How Intel's 8086 microprocessor access in 16-bit word? Explain with example.

The 8086 is a l6-bit microprocessor chip designed by Intel All internal registers, as well as internal and external data buses, are 16-bits wide, which firmly established the “16-bit microprocessor” identity of the 8086. A 20-bit external address bus provides a 1 MB physical address space ( = 1,048,576). This address space is addressed by means of internal memory "segmentation". The data bus is multiplexed with the address bus in order to fit all of the control lines into a standard 40-pin dual inline package. It provides a 16-bit I/O address bus, supporting 64 KB of separate I/O space. The maximum linear address space is limited to 64 KB, simply because internal address/index registers are only 16-bits wide. Programming over 64 KB memory boundaries involves adjusting the segment registers (see below); this difficulty existed until the 80386 architecture introduced wider (32-bit) registers (the memory management hardware in the 80286 did not help in this regard, as its registers are still only 16-bits wide).