Question

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 1

The value is
`V =168.75\ millivolt`

Step-by-step explanation:

From the question we are told that

The workdone is
`W= 2.70 * 10^(-20 ) \ J`

Generally charge on the positive sodium ion is equivalent to the charge on a proton, the value is
`e = 1.60 *10^(-19) \ C`

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

`V = (W)/(e)`

=>
`V  =  (2.70 * 10^(-20 ) )/(1.60 *10^(-19) )`

=>
`V  = 0.16875 \  V`

converting to millivolt

`V  = 0.16875 * 1000`

=>
`V  =168.75\ millivolt`