Chapter 18, Problem 3P

To determine

The current in amperes, if ${\text{1200 Na}}^{\text{+}}$ ions flow across a cell membrane in $\text{3}\text{.5}\text{μs}$ .

Answer to Problem 3P

The current in ampere: $\text{5}{\text{.5×10}}^{\text{-11}}\text{A}$

Explanation of Solution

Given info:

Number of ion, ${\text{n=1200 Na}}^{\text{+}}$ , time $\text{t=3}\text{.5}\text{μs}$ and charge $\text{e=}\text{1}{\text{.6×10}}^{\text{-19}}$

Formula used:

The current is given by

  $\text{I}\text{=}\frac{\Delta \text{Q}}{\Delta \text{t}}$

Where, $I=$ current, $Q=$ amount of charge, and $\Delta t=$ change in time.

Calculation:

Substituting the given values in the formula, $\text{I}\text{=}\frac{\Delta \text{Q}}{\Delta \text{t}}$

  $\text{=}\frac{\text{1200×1}{\text{.6×10}}^{\text{-19}}}{\text{3}{\text{.5×10}}^{\text{-6}}}$

  $\text{=}\text{5}{\text{.5×10}}^{\text{-11}}\text{A}$

Conclusion:

The current flow across the cell membrane: $\text{5}{\text{.5×10}}^{\text{-11}}\text{A}$