During an action potential, Na^+ ions move into the cell at a rate of about 3x10^-7 mol/m^2*s.

How much power must be produced by the "active Na^+ pumping" system to produce this flow against a +30-mV potential difference? Assume that the axon is 40cm long and 30um in diameter.

Please help me out I am lost. Thanks

I am not familiar with the terminology. How can a "potential" be an event or a time interval?

Power = (current) x (voltage)

The voltage is apparently 30*10^-3 V

Convert the 3x10^-7 mol/(m^2*s) to
coulombs/(m^2*s)
1 mole of Na+ ions contains 6.02*10^23 ions with a charge of 9.63*10^4 Coulombs. Your ion flux is 2.89*10^-2 Amps/m^2

Multiply that by the appropriate axon area.

Can you please help out am still stuck one this

During an action potential, Na^+ ions move into the cell at a rate of about 3x10^-7 mol/m^2*s.

How much power must be produced by the "active Na^+ pumping" system to produce this flow against a +30-mV potential difference? Assume that the axon is 40cm long and 30um in diameter.

Please help me out I am lost. Thanks
Physics please help - drwls, Sunday, February 10, 2013 at 12:52pm
I am not familiar with the terminology. How can a "potential" be an event or a time interval?

Power = (current) x (voltage)

The voltage is apparently 30*10^-3 V

Convert the 3x10^-7 mol/(m^2*s) to
coulombs/(m^2*s)
1 mole of Na+ ions contains 6.02*10^23 ions with a charge of 9.63*10^4 Coulombs. Your ion flux is 2.89*10^-2 Amps/m^2

Multiply that by the appropriate axon area.
Physics please help - John, Tuesday, February 12, 2013 at 1:25am

Can you please help out am still stuck one this!!!

I am really

confue on this question so, pliz could you help answer me this question?

To determine the power required to produce the flow of Na+ ions against the +30 mV potential difference, we first need to calculate the total number of Na+ ions that move into the cell per second.

To calculate the number of Na+ ions that move into the cell per second, we need to know the area of the cell membrane and the rate of Na+ ions flowing per unit area.

Area of the cell membrane = 2 * π * radius * length
= 2 * π * (30μm/2) * 40cm
= 2 * π * 15μm * 40cm
= 1200 * π μm^2
= 1200 * π * 10^-8 m^2 (since 1 μm = 10^-6 m)

The flow rate of Na+ ions per unit area is given as 3x10^-7 mol/m^2*s.

Now, we can calculate the number of Na+ ions moving into the cell per second:

Number of Na+ ions/s = flow rate of Na+ ions * area of the cell membrane
= (3x10^-7 mol/m^2*s) * (1200 * π * 10^-8 m^2)

Next, we need to calculate the electrical work required to move these ions against the potential difference.

The electrical work can be calculated using the equation:

Work = charge * potential difference

Since we know the flow rate of Na+ ions and the elementary charge (e=1.6 x 10^-19 C), we can determine the charge moved per second:

Charge moved/s = flow rate of Na+ ions * elementary charge
= (3x10^-7 mol/m^2*s) * (1200 * π * 10^-8 m^2) * (6.022 x 10^23 ions/mol) * (1.6 x 10^-19 C/ion)

Now, we can calculate the power required to produce this flow against the +30 mV potential difference:

Power = Work / time
= (Charge * potential difference) / time

Since we've already calculated the charge moved per second, we can substitute it in:

Power = (Charge moved/s * potential difference) / time

Remember to convert the potential difference from millivolts to volts before calculating the power.