Question

A na+ ion moves from inside a cell, where the electric potential is -63 mv, to outside the cell, where the potential is 0 v. what is the change in the ion's electric potential energy as it moves from inside to outside the cell?

Answer 1

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

Step-by-step explanation:

The change in the ion's electric potential charge as it moves from inside to outside the cell is:

`\Delta U= q \Delta V`

where

q is the charge of the ion

`\Delta V` is the potential difference

- The charge of the ion is +e, since it is positively charged, so
`q=1.6\cdot 10^(-19) C`

- The potential difference is

`\Delta V=V_f - V_i = 0 V-(-63 mV)=+63 mV=+0.063 V`

So, the change in the ion's electric potential energy is

`\Delta U=(1.6\cdot 10^(-19)C)(0.063 V)=1.0\cdot 10^(-20) J`