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A potential difference Δ V ΔV exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 2.20 × 10 − 20 J 2.20×10−20 J of work is required to eject a positive potassium ion (K + ) (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?

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Answer:

Explanation:

Given

Work required to eject a positive Potassium ion [tex]W=2.20\times 10^{-20}\ J[/tex]

Charge on the Potassium ion [tex]q=1.6\times 10^{-19}\ C[/tex]

During ejection Potential difference between the inner and outer membrane exert a force opposite to the direction of motion so some amount of work is required to over come this force

Work done=Potential difference\times charge

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

[tex]\Delta V=\frac{W}{q}[/tex]

[tex]\Delta V=\frac{2.20\times 10^{-20}}{1.6\times 10^{-19}} [/tex]

[tex]\Delta V=1.375\times \times 10^{-1}[/tex]

[tex]\Delta V=0.1375\ V[/tex]

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