Question

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 70 cm long and 30 μm in diameter.

Answer 1

Answer: The correct question is

During an action potential, Na+ ions move into the cell at a rate of about 3×10−7mol/m2⋅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 70 cm long and 30 μm in diameter.

Explanation:

length of axon is 70cm= 0.7m

potential difference= 30mv = 30/1000= 0.03V

Diameter= 30цm

Radius= 30цm/2= 15Цm= 15 x 10⁻⁶m

area= 2πrl= 2 x 3.14 x 15x 10⁻⁶m x 0.7 =0.0000659

P= QV/t where P= power, Q is charge, V is potential difference and t is time.

Q/t = avogadro*number of ions*area*e⁻charge

=(6.02 x 10²³)(3 x 10⁻⁷)(0.0000659)(1.6 x 10⁻¹⁹) =0.0019 x 10⁻³c/s

Power = 0.0019 x 10⁻³c/s x 0.03V = 5.7 X 10⁻⁸W