8085-style microprocessor basics: In a typical 8-bit microprocessor system (such as Intel 8085 family), how many bits wide is the data bus used for reading/writing data?

Introduction / Context:
The data bus width indicates how many data bits a processor can transfer in one operation. In classic 8-bit microprocessors (e.g., 8085/6502/Z80), the data bus is central to memory and I/O transfers. This question focuses on identifying the data bus width in such systems.

Given Data / Assumptions:

• Target class: 8-bit microprocessor architecture (e.g., 8085).
• We refer to external data path width (bus), not internal micro-operations.
• Standard memory-mapped I/O assumed.

Concept / Approach:
An ”8-bit” processor label historically denotes an 8-bit ALU and an 8-bit external data bus. This allows byte-wide transfers in a single bus cycle. Address buses are often wider (e.g., 16 bits) to select many memory locations, but data bus width defines the size of each atomic data transfer.

Step-by-Step Solution:
Recognize the 8-bit processor class: byte-level operations.Match bus to processor width: external data bus = 8 bits.Confirm by device documentation: 8085 data bus D0–D7 (8 lines).Therefore, the correct choice is 8 bits.

Verification / Alternative check:
Look at pinouts: the 8085 has D0 through D7 pins, clearly indicating an 8-bit data bus.

Why Other Options Are Wrong:
7 and 9: nonstandard widths for classic 8-bit MCUs/MPUs.16: corresponds to 16-bit processors (e.g., 8086 external data bus is 16 bits), not an 8-bit device.

Common Pitfalls:
Confusing address bus width (often 16 bits in 8-bit systems) with data bus width (8 bits). These serve different roles.

Final Answer:
8