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The electric potential inside a living cell is lower than the potential outside. Suppose the electric potential difference between the inner and the outer cell wall is 0.095 V, a typical value. To maintain the internal electrical balance, the cell pumps out sodium ions. How much work must be done to remove a single sodium ion (charge e)

Sagot :

Answer:

1.52 × 10⁻²⁰ J

Explanation:

The electrical potential difference is defined as the amount of work done in carrying a unit charge from one point to another point in an electric field. Electric potential difference is measured in volts. It is given by the formula:

ΔV = ΔU / q

ΔV is electric potential difference between the two points, ΔU is the work done and q is the unit charge.

Given that ΔV = 0.095 V, q = 1.6 × 10⁻¹⁹ C. Hence:

ΔU = ΔV.q

ΔU = 0.095 V * 1.6 × 10⁻¹⁹ C

ΔU = 1.52 × 10⁻²⁰ J

The electric potential difference is the amount of effort done in an electrical field to shift a unit charge from one spot to another.The electric potential difference will be 1.52×10⁻²⁰J.

What is the electric potential difference ?

The electrical potential difference is the amount of effort done in an electrical field to shift a unit charge from one spot to another.

Traditional current flows from positive to negative terminals, signifying positive charge transfer in that direction.

The given data in the problem is

[tex]\triangle V[/tex] is the electric potential difference between the inner and the outer cell wall = 0.095 V

[tex]\triangle U[/tex] is the amount of work done

q is the charge on the electron =1.6×10⁻¹⁹C

Electric potential difference is given by the formula

[tex]\triangle V=\frac{\triangle U}{q} \\\\ \triangle U=\triangle Vq\\\\\triangle U=0.095 V\times1.6\times10^{-19}\\\\ \triangle U=1.52\times10^{-20}J[/tex]

Hence the electric potential difference will be 1.52×10⁻²⁰J.

To learn more about the electric potential difference refer to the link;

https://brainly.com/question/9383604

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