How Many Potassium Ions Pass Through a 0.30-nm Channel in 1.0 ms?

[kyang002]

Measurements with microelectrodes have shown that a 0.30-nm-diameter potassium ion (K+) channel carries a current of 1.8 pA.

How many potassium ions pass through if the ion channel opens for 1.0 ms?

What is the current density in the ion channel?

I am completely lost for this one. Anyone know of any equations that I can use?

Current is, its I = dQ/dt, the flow of charge through a point, through time. Like measuring the flow of water through a pipe.
The area the flow is going through (the inner area of the "pipe") because they give you the diameter. area = pi*r^2, 2*r = diameter.

I = dQ/dt = n*A*q*dx/dt = n*A*q*Velocity
so first solve I = n*A*q*Velocity

then use that with velocity = dx/dt to get dx.
Then use dQ = (n*A*dx)*q
to get dQ.

But I am stuck on how to solve for the variables and how to complete the problem. Can anyone help me out with the answer?

 

[chroot]

Needless to say, you're making this more difficult than you need to. What's the definition of an ampere? One ampere is one coulomb of charge moving past a fixed point each second. Each potassium ion has a fixed charge, +e. One coloumb of charge contains 6.25 * 10^18 such ions.

1.8 pA = \frac{1.8 \cdot 10^{-12} C}{s} \cdot \frac{6.25 \cdot 10^{18} \textrm{ ions}}{C} \cdot 1 \cdot 10^{-3} s

Perform the multiplication and get your answer in number of ions.

- Warren

 

[wais]

To solve for the number of potassium ions passing through in 1.0 ms, we can use the equation dQ = n*A*q*dx, where dQ is the charge passing through, n is the number of ions, A is the area of the ion channel, q is the charge of each ion, and dx is the distance the ions travel in 1.0 ms. We can rearrange this equation to solve for n: n = dQ/(A*q*dx).

Plugging in the given values, we get n = (1.8 pC)/(π*(0.30/2)^2*1.6*10^-19 C*0.30 nm*10^-9 m/nm) = 3.0*10^5 ions.

To calculate the current density in the ion channel, we can use the formula J = I/A, where J is the current density, I is the current, and A is the cross-sectional area of the channel. Plugging in the given values, we get J = (1.8 pA)/(π*(0.30/2)^2) = 1.2*10^-6 A/m^2.