Question

Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions, each with charge +e, into the axon. Measurements have revealed that typically about 5.7×10^11 Na+ ions enter each meter of the axon during a time of 7.0 ms. What is the current during this inflow of charge in a meter of axon?

Answer 1

Answer:

$I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A$

Explanation:

For this case we have the following info given:

Number of Na+ ions $5.7 x10^{11} ions$

Each ion have a charge of +e and the crage of the electron is $1.6 x10^{-19}C$

The time is given $t = 7 ms$ if we convert this into seconds we got:

$t = 7ms * \frac{1s}{1000 ms}= 0.007s$

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

$Q = Ne$

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

$Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C$

And from the general definition of current we know that:

$I =\frac{Q}{t}$

And since we know the total charge Q and the time we can replace:

$I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A$

The current during the inflow charge in the meter axon for this case is $13.02 \mu A$