Question

1. Assume the axon had a diameter of 30um and was 5 cm long. The capacitance of this axon would be ____ nF.

2. Number of Na⁺ ions that are present before the action potential is ____ x 10¹²

3. Number of Na⁺ ions that flow during one action potential is ____ x 10⁸ .

4. Fractional change in the concentration of Na⁺ ions inside the axon during one action potential is ____ x 10^-5 .

Answer 1

1. Given
An axon has d = 30*10^-6 m
length, l = 0.05 m
now, for a cylinderical capacitor,
inner radius, ri, outer radius r2, then
hence, capacitance is C = 2*pi*epsilon/ln(ro/ri)
now for axon, thickness of mylein sheath is 0.3*10^-6 m
hence, ri = d/2, ro = d/2 + 0.3*10^-6 m
hence
C = 2*pi*8.854*10^-12 /ln((15 + 0.3)/(15))
C = 2.80928999*10^-9 F = 2.8 nF

2. Before action potential, the neuron has a resting potential
resting potential = -70 mV
using Q = CV
we have
number of Na+ ions = 2.8*10^-9*70*10^(-3)*(-1)/1.6*10^-19 = -0.00122906437219*10^12

3. acting potential = 30 mV
hence, Na+ inos = 2.8*10^-9*30*10^(-3)/1.6*10^-19 = 0.000525*10^12 ions
number of ions those flow is 0.000525*10^12 ions + 0.00122906437219*10^12 = 1.75406437219 *10^9 ions

4. fractional change in concentration is
df = 1.75406437219/0.000525*1000 = 3.3410749 %
hence, in fraction, 0.0334107499464761904761904761904 = 3341.07 *10^-5