Question

A na+ ion moves from inside a cell, where the electric potential is -72 mv, to outside the cell, where the potential is 0 v. what is the change in the ion's electric potential energy as it moves from inside to outside the cell?

Answer 1

The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:
$\Delta U = q \Delta V = q (V_f - V_i)$
where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge: $q=1.6 \cdot 10^{-19} C$, and Vf and Vi are the final and initial voltages.

Substituting the numbers, we find:
$\Delta U = (1.6 \cdot 10^{-19}C)(0 V-(-0.072 V))=1.15 \cdot 10^{-20} J$

Answer 2

The change in electric potential energy of the ion is $\fbox{\begin\\1.152\times{10^{-20}}\,{\text{J}}\end{minispace}}$ or $\fbox{0.072\text{ eV}}$.

Further Explanation:

Cell is composed of many ions. It has high concentration of Potassium ions inside and a low concentration outside the membrane. It has high concentrations of Sodium and Chloride ions in the extracellular region, and low concentrations in the intracellular regions.

When an ion moves from one point to other in a cell, the potential of ion changes because two different points in a cell are at different potential. The change in potential of ion will change the potential energy of ion.

Given:

The initial potential of ion is $- 72\,{\text{mV}}$.

The final potential of ion is $0\,{\text{V}}$.

Concept:

The initial potential of ion in volts is $-72\times{10^{-3}}\,{\text{V}}$.

The equation of change in potential energy given as:

$\fbox{\begin\\\Delta U = q({V_f} - {V_i})\end{minispace}}$

Here, $\Delta U$ is the change in potential energy, $q$ is the charge of positive ion, ${V_i}$ is the initial potential and ${V_f}$ is the final potential of ion.

Substitute $1.6\times{10^{-19}}\,{\text{C}}$ for $q$ , $-72\times{10^{-3}}\,{\text{V}}$ for ${V_i}$ and $0\,{\text{V}}$ for ${V_f}$ in above equation.

$\begin{aligned}\\\Delta U&=\left({1.6\times{{10}^{-19}}\,{\text{C}}}\right)\left[{0\,{\text{V}}-\left({-72\times{{10}^{-3}}\,{\text{V}}}\right)}\right]\\&=\left({1.6\times{{10}^{-19}}\,{\text{C}}}\right)\left({72\times{{10}^{-3}}\,{\text{V}}}\right)\\&=1.152\times{10^{-20}}\,{\text{J}}\\\end{aligned}$

The above energy can be expressed in electron volts as:

$\begin{aligned}\Delta U&=\dfrac{1.152\times10^{-20}\text{ J}}{1.6\times10^{-19}\text{ C}}\\&=0.072\text{ eV}\end{aligned}$

Thus, the change in electric potential energy of the ion is $\fbox{\begin\\1.152\times{10^{-20}}\,{\text{J}}\end{minispace}}$ or $\fbox{0.072\text{ eV}}$.

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Answer Details:

Grade: College

Subject: Physics

Chapter: Electrostatics

Keywords:

Na+ ion, electric potential, cell, potential energy, -72mV, -72*10^-3V, 0V, charge and change in electric, electron volt, Joule, 0.072 eV, ions inside a cell.