Answer:
Current, [tex]I=6.78\times 10^{-11}\ A[/tex]
Explanation:
In this case, we need to find the current in amperes if 1400 Na+ ions flow across a cell membrane in 3.3 μs.
Charge, [tex]q=1400\times 1.6\times 10^{-19}=2.24\times 10^{-16}\ C[/tex]
Time taken, [tex]t=3.3\ \mu s=3.3\times 10^{-6}\ s[/tex]
Let I is the current. It is given by total charge per unit time. It is given by :
[tex]I=\dfrac{q}{t}[/tex]
[tex]I=\dfrac{2.24\times 10^{-16}}{3.3\times 10^{-6}}[/tex]
[tex]I=6.78\times 10^{-11}\ A[/tex]
So, the current of [tex]6.78\times 10^{-11}\ A[/tex] is flowing across a cell membrane. Hence, this is the required solution.
