A potential difference of 102 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative relative to the outer surface. How much work is required to eject a positive sodium ion (Na+) from the interior of the cell? (in J)
W = q•Δφ = 11•e•102•10^-3 =
=11•1.6•10^-19•102•10^-3=1.8•10^-19 J
To find the work required to eject a positive sodium ion (Na+) from the interior of the cell, you can use the formula:
Work = q * V
where q is the charge of the ion and V is the potential difference across the cell membrane.
In this case, the charge of a sodium ion is +1 (since it has a positive charge) and the potential difference is 102 mV.
First, convert the potential difference to volts:
102 mV = 0.102 V
Now, you can calculate the work:
Work = (1) * (0.102)
Work = 0.102 J
Therefore, the work required to eject a positive sodium ion (Na+) from the interior of the cell is 0.102 Joules.