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A potential difference Δ V exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 1.00 × 10 − 20 J of work is required to eject a positive potassium ion (K + ) from the interior of the cell, what is the magnitude of the potential difference (in millivolts) between the inner and outer surfaces of the cell?

Respuesta :

Answer:

The magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0625 V

Explanation:

Given that,

Work done [tex]W= 1.00\times10^{-20}\ J[/tex]

We need to calculate the magnitude of the potential difference between the inner and outer surfaces of the cell

Using formula of wok done

[tex]W=V\times q[/tex]

[tex]V=\dfrac{W}{q}[/tex]

Where,

V = volt

W = work done

q = charge

Put the value into the formula

[tex]V=\dfrac{1.00\times10^{-20}}{1.6\times10^{-19}}[/tex]

[tex]V=0.0625\ V[/tex]

Hence, The magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0625 V

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