Answer:
[tex]1.0\cdot 10^{-20} J[/tex]
Explanation:
The change in the ion's electric potential charge as it moves from inside to outside the cell is:
[tex]\Delta U= q \Delta V[/tex]
where
q is the charge of the ion
[tex]\Delta V[/tex] is the potential difference
- The charge of the ion is +e, since it is positively charged, so [tex]q=1.6\cdot 10^{-19} C[/tex]
- The potential difference is
[tex]\Delta V=V_f - V_i = 0 V-(-63 mV)=+63 mV=+0.063 V[/tex]
So, the change in the ion's electric potential energy is
[tex]\Delta U=(1.6\cdot 10^{-19}C)(0.063 V)=1.0\cdot 10^{-20} J[/tex]
