How many potassium ions pass through if the ion channel opens for 1.0 {\rm ms}?

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SUMMARY

The discussion centers on calculating the number of potassium ions that pass through a potassium ion channel when it opens for 1.0 ms, given a current of 1.8 pA. Using the formula Delta charge (Q) = Current (I) * Delta time, the total charge was calculated as 1.8E-15 C. Dividing this by the charge of a single ion (1.6E-19 C) results in approximately 11,250 ions. The final answer was adjusted to "1.1e4" to meet the significant figures requirement of the platform used, Mastering Physics.

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Homework Statement


The biochemistry that takes place inside cells depends on various elements, such as sodium, potassium, and calcium, that are dissolved in water as ions. These ions enter cells through narrow pores in the cell membrane known as ion channels. Each ion channel, which is formed from a specialized protein molecule, is selective for one type of ion. Measurements with microelectrodes have shown that a 0.30-nm-diameter potassium ion ({\rm{{\rm K}}}^{\rm{ + }} ) channel carries a current of 1.8 {\rm pA}.

How many potassium ions pass through if the ion channel opens for 1.0 {\rm ms}?

Homework Equations



Delta charge (Q)=Current (I) * delta time
1 ion = 1.6E-19 C



The Attempt at a Solution


Delta Q=1.8E-12*1E-3=1.8E-15C

1.8E-15C/1.6E-19=11,250

Mastering physics wants 2 sig figs, so entered "11,000" which is wrong.
 
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deborahcurrie said:
1.8E-15C/1.6E-19=11,250

Mastering physics wants 2 sig figs, so entered "11,000" which is wrong.

It looks right to me. Is it possible to enter "1.1e4"?
 
That did it. Thanks for thinking of the tip!
 

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