How does the conduction of a nerve impulse occur?

(N/A) $ \Rightarrow $ Neurons are excitable cells because their membranes are in a polarised state.
Different types of ion channels are present on the neural membrane. These ion channels are selectively permeable to different ions.
When a neuron is not conducting any impulse, i.e., at rest, the axonal membrane is comparatively more permeable to $ K^{+} $ and nearly impermeable to $ Na^{+} $.
- Similarly, the membrane is impermeable to negatively charged proteins present in the axoplasm.
The axoplasm inside the axon contains a high concentration of $ K^{+} $ and negatively charged proteins and a low concentration of $ Na^{+} $.
In contrast, outside the axon, a low concentration of $ K^{+} $ and a high concentration of $ Na^{+} $ form a concentration gradient.
These ionic gradients across the resting membrane are maintained by the active transport of $ 3 Na^{+} $ outwards and $ 2 K^{+} $ inside the cell.
As a result, the outer surface of the axonal membrane is positively charged and its inner surface becomes negatively charged; therefore, it is polarised.
The electrical potential difference across the resting membrane is called the resting potential.
When a stimulus is applied at point $ A $ on the polarised membrane, it becomes freely permeable to $ Na^{+} $. $ Na^{+} $ influx is followed by the reversal of the polarity, i.e., the outer membrane is negatively charged and the inner side is positively charged.
The polarity at the site $ A $ is reversed and hence depolarised. The electrical potential difference across the plasma membrane at the site $ A $ is called the action potential, which is in fact termed as a nerve impulse.
At sites immediately ahead, the axon membrane has a positive charge on the outer surface and a negative charge on its inner surface. As a result, a current flows on the inner surface from site $ A $ to site $ B $. On the outer surface, current flows from site $ B $ to site $ A $ to complete the circuit of current flow. Hence, the polarity at the site is reversed and an action potential is generated at site $ B $.
Thus, the impulse (action potential) generated at site $ A $ arrives at site $ B $.
The sequence is repeated along the length of the axon, and consequently, the impulse is conducted.
The rise in the stimulus-induced permeability to $ Na^{+} $ is short-lived. It is quickly followed by the rise in permeability to $ K^{+} $. Within a fraction of a second, $ K^{+} $ diffuses outside and restores the resting potential of the membrane.