Final answer:
Sodium ions entering the node of Ranvier cause membrane depolarization by shifting the internal charge of the neuron to be more positive. Following depolarization, potassium ions exit the cell restoring the negative internal charge (repolarization). This process facilitates rapid signal transmission through saltatory conduction in myelinated neurons.
Step-by-step explanation:
What Happens Once Na+ Enters the Node of Ranvier?
When sodium ions (Na+) enter the node of Ranvier, they initiate a crucial phase of neuronal signal transmission known as the action potential. Upon entering the node, the Na+ ions cause a change in the electrical charge across the neuron's membrane. Sodium ions, due to a high outside-to-inside concentration gradient, rush into the neuron, moving the membrane potential from its resting state of about -70 mV towards zero, and subsequently, up to about +30 mV. This rapid influx of positively charged ions causes the membrane to depolarize, shifting the internal charge of the axon segment at the node to be more positive.
As the membrane potential reaches the peak at +30 mV, another event occurs - the opening of potassium ion (K+) channels. These channels allow K+ to exit the cell, leading to the repolarization phase where the inside of the membrane becomes negative again. Both Na+ and K+ channels eventually close, and the sodium-potassium pump works to restore the resting potential.
The crucial aspect of this process in myelinated neurons is that it allows the action potential to propagate quickly down the axon through a mechanism called saltatory conduction. The action potential 'jumps' from one node of Ranvier to the next due to the myelin sheath covering other segments of the axon and concentrating ion flow at the nodes. The efficacious distance between nodes of Ranvier ensures that the depolarization remains above the threshold needed to activate the next node, facilitating rapid signal propagation.