An unknown potential difference exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 2.5 x 10-20 J of work is needed to eject a positive Calcium ion (Ca2+) from the interior of the cell, what is the value of the potential difference?
V*q=work
q= 2*1.6E-19 coulombs
doesnt vq = - work ?...signs are important
since work = -u, and u = vq
To find the value of the potential difference, we can use the equation:
ΔV = W/q
where ΔV is the potential difference, W is the work done, and q is the charge. In this case, the charge is the charge of a single positive Calcium ion (Ca2+).
First, let's convert the work done to its SI unit (Joules):
W = 2.5 x 10^(-20) J
Next, we need to determine the charge of a single Ca2+ ion. Calcium ions have a charge of +2 (since they have two extra positive charges compared to a neutral atom). The charge q is therefore:
q = +2
Now, we can substitute these values into the equation:
ΔV = W/q
ΔV = (2.5 x 10^(-20) J) / (+2)
Calculating this, we get:
ΔV = 1.25 x 10^(-20) J / (+2)
ΔV = 6.25 x 10^(-21) J
Therefore, the potential difference between the inner and outer surfaces of the cell membrane is 6.25 x 10^(-21) J.