Final answer:
The depletion layer width at Junction A decreases with light doping, which is true. A lightly dopd junction has fewer charge carriers, resulting in a narrower depletion region (option a).
Step-by-step explanation:
The statement that the depletion layer width of Junction A decreases with light doping is True. When a semiconductor junction, such as a p-n junction, is lightly dopd, it means that there are fewer dopant atoms to contribute free charge carriers (electrons and holes). Consequently, there's a lower level of injected carriers on each side of the junction, which results in a narrower depletion region. The depletion layer is where the free charge carriers have been depleted, and an electric field is established due to the uncovered ions. This field opposes further charge carrier motion, resulting in the balance between drift and diffusion currents, and the existence of a potential barrier.
Hence, the answer is option a.