Question

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 1.35 × 10 − 20 J 1.35×10−20 J of work is required to eject a positive sodium ion (Na + ) (Na+) from the interior of the cell, what is the magnitude of the potential difference between the inner and outer surfaces of the cell?

Answer 1

Answer:

Explanation:

Given

Work required $W=1.35\times 10^{-20}\ J$

Work done to eject the sodium ion from interior of the cell is given by the product of charge and Potential difference between inner and outer surface of the cell.

$W=q\times V$

Charge on sodium ion $q=1.6\times 10^{19}\ C$

$V=\frac{W}{q}$

$V=\frac{1.35\times 10^{-20}}{1.6\times 10^{19}}$

$V=0.0718\ V$            

Answer 2

The magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0844 volt.

Potential difference:

When potential difference is V and charge of ion is q. Then, work required to eject a postive charge from the interior of cell is computed as,

                   Work  = q * V

It is given that, work W = $1.35*10^{-20}J$

Charge on sodium ion , $q=1.6*10^{-19}C$

Substituting above values in above relation.

                   $V=\frac{W}{q}\\ \\V=\frac{1.35*10^{-20} }{1.6*10^{-19} }\\ \\V=0.0844Volt$

Hence, the magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0844 volt.

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