Question

Animal cells have a membrane that separates the interior of the cell from the outside environment. Typically, an electric potential difference exists between the inner and outer surfaces of the membrane. Consider one such cell where the magnitude of the potential difference is 66 mV, and the outer surface of the membrane is at a higher potential than the inner surface. A sodium ion (Na+) is initially just inside the cell membrane (initially at rest). How much work (in J) is required for a cell to eject the ion, so that it moves from the interior of the cell to the exterior?

Answer 1

Answer:

1.05 × 10⁻²¹J

Explanation:

The potential difference between the exterior and the interior of the cell is important for the conduction of nerve impulse in the body.

The work required to eject the electron can be calculated as follows:

w = q × v

Here, w is work done, q is the charge on sodium ion = 1.6 × 10⁻¹⁹J and v is the voltage = 66 mV.

W =  1.6 × 10⁻¹⁹ × 66 mV

W = 1.05 × 10⁻²¹J.

Thus, the answer is 1.05 × 10⁻²¹J.