Question

Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.09 How much work is done by the clectric force when a sodium ion charge d e s brore the outside to the inside Number Units the tolerance is */-2%

Answer 1

Answer:

$1.44\cdot 10^{-20} J$

Explanation:

The difference in electric potential between the inside and the outside of the cell is

$\Delta V = 0.09 V$

The work done by the electric force to bring the ion from outside to inside is

$W=q\Delta V$

where

q is the charge of the ion

A sodium ion, $Na^+$, has given off an electron: so its electric charge is

$q=+e=+1.6\cdot 10^{-19} C$

Therefore, the work done is:

$Q=(1.6\cdot 10^{-19})(0.09)=1.44\cdot 10^{-20} J$