A potential difference exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 1.15 x 10-20 J of work is required to eject a positive sodium ion (Na ) from the interior of the cell, what is the magnitude of the potential difference between the inner and outer surfaces of the cell?

0.718* 10^-1 V volt is the potential difference required.

Explanation:

From the formula, W (input) =q( charge on the ion) *Potential difference

Work input in the form of energy is given = 1.15X10^-20 J

since Na^+ is positively charged with +1 so charge on it will be

1.6X10^-19 C

so,

Potential difference= work input/charge on the sodium ion

                                  =   1.15*10^-20 / 1.6*10^-19

                                    =  0.718* 10^-1 V

  Potential difference required between the inner and outer surfaces of the cell is 0.718 *10^-1

50% lit on the left side

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