Consider a short section to an unmyelinated axon of length l

Consider a short section to an unmyelinated axon of length l_R = 1 mu m. The radius of the axon is a = 1 mu m, and the membrane thickness b = 6 nm. At activation, Na^+ ions move from outside to inside the axon causing the potential change tromp -7mV to + 40 mV. Considering that the potential change is caused by the changes on the surface charge of the membrane (due to the excess Na^+ ions inside), calculate the number of the excess Na^+ ions that move from outside to inside at activation. At rest, the concentration of Na^+ ions is 15 m mole/liter in the inside of the axon. Calculate the number of Na^+ ions in the inside of the axon at rest for the section of the axon caused by the calculate the fractional change in the number of Na ions in the inside the activation.

Solution

a)The number of excess ions is 70mv+40mv=110

b)Number of ions inside the axon=number of excessions-concentration of the ions=>110-15=95

fractional change =number of ions inside/number of excess ions here=95/110=0.86