Question

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?

Answer 1

Answer: W =
`1.04.10^(-20)` J

Step-by-step 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.