Question

When a singly ionized potassium ion moves through a channel in the membrane passing from the outside to the inside, what is the magnitude of the work done by the electric field of the membrane?.

Answer 1

The magnitude of the work done by the electric field of the membrane is W = 1.28 × 10⁻²⁰ Joules.

We start with the necessity to take into account a value for the voltage present there in order to solve this problem by first considering that the membranes have two layers, one internal and one external, each responsible for producing a potential difference between the two levels.

As a result, in order to find a solution, it is necessary to take into account the potential difference between the two surfaces. In this instance, we'll assume a particular value for the load, but the recipient is free to substitute a different value if they prefer.

The product of the potential difference and the charge is used to define the work that an electric field performs. The charge of the potassium ion will be equal to that of its electron, so,

q = 1.6 × 10⁻¹⁹ Coulombs

Then the Work would be:

W = Vq

Here,

v = Potential difference

q = Charge

The 80mV potential difference we will have is quantified as follows:

W = (80mV (1V/1000mV))( 1.6 × 10⁻¹⁹ C)

W = 1.28 × 10⁻²⁰ Joules is the amount of work that the membrane's electric field has produced.

Find more on work done at : brainly.com/question/25573309

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