Question

Consider one such cell where the magnitude of the potential difference is 40 mV, and the inner surface of the membrane is at a higher potential than the outer surface. A sodium ion (Na ) 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 = 6.4x10⁻²¹J

Explanation: Work is the energy transferred to or from an object due to the application of a force along a displacement.

The ion Na is gaining energy to move from a lower potential to a higher one.

For this to happen, the work done is calculated as

$W=q.\Delta V$

in which

W is work in Joules (J)

q is charge in Coulomb (C)

$\Delta V$ is potential difference in Volts (V)

The ion Na has one positive charge. In coulombs, the charge is

q = 1.6x10⁻¹⁹C

The potential difference in Volts is

$\Delta V$ = 40x10⁻³ or 4x10⁻²V

Calculating

$W=q.\Delta V$

W = 1.6x10⁻¹⁹. 4x10⁻²

W = 6.4 x 10⁻²¹ J

The work done to move an ion of sodium from the exterior to the interior is 6.4 x 10⁻²¹ J.