Calculating Chlorine Ion Acceleration & Speed of Light:Why?

[casualdehyde]

Question

Assuming the electric potential difference between the inside and outside of a cell is 70 mV and the thickness of the region across which this exists is 7 nm, calculate the acceleration a chlorine ion would experience in the absence of other forces. In the absence of other forces how long would it be before it was traveling at the speed of light? Why does this not happen? Chlorine has a molecular weight of around 35 g/mole, Avogadro’s number is 6.022 × 10^23 per mole, and the speed of light, c, is 3×108 m/s.

Equations
Conservation of energy
KE=1/2mv^2
E=qV=Work
1/2mv^2=qV -

Attempt
Am I finding the kinetic energy generated and then the mass through that? I'm a bit stuck. I assume that the speed of light is only given to compare the velocity of the chlorine once I get it.
Is the charge of the chlorine just -1e? = -1.6x10^-19C As it's a negative charge does that mean the ion is slowing down as it moves through the membrane (I assume this is wrong as the question talks about it nearing the speed of light). I also don't know how to bring the cell thickness into it - I can use it to calculate the electrical field, is that what it's for 7.0x10^-3V / 7x10^-9m = 1x10^6V/m?
The mass of the ion is 35g/mole /6.022x10^23 = 5.8x1-^23g.
v^2 = (-1.6x10^-19C x 0.7V)/(0.5 x 5.8x1-^23g) - Definitely wrong as that would give me a negative number and I need to find the square root.

It's probably a really easy question but I'm stumped. I'd appreciate some help just to get me going :) and to better understand what is going on.

 

[gneill]

Question

Assuming the electric potential difference between the inside and outside of a cell is 70 mV and the thickness of the region across which this exists is 7 nm, calculate the acceleration a chlorine ion would experience in the absence of other forces. In the absence of other forces how long would it be before it was traveling at the speed of light? Why does this not happen? Chlorine has a molecular weight of around 35 g/mole, Avogadro’s number is 6.022 × 10^23 per mole, and the speed of light, c, is 3×108 m/s.

Equations
Conservation of energy
KE=1/2mv^2
E=qV=Work
1/2mv^2=qV -

Attempt
Am I finding the kinetic energy generated and then the mass through that? I'm a bit stuck. I assume that the speed of light is only given to compare the velocity of the chlorine once I get it.

You're given the molar mass of chlorine and Avogadro's Number, so you can determine the mass of a single chlorine atom by that route. I think the speed of light is given so you might determine how long it would take, given the acceleration of the ion, for it to reach that speed. The question doesn't say anything about requiring Relativity theory, so I suppose you're to assume Classical physics for the motion of the ion.

Is the charge of the chlorine just -1e? = -1.6x10^-19C As it's a negative charge does that mean the ion is slowing down as it moves through the membrane (I assume this is wrong as the question talks about it nearing the speed of light).

I think you can assume that the chlorine ion has a single -1e charge. The problem doesn't state the direction of travel for the ion, so no need to fret about that. No inital velocity was given either, so you can assume it starts from rest and accelerates (in whatever direction... you really only care about the magnitudes here).

I also don't know how to bring the cell thickness into it - I can use it to calculate the electrical field, is that what it's for 7.0x10^-3V / 7x10^-9m = 1x10^6V/m?

Sure. Having E, the electric field, will let you determine the force on the ion, hence the acceleration. Check your value for the potential in the above calculation. You've slipped an order of magnitude in converting mV to V.

The mass of the ion is 35g/mole /6.022x10^23 = 5.8x1-^23g.
v^2 = (-1.6x10^-19C x 0.7V)/(0.5 x 5.8x1-^23g) - Definitely wrong as that would give me a negative number and I need to find the square root.

You can work with magnitudes here for simplicity. Or if you want to be picky, consider the direction that the atom will be accelerated through the field, and hence what sign should be assigned to the potential change.

 

[casualdehyde]

Okay so can I just get away with using F=ma and F=q0E?
Seeing as we're not worrying about direction, just acceleration q=1.6x10^-19 C. Find E: 0.07 volts/ 7.0 × 10-9 metres = 10000000V/m.
F= 1.6x10^-19 C x 10000000V/m = 1.6x10^-12N

Now I can put this into F/m = a
1.6x10^-12N/5.8^-26kg = 2.8x10^13 m/s^2 Is this right?

If the speed of light is 3x10^8 m/s I can find the time to reach it by: 3x10^8 m/s / 2.8x10^13 m/s^2 = 0.00001089754 seconds (if no external forces were acting on it)?
As for the reason why this does not happen I'm sure its due to the external force but I'm not sure which one the answer is looking for specifically as I've taken into consideration the electric field (this encompasses charge and the thickness of the cell) and the ion mass I can't use them in my answer can I? Would stuff like drag (air resistance) come into play or am I looking for an answer that utilities something about the potential 100mV? Hopefully I got all the working out right this time. Thanks for the help by the way.

 

[gneill]

Okay so can I just get away with using F=ma and F=q0E?

I believe so. Although I don't know what course this is for, to me it just has the feel of a Newtonian realm question. Correct me if I'm wrong.

Seeing as we're not worrying about direction, just acceleration q=1.6x10^-19 C. Find E: 0.07 volts/ 7.0 × 10-9 metres = 10000000V/m.
F= 1.6x10^-19 C x 10000000V/m = 1.6x10^-12N

Now I can put this into F/m = a
1.6x10^-12N/5.8^-26kg = 2.8x10^13 m/s^2 Is this right?

Looks good.

If the speed of light is 3x10^8 m/s I can find the time to reach it by: 3x10^8 m/s / 2.8x10^13 m/s^2 = 0.00001089754 seconds (if no external forces were acting on it)?

Sure. You might want to use scientific notation and round to a reasonable number of digits.

As for the reason why this does not happen I'm sure its due to the external force but I'm not sure which one the answer is looking for specifically as I've taken into consideration the electric field (this encompasses charge and the thickness of the cell) and the ion mass I can't use them in my answer can I? Would stuff like drag (air resistance) come into play or am I looking for an answer that utilities something about the potential 100mV? Hopefully I got all the working out right this time. Thanks for the help by the way.

You might consider the actual speed the ion achieves as it crosses the boundary. How much actual KE does it pick up traversing the field just 7 nm wide? Also consider what might happen to the field once outside the cell; is it likely to remain uniform?

 

[casualdehyde]

Wow! Never thought I would even get half way through this calculation, thank you so much! Seeing as I'm going of the assumption that the particle is not moving at the start should I state that at the beginning of my answer?
Would a reason that the particle doesn't keep accelerating until it reaches the speed of light because once it cross the cell membrane and reaches the other side (I assume extracellular as the the high Na+ concentration makes it more positive, doesn't matter though it's positive regardless) the potential difference is reduced (the main driving force of its acceleration (opposite charges attract)) is diminished leading to a loss of acceleration and therefore it not reaching the speed of light? Is that what you mean when you say the field is no longer uniform.

So qV/0.5m = v^2
(1.6x10^-19 C x 0.07 volts)/0.5 x5.8^-26kg = 386206.8966 sqrt = 621.455 m/s. 621.455 m/s /2.8x10^13 m/s^2
I get that it takes 2.2x10^-11 seconds for the ion to accelerate to that speed i.e. Cross the membrane?
I don't really now where to go from there, assuming I'm on the right track at all.

 

[gneill]

Wow! Never thought I would even get half way through this calculation, thank you so much! Seeing as I'm going of the assumption that the particle is not moving at the start should I state that at the beginning of my answer?

Sure. It's always a good idea to state your assumptions.

Would a reason that the particle doesn't keep accelerating until it reaches the speed of light because once it cross the cell membrane and reaches the other side (I assume extracellular as the the high Na+ concentration makes it more positive, doesn't matter though it's positive regardless) the potential difference is reduced (the main driving force of its acceleration (opposite charges attract)) is diminished leading to a loss of acceleration and therefore it not reaching the speed of light? Is that what you mean when you say the field is no longer uniform.

The electric field across the membrane is being maintained by the cell actively maintaining a charge imbalance via the ions it concentrates. So the inner and outer walls, or surfaces of the membrane look sort of like a capacitor's plates. Where do you find the field of a parallel plate capacitor? What does it look like outside of the capacitor?

So, once across the boundary membrane, what does the ion "feel" in the way of forces?

So qV/0.5m = v^2
(1.6x10^-19 C x 0.07 volts)/0.5 x5.8^-26kg = 386206.8966 sqrt = 621.455 m/s. 621.455 m/s /2.8x10^13 m/s^2
I get that it takes 2.2x10^-11 seconds for the ion to accelerate to that speed i.e. Cross the membrane?
I don't really now where to go from there, assuming I'm on the right track at all.

You don't need to say much after you state that the total speed of the ion after it crosses the boundary is a paltry few hundred meters per second and argue that no further acceleration takes place.

 

[casualdehyde]

So am I saying that once the ion has left the 7nm region of parallel plate capacitor it is no longer being attracted to (driven) by an opposing charge which is why it's actual velocity is no where near the speed of light 600m/s? Will it actually be feeling a pulling force back towards the positive plate (slowing it down) or am I over thinking it?

 

[gneill]

So am I saying that once the ion has left the 7nm region of parallel plate capacitor it is no longer being attracted to (driven) by an opposing charge which is why it's actual velocity is no where near the speed of light 600m/s? Will it actually be feeling a pulling force back towards the positive plate (slowing it down) or am I over thinking it?

Thinking is good. Do a web search and look up how the field behaves for a parallel plate capacitor. That should answer your question. If not, I'll be here

 

[mfb]

So am I saying that once the ion has left the 7nm region of parallel plate capacitor it is no longer being attracted to (driven) by an opposing charge which is why it's actual velocity is no where near the speed of light 600m/s?

Right.

Will it actually be feeling a pulling force back towards the positive plate (slowing it down) or am I over thinking it?

That we don't know, it depends on the unknown environment. You don't have to care about that.
The important point: it won't keep accelerating like that for many meters and microseconds that would be required to get close to the speed of light.