Ionic Potentials across Cell Membranes

[mtbgymnast]

Homework Statement

In its resting state, the membrane surrounding a neuron is permeable to potassium ions but not permeable to sodium ions. Thus, positive K ions can flow through the membrane in an attempt to equalize K concentration, but Na ions cannot. This leads to an excess of Na ions outside of the cell. If the space outside the cell is defined as zero electric potential, then the electric potential of the interior of the cell is negative. This resting potential is typically about -80 \rm mV. A schematic of this situation is shown in the figure.

Image 1 http://session.masteringphysics.com/problemAsset/1011446/22/1011446A.jpg

Image 2 http://session.masteringphysics.com/problemAsset/1011446/22/1011446B.jpg

There are 4 parts to this question.

A) Question: During the resting phase, what is the electric potential energy of a typical Na ion outside of the cell?
Answer: 0 meV

B) Question: During the resting phase, what is the electrical potential energy of a typical K ion inside of the cell?
Answer: -80meV

C) :uhh: During depolarization, what is the work done (by the electric field) on the first few Na ions that enter the cell?
The answer is one of the following:
-40 {\rm meV}
+40 {\rm meV}
-80 {\rm meV}
+80 {\rm meV}
-120 {\rm meV}
+120 {\rm meV}
0 {\rm meV}

D) :uhh: During repolarization, what is the work done (by the electric field) on the first few K ions that exit the cell?
The answer is one of the following:
-40 {\rm meV}
+40 {\rm meV}
-80 {\rm meV}
+80 {\rm meV}
-120 {\rm meV}
+120 {\rm meV}
0 {\rm meV}

Homework Equations

There are 4 parts to this question. I already answered parts A and B, but can't figure out Parts C and D
How do I find the magnitude?
How do I figure out the sign?

The Attempt at a Solution

The work done on a force is positive if the force and displacement are parallel, but negative if they are in opposite directions. For this problem, the magnitude of the work done is equal to the change in the ions electric potential energy.
I thought for part C the answer was either +40 or -40, but I tried those answers and they were wrong. I thought the magnitude was 40 because of a change from 0 to 40 meV so I'm confused on what the change is.
For part C and D, I do not know the magnitude or the sign of the answer.
Where is the force? I am having trouble figuring out where the force is in relation to the displacement, so I don't know if the answers are positive or negative.
How do I find the magnitude?
How do I figure out the sign?

 

[gtmh3]

The work done on an object is equal to the magnitude of the object’s change in energy. The energy is the potential energy. For depolarization, the difference in potential energy can only be the initial values because nothing else has occurred yet--->-80meV inside and 0 outside => delta U= 80meV=magnitude of work. The sign is dependent on whether the object is moving in the direction of the force. It is positive, since in an effort to reach equilibrium the Na ions are rushing in, so they are going with the force.

 

[baba89]

Dear student,

Thank you for your question. I understand that you are having trouble with parts C and D of the homework question. Let me try to explain them in more detail to help you understand the solution better.

Part C is asking for the work done on the first few Na ions that enter the cell during depolarization. To find the work done, we need to know the change in the ions' electric potential energy (U). This change is equal to the final electric potential energy (Uf) minus the initial electric potential energy (Ui). In this case, the final electric potential energy is -80 meV (from part B) and the initial electric potential energy is 0 meV (from part A). Therefore, the change in electric potential energy is -80 meV - 0 meV = -80 meV.

Now, we need to find the work done, which is equal to the change in electric potential energy multiplied by the number of ions (N) that enter the cell. The number of ions is not given in the question, but we can assume that it is a small number. Therefore, the work done is approximately equal to -80 meV x N. We can see from this equation that the sign of the work done will be negative because the change in electric potential energy is negative and the number of ions is positive.

To find the magnitude of the work done, we need to know the value of N. Unfortunately, we do not have this information, so we cannot calculate the exact magnitude of the work done. However, we can make some assumptions to get an approximate answer. If we assume that the number of ions is 1, the magnitude of the work done will be -80 meV x 1 = -80 meV. If we assume that the number of ions is 10, the magnitude of the work done will be -80 meV x 10 = -800 meV. As you can see, the magnitude of the work done will depend on the number of ions that enter the cell.

Similarly, for part D, we need to find the work done on the first few K ions that exit the cell during repolarization. The process of finding the work done is the same as in part C. The change in electric potential energy is now 0 meV - (-80 meV) = 80 meV (note the change in sign). The number of ions is again not given in the