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otential difference ΔVΔV exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 2.70×10−20 J2.70×10−20 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 (in millivolts) between the inner and outer surfaces of the cell?

Answer :

okpalawalter8

Answer:

The value is [tex]V =168.75\ millivolt [/tex]

Explanation:

From the question we are told that

The workdone is [tex]W= 2.70 * 10^{-20 } \ J[/tex]

Generally charge on the positive sodium ion is equivalent to the charge on a proton, the value is [tex]e = 1.60 *10^{-19} \ C[/tex]

Generally the potential difference between the inner and outer surfaces of the cell is mathematically represented as

[tex]V = \frac{W}{e}[/tex]

=>      [tex]V  =  \frac{2.70 * 10^{-20 } }{1.60 *10^{-19} }[/tex]

=>     [tex]V  = 0.16875 \  V[/tex]

converting to millivolt

         [tex]V  = 0.16875 * 1000 [/tex]

    =>   [tex]V  =168.75\ millivolt [/tex]

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