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Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a higher potential than the outer surface. A potassium ion (K+) is initially just outside the cell membrane (initially at rest). How much work (in J) is required for a cell to absorb the ion, so that it moves from the exterior of the cell to the interior?

Respuesta :

Answer: W = [tex]1.04.10^{-20}[/tex] J

Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.

The work to transport an ion from a lower potential side to a higher potential side is calculated by

[tex]W=q.\Delta V[/tex]

q is charge;

ΔV is the potential difference;

Potassium ion has +1 charge, which means:

p = [tex]1.6.10^{-19}[/tex] C

To determine work in joules, potential has to be in Volts, so:

[tex]\Delta V=65.10^{-3}V[/tex]

Then, work is

[tex]W=1.6.10^{-19}.65.10^{-3}[/tex]

[tex]W=1.04.10^{-20}[/tex]

To move a potassium ion from the exterior to the interior of the cell, it is required [tex]W=1.04.10^{-20}[/tex]J of energy.

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