# Approximating number of collisions in canal ray tube

1. Mar 26, 2017

### vishnu 73

1. The problem statement, all variables and given/known data
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.
2. Relevant equations
not really fure which equations will work
f = ma?
mvi = mvf
pv = nrt????
3. The attempt at a solution
someone get me started give me a hint to where to start from

2. Mar 26, 2017

### Staff: Mentor

Mean free path is probably a useful concept here.

3. Mar 26, 2017

### haruspex

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.

4. Mar 26, 2017

### Staff: Mentor

I guess we have to assume room temperature.

5. Mar 28, 2017

### vishnu 73

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

6. Mar 28, 2017

### Staff: Mentor

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.

7. Mar 30, 2017

### vishnu 73

well in the formula d is used because the effective diameter of collision zone was d but here it is rion + rfiller gas i guess so we have substitute d with the new effective radius am i right

8. Mar 30, 2017

### Staff: Mentor

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

9. Mar 30, 2017

### vishnu 73

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

10. Apr 1, 2017

### vishnu 73

sir no matter how hard i try i just dont 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

11. Apr 1, 2017

### Staff: Mentor

The tube is 1 meter long.

12. Apr 2, 2017

### vishnu 73

ok this what i got so far
N/v = n/v * Na = p/RT * Na
N:number of molecules
n : number of moles
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

13. Apr 2, 2017

### Staff: Mentor

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?

Last edited: Apr 3, 2017
14. Apr 3, 2017

### vishnu 73

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 (diameterion + diameterfillergas)/2 which is just radiusion + radiusfillergas 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

15. Apr 3, 2017

### Staff: Mentor

I meant diameter, not distance. In post 5 you said d is the diameter.
The question asks for the total number of collisions.

16. Apr 3, 2017

### vishnu 73

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?

17. Apr 3, 2017

### Staff: Mentor

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.

18. Apr 3, 2017

### vishnu 73

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

19. Apr 4, 2017

### Staff: Mentor

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

20. Apr 5, 2017

### vishnu 73

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