Electric Fields: Sodium Ions pumped

AI Thread Summary
Living cells actively pump sodium ions (Na+) from inside to outside the membrane, requiring significant energy, estimated at 20% of resting energy consumption. The inside of the membrane is at a lower potential compared to the outside, which is confirmed as the answer to part a of the homework. To calculate the work done to move one sodium ion, the equation W=qΔV can be used, where q is the charge of the ion and ΔV is the potential difference. If the membrane thickness is doubled, the work required to pump out one sodium ion would also increase, as the electric field strength is inversely related to thickness. The discussion emphasizes the importance of understanding the relationship between voltage, charge, and work in the context of cellular processes.
Madison029
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Homework Statement


Living cells actively “pump” positive sodium ions (Na+) from inside the cell to outside
the cell. It’s called pumping because it requires work to move the ions from the
negatively charged inner surface of the cell membrane to the positively charged outer
surface. It is estimated that as much as 20% of the energy we consume in a resting state
is used in this sodium pumping. The potential difference across the membrane is 0.070V,
and the membrane is 0.10μm thick.
a.) Which is at higher potential, the inside or outside of the membrane?
b.) How much work has to be done to move one sodium ion from inside the
membrane to outside?
c.) If the thickness of the membrane were doubled, how much work would it require
to pump out one sodium ion?
d.) Assuming the E field is constant in the membrane, what is the magnitude and
direction of the E field between the inside and outside of the membrane?



Homework Equations


None Given


The Attempt at a Solution


I have attempted this but I am having trouble starting out correctly. Like which is the proper equation to use...my general understanding is not jump starting this problem.
 
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First of all they give you the answer to a) don't they?
So what is that?

Second of all if you know the voltage change and you know the charge don't you know the work required to affect the ion transfer?
 
Yea I now know A, which is outside. And the equation I am using is the work function W=qEd...assuming q is the charge of the ion,1. then E=V/d...plugging all that in gets me back to the original voltage of .070V.
 
More to the point, don't you think, work is q*ΔV, since work is over the distance?
 
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