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

`1.44\cdot 10^(-20) J`

Step-by-step 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`