I'm having some trouble with a problem on my physics pre-exam, if there is any formula please let me know so i can do a bit of research on it. any help is greatly appreciated in advance.

The inner membrane of a cell membrane is negative relative to the outer surface. if 1.6 X 10^-20 J of work is required to eject a positive sodium ion (Na+) from the interior of the cell, what is the potential difference between the inner and outer surface? Hint: Na+ has one electron deficiency.

by the way i already know the answer, i dont need it. (100 mv) i need to know how to do it or why it is 100 mv

woltage*q=work

q is the charge on an electron: 4.35516759 × 10-19 coulombs

To solve this problem, you need to understand the concept of electrical potential difference and how it relates to the work done. The electrical potential difference between two points is a measure of the work needed to move a unit positive charge from one point to the other. In this case, the potential difference between the inner and outer surface of the cell membrane can be calculated using the given work done to eject a sodium ion.

First, let's assume that the charge on the sodium ion (Na+) is q. Since it is mentioned that Na+ has one electron deficiency, the charge on Na+ would be +1e, where e is the elementary charge (1.6 x 10^-19 C).

Now, we know that the work done (W) is given by:
W = q * ΔV,
where q is the charge and ΔV is the potential difference.

We can rearrange this equation to solve for ΔV:
ΔV = W / q.

Substituting the given values:
W = 1.6 x 10^-20 J, q = +1e = +1 * 1.6 x 10^-19 C,

ΔV = (1.6 x 10^-20 J) / (1 * 1.6 x 10^-19 C).

Now, cancel out the common factors:
ΔV = (1.6 x 10^-20 J) / (1.6 x 10^-19 C) = 0.1 V.

So, the potential difference between the inner and outer surface of the cell membrane is 0.1 Volts.

Remember to double-check the units and perform any necessary unit conversions if required.