How reverse current suddenly increase at the breakdown voltage in case of zener diode?

Explanation

We know that the reverse current through the junction diode is due to flow of minority carriers (i.e., flow of electrons from p to n side and holes from n to p-side of p-n junction diode). As the reverse bias voltage across the junction is increased, the electric field at the junction becomes significant. When the reverse bias voltage becomes equal to zener voltage (i.e. `V=V_(z))`, then the electric field strenth across the junction becomes quite high. This electric field across the junction is sufficient to pull valence electrons from the host atoms on the p-side and accelerate them towards n-side. The movement of these electrons across the junction accounts for high current which is observed at the break down reverse voltage.