Question

If the action potential starts with a potential difference of +40.0 mV, and the length constant of the axon is 2.00 mm, how far from the site of the stimulus will the potential difference across the membrane drop to 1.00 mV (assuming the action potential is not regenerated)?

Answer 1

Answer:

Answer is given in the attachment.

Explanation:

Answer 2

Answer:

= 7.38mm

Explanation:

The decrease in potential difference across membrane in the distance is given by the equation

$V(x) = V_{max}(e^{-x/l})$

where

$V_{max}$ = max voltage attained in action potential

l = length constant

x = the distance

$V(x) = V_{max}(e^{-x/l})$

$1mv=40mv(e^{-x/2})$

taking natural log on both side

$In(\frac{1}{40} )=\frac{-x}{2} \\\\3.69=\frac{x}{2} \\\\x = 7.38mm$