variable from main memory.

Some older chips went even further. Back in the day, if your chip had 16-bits of addressable memory, but the external address bus is only 8-bits, then you would be familiar with the Address Latch Enable. One clock cycle sends the upper 8-bits of the 16-bit address, and the next clock cycle sends the lower 8-bits. A third cycle could then read/write the variable from/to memory.

There are other, better reasons to have an instruction cycle that is multiple clock cycles in length. One of the best reasons is pipelining. This is a trick that modern processors use to more fully exploit all the execution units available in a chip; for example, while one instruction is being executed, the next is being fetched at the same time.
Clock is nothing but a moment when you want an event to occur. Now, we do not prefer level triggering in digital circuits because when more time is given, interaction may occur between different circuits within the microprocessor leading to short circuits. So we GO FOR EDGE TRIGGERING.
Now, the microprocessor doesn't know when edge triggering has occured. It only understands 1 or 0. So a flip flop is required to produce 1 and 0 whenever a clock edge is detected. Thus , it becomes necessary to use a flip flop. As a result the clock frequency is divided by two.