Question

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 $0.30nm$diameter potassium ion $\left({\mathrm{K}}^{+}\right)$channel carries a current of role="math" localid="1649134390247" $1.8\mathrm{pA}$.

a. How many potassium ions pass through if the ion channel opens for $1.0\mathrm{ms}$?

b. What is the current density in the ion channel?

Short answer

a. There are $11250$potassium ions will pass through if the ion channel opens for $1.0ms$

b. The current density in the ion channel is$2.55\times {10}^7\mathrm{A}/{\mathrm{m}}^2.$

Step-by-step solution

Given Information (Part a) 

Potassium ion diameter$=0.30nm$

Current$=1.8\mathrm{pA}$

Explanation (Part a)

The current is the charge that goes through a circuit per unit of time. Consequently, when time $t$comes, the charge will be:

role="math" localid="1649137054592" $q=It$

Here, charge is a multiple of the elementary charge,

$q=ne$

Modify the expression,

role="math" localid="1649134772412" $ne=It$

The equation for find the number of charges,

$n=\left(\frac{It}{e}\right)$

Substitute the values,

$n=\frac{{\displaystyle 1.8·{10}^{-12}·{10}^{-3}}}{{\displaystyle 1.6·{10}^{-19}}}$

$n=11250.$

Final Answer (Part a)

Hence, the number potassium ions will pass through if the ion channel opens for$1.0ms$is$11250.$

Given Information (Part b) 

Potassium ion diameter

Current$1.8pA$

Explanation (Part b) 

Density is calculated by dividing the current flow by the cross-sectional area at which it flows, which in our situation is a circle.

So, we have

$J=\frac{I}{A}=\frac{4I}{\pi D^2}$

Substitute the values,

role="math" localid="1649137777099" $J=\left(\frac{4\times 1.8\times {10}^{-12}A}{\pi \times {\left(3\times {10}^{-10}m\right)}^2}\right)$

$=2.55\times {10}^7\mathrm{A}/{\mathrm{m}}^2$

Final Answer (Part b) 

Hence, the current density in the ion channel is$2.55\times {10}^7\mathrm{A}/{\mathrm{m}}^2$.