Question

The inside of a cell membrane is at a potential of \(90.0 \mathrm{mV}\) lower than the outside. How much work does the electric field do when a sodium ion (Na') with a charge of \(+e\) moves through the membrane from outside to inside?

Short answer

Answer: The work done by the electric field on a sodium ion is -1.44 x 10^-20 J.

Step-by-step solution

Identify the given quantities

We are given the following information: - Potential difference between inside and outside of the cell membrane (\(\Delta V\)): \(-90.0 \mathrm{mV}\) - Charge of a sodium ion (\(q\)): \(+e\)

Convert the potential difference to volts (V)

The potential difference value is given in millivolts (mV). Convert it to volts (V) by dividing by 1000: \(\Delta V = -90.0 \mathrm{mV} \times \frac{1 \mathrm{V}}{1000 \mathrm{mV}} = -0.090 \mathrm{V}\)

Calculate the work done by the electric field

Use the formula for work done by an electric field, \(W=q\Delta V\), where \(q\) is the charge of the sodium ion and \(\Delta V\) is the potential difference: \(W = (+e)(-0.090 \mathrm{V})\) Here, \(e\) represents the elementary charge, which is approximately \(1.6 \times 10^{-19} \mathrm{C}\). Substitute this value into the equation: \(W = (1.6 \times 10^{-19}\mathrm{C})(-0.090 \mathrm{V})\)

Calculate the final answer

Multiply the values to get the work done by the electric field: \(W = -1.44 \times 10^{-20} \mathrm{J}\) The electric field does \(-1.44 \times 10^{-20} \mathrm{J}\) of work on the sodium ion as it moves through the cell membrane from outside to inside. The negative sign indicates that the work is done against the electric force, which means that the work is done on the sodium ion.