Under REVERSE BIAS, an IDEAL p-n junction diode would carry zero current. A real diode, however, carries a very small current. The carriers responsible for this current are

1. Under reverse bias, the external field ADDS to the internal junction field, sweeping majority carriers AWAY from the junction. So majority-carrier diffusion current essentially STOPS. 2. But MINORITY carriers are present in both regions due to thermal generation of electron-hole pairs: a few electrons exist in the p region, a few holes in the n region. 3. The strong reverse field near the junction collects these minority carriers as soon as they arrive at the junction edge, generating a small reverse current. 4. This is called the REVERSE SATURATION CURRENT, $I_0$. It is small (typically nA to µA), strongly dependent on temperature (since minority carrier generation is thermal), and roughly independent of the reverse voltage until breakdown. 5. Option A is the FORWARD-bias mechanism. Option B mixes up carrier generation locations. Option C is wrong — ions are fixed in the lattice. 6. Symbolically, the diode equation captures both directions: $I = I_0(e^{eV/kT} - 1)$ — under reverse bias ($V < 0$), the exponential collapses and $I \to -I_0$. _Source: NCERT Class 12 Physics Part 2, Ch 14, §14.6.2 (p-n junction diode under reverse bias), p. 12–13._