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ch4aika [34]
3 years ago
5

Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a h

igher 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?
Physics
1 answer:
Gelneren [198K]3 years ago
8 0

Answer: W = 1.04.10^{-20} 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

W=q.\Delta V

q is charge;

ΔV is the potential difference;

Potassium ion has +1 charge, which means:

p = 1.6.10^{-19} C

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

\Delta V=65.10^{-3}V

Then, work is

W=1.6.10^{-19}.65.10^{-3}

W=1.04.10^{-20}

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

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