Approximating number of collisions in canal ray tube

[timetraveller123]

Homework Statement

Canal rays, i.e., positive ion rays are generated in a gas discharge tube. How often does an ion (r= 0.05 nm) collide with an atom of the ideal filler gas(r= 0.1 nm) if it travels 1 m in a straight path through the discharge tube and if the pressure in the tube is 1 mbar? 1 mbar corresponds to 10²Pa. All the particles are assumed to have the same velocity.

Homework Equations

not really fure which equations will work
f = ma?
mv_i = mv_f
pv = nrt?

The Attempt at a Solution

someone get me started give me a hint to where to start from

 

[mfb]

Mean free path is probably a useful concept here.

 

[haruspex]

Homework Statement

Canal rays, i.e., positive ion rays are generated in a gas discharge tube. How often does an ion (r= 0.05 nm) collide with an atom of the ideal filler gas(r= 0.1 nm) if it travels 1 m in a straight path through the discharge tube and if the pressure in the tube is 1 mbar? 1 mbar corresponds to 10²Pa. All the particles are assumed to have the same velocity.

Homework Equations

not really fure which equations will work
f = ma?
mv_i = mv_f
pv = nrt?

The Attempt at a Solution

someone get me started give me a hint to where to start from

There does not seem to be enough information.
You are given a pressure (dimensionally ML^-1T^-2), two distances (the radii, dimensionally L) and are asked for the mean free path, another L.
Since M and T only occur in the pressure, there is no way to combine that with other variables to end up with just an L. Hence it cannot be used. That leaves us with two Ls. While that could be enough based on dimensional analysis, it is clearly not enough in practice. There is no limit on how closely the gas molecules may be packed.

A temperature could be useful.

 

[mfb]

I guess we have to assume room temperature.

 

[timetraveller123]

ok sorry for the late reply i had completely forgotten about mean path just revised up on it
so apparently this was the equation
λ=1/(√2 π d² N/v)
where λ is distance between collision
d is diameter of particle
N is number of particles and v is volume to tube
ok so how does the radius of filler gas play a part and which diameter should i use and how am i supposed to know N help! thanks

 

[mfb]

How close to ion and gas atoms have to come for a collision?

Your formula has a particle diameter in it, that will be related to this distance.

N/v comes from thermodynamics.

 

[timetraveller123]

well in the formula d is used because the effective diameter of collision zone was d but here it is r_ion + r_{filler gas} i guess so we have substitute d with the new effective radius am i right
sorry for the late reply had a lot of school work

 

[mfb]

Be careful with the difference between radius and diameter. Apart from that: Right.

 

[timetraveller123]

ok then i will give it a go reply soon as possible my entire of tomorrow is not available
thanks

 

[timetraveller123]

sir no matter how hard i try i just don't know how to apply the formula

what does the question mean by 1m in a straight tube does it mean the ion travels 1m before encountering collision(sounds improbable) or is the tube 1m long

 

[mfb]

The tube is 1 meter long.

 

[timetraveller123]

ok this what i got so far
N/v = n/v * N_a = p/RT * N_a
N:number of molecules
n : number of moles
N_a:Avogadro constant
p pressure
T temperature assumed 293 kelvin
so N/v i got to be 2.26 * 10²²
then for d² i got (0.15nm)² = 0.0225 * 10^-18
then pluggin in the formula i got λ = 0.000406646 m
is that correct?
if not where did i go wrong?
if yes then how do i procceed from there

 

[mfb]

DistanceDiameter and radius are not the same.

If you have a collision every x mm, and the total length is 1 meter, how many collisions do you have?

 

[timetraveller123]

no wait what do you mean by distance and radius are not the same as in the original formula the d stands for radius of collision zone so in my problem the radius of collision zone isn't it (diameter_ion + diameter_fillergas)/2 which is just radius_ion + radius_fillergas which = 0.1nm + 0.05nm = 0.15nm
and furthermore the question asks how often does the ions collide is it asking for number of collisions per second, or total number of collisions

 

[mfb]

I meant diameter, not distance. In post 5 you said d is the diameter.

d is diameter of particle

The question asks for the total number of collisions.

 

[timetraveller123]

no but in that case the particles are all identical hence have same radius thus the radius of collision zone was the diameter of one particle but here the radius is not identical thus here the effective radius is the radius combined together it is mistake on my part shouldn't have used d for effective radius
am i right?

 

[mfb]

The effective radius is the sum of the radius of the target and the radius of the projectile.
The effective diameter is twice that radius.

 

[timetraveller123]

found this picture in hyperphysics
http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/menfre.html
so based on this picture what i perceive is that the πd² part is calculating the area of collision which has a radius equivalent to that of the diameter of the particle so hence the d is not the diameter of zone but the diameter of individual particles but in our case the particles are not similar hence the radius of zone is the sum of radii of the two particles

 

[mfb]

Ah, the initial formula is for non-point particles already.
Okay, your result is right. I misinterpreted the formula.

 

[timetraveller123]

so now is the number of collisions 4000 along the 1m?

 

[mfb]

Looks right.

 

[timetraveller123]

but i just checked the answer and i was off by a factor of ten the acceptable answers was 400 to 1000 :( not sure where i went wrong can you help me find i can't find any miscalculations or is the answer wrong

 

[mfb]

Let's see.
1mbar/(293 kelvin * Boltzmann constant) = 2.47E22/m³ (WolframAlpha)

Plugging that into the formula for the mean free path we get 405 micrometers. Dividing 1 meter by 400 micrometers gives 2500 collisions. That is outside the given range.

To get about 700 collisions, we would have to increase the mean free path by a factor of about 4. I don't see a reasonable change that could lead to such a factor. The radius/diameter discussion, if changed, would decrease the mean free path.

Forgetting the ion radius would lead to a factor ~2, forgetting the filler gas radius would lead to a factor 9, both don't work.

I don't know.

 

[timetraveller123]

ok then never mind i still i got the concept right maybe the answer is wrong thanks for your help :)