Question

A new microorganism is isolated from a lake and is placed into a solution of KCl. The voltage difference across its membrane is measured at 120 mV. How much energy is required to move a proton from the negative side of the membrane to the positive side?

Answer 1

Answer: $\Delta U=$ 1.922x10⁻²⁰ Joules

Explanation: Electric Potential Energy (U) is the energy a charged object has due to its location in an electric field and it will only exist with the object is charged.

Voltage or Electric Potential Difference (V) is external work done to move a charge from one point to another in a electric field.

These terms have a relationship, which is given by:

$\Delta U=q\Delta V$

where

q is the charge

Proton is positive and has a charge of 1.6x10⁻¹⁹C.

Unit for potential energy is Joule (J). The relation between mV and J is

1mV = 10⁻³J

Then:

V = 120x10⁻³

V = 0.12

So, for a proton to move from the negative side of a membrane to the positive:

$\Delta U=q\Delta V$

$\Delta U=1.6.10^{-19}.0.12$

$\Delta U$ = 1.922 x 10⁻²⁰

Energy necessary to transport a proton from negative side of the membrane to the positive is 1.922 x 10⁻²⁰J.