Action Potential with Nerve Cell and Finding Power

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During an action potential, Na+ ions enter the cell at a rate of approximately 3 x 10^-7 mol/m^2 s, requiring power from the active Na+ pumping system to overcome a +30 mV potential difference. To calculate the power, the formula P = IV is used, where I represents the current and V the voltage. The dimensions of the axon, 40 cm in length and 30 µm in diameter, are essential for estimating the cross-sectional area needed for current calculations. The conversion of moles of Na+ ions to charge in coulombs is necessary to determine the actual current. Understanding these relationships is crucial for solving the problem effectively.
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Homework Statement



During an action potential, Na + ions move into the cell at a rate of about 3 x 10^-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 30 mu m in diameter.

Homework Equations



P = IV
maybe R = ρL / A

The Attempt at a Solution



I understand I need to use P=IV but I don't understand how to incorporate the measurements of the axon without knowing the resistivity of the liquid. Should I assume water?

P = (3.7 * 10^-7 mol/m^2 s)(0.03 V)
P ≈ 9.09 * 10^-9 W
 
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The diffusion rate is given in terms of moles per unit area per second. Use the axon dimensions to estimate area. You'll also have to convert the moles of charges into actual charge units (what's the unit charge of an Na+ ion in coulombs?).
 
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