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: 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 = 30 цm

Radius = 30 цm/2 = 15 Цm = 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