A sodium ion (mass 3.84 times 10^-26 kg, charge +1.60 times 10^-19 C) sits on a
ID: 1836199 • Letter: A
Question
A sodium ion (mass 3.84 times 10^-26 kg, charge +1.60 times 10^-19 C) sits on a membrane between a calcium ion (mass 6.68 times 10^-26 kg, charge +3.20 times 10^-19 C) and a chlorine ion (mass 5.85 times 10^-26 kg, charge-1.60 times 10^-19 C) configured as shown. (a) Would I have to do positive work to move the sodium ion across the membrane, or would the other two ions do positive work on it as it crosses the membrane? (b) How much positive work is done on the sodium ion if it crosses the membrane perpendicularly? (c) Is there any other path that the sodium ion could take to get from its initial location to the point directly across the membrane with less expenditure of energy? If so, what is it? If not, explain why not.Explanation / Answer
a) The electric field is from left to right and Na ion carrying a +ve charge and hence the force on it is to the right. Hence work is one by the electric field and work is done bythe field of the two charges.
b) work done is equal to the difference in PE of the system of charges before and after moving the charge.
Initial PE of the system
Ui = k(3.2e-19 *1.6e-19 /4.5e-9 - 1.6e-19*1.6e-19/1.5e-9 - 3.2e-19 *1.6e-19 /6.0e-9)
when the Na ion has moved perpnedicualr to the memebrane.
the distance between Ca and Na inons = sqrt(4.5^2 +2.0^2) = 5.2 nm
distance between Na and Cl inons = 2.5 nm
Uf = k(3.2e-19 *1.6e-19 /5.2e-9 - 1.6e-19*1.6e-19/2.5e-9 - 3.2e-19 *1.6e-19 /6.0e-9)
work done = Uf -Ui = 5.404 J
c) It is a conservative force an hecne work done is independent of the path. It is the difference in the PE at the tow points hence net work done will remain same in any path
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