How do nerve impulses travel within a nerve cell?

The membrane of all nerve cells maintains a potential difference of -70mV by pumping positive Na+ ions out of the cell. This is called the resting potential. A nerve impulse is a change in this potential - an increase up to +30mV - caused by an influx of these Na+ ions, this is called depolarisation. When a new area of the nerve cell membrane is reached by this depolarisation, if the potential change is large enough (above the threshold potential ~45mV) , it allows many voltage-gated Na+ protein channels to open. This allows a new influx of Na+ ions, which increases the potential in this region. In myelinated motor neurons, these regions are the Nodes of Ranvier which lie between Schwann cells, leading to large gaps between nodes, while in unmyelinated relay neurons, these regions may simply lie next to each other continuously.
Once a region of a nerve cell has been depolarised, it then uses Na+/K+ pumps (carrier proteins) to return to the resting potential, such that it can transmit another impulse. The time between depolarisation, and the readiness to transmit another signal is called the refractory period.

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Answered by Felix W. Biology tutor

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