Question

{"given":"An unbiased (isolated) p-n junction diode — no external voltage applied.","key_observation":"When the p-n junction is formed, electrons diffuse from n to p and holes from p to n, creating a depletion region. The n-side loses electrons and becomes positively charged; the p-side gains electrons and becomes negatively charged. This charge separation produces a built-in electric field pointing from n-side (positive) to p-side (negative).","option_analysis":[{"label":"(A)","text":"the potential is the same everywhere","verdict":"incorrect","explanation":"The depletion region creates a contact potential difference (typically ~0.3–0.7 V), so the potential is NOT the same everywhere; there is a built-in potential barrier."},{"label":"(B)","text":"the p-side is at a higher potential than the n-side","verdict":"incorrect","explanation":"The n-side (positively charged due to ionized donors) is at a higher potential; the p-side (negatively charged due to ionized acceptors) is at a lower potential."},{"label":"(C)","text":"there is an electric field at the junction directed from the n-side to the p-side","verdict":"correct","explanation":"The built-in electric field points from the positively charged n-side to the negatively charged p-side, opposing further diffusion of majority carriers."},{"label":"(D)","text":"there is an electric field at the junction directed from the p-side to the n-side","verdict":"incorrect","explanation":"This is the opposite of the actual field direction; the field points from n to p (not p to n) due to the positive space charge on the n-side and negative space charge on the p-side."}],"answer":"(C)","formula_steps":[]}