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Homework Help: Electric Potential, Capacitance and charge of a Geiger Counter

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the potential difference, capacitance of the Geiger Counter and the charge on the anode.
    So, we have a Geiger Counter with low-pressurized neon inside. What we want is to have an alpha particle or electron to travel into the Geiger Counter with enough energy to collide with a neon atom and ionize it. then we will cause that free electron to accelerate through the potential difference in the G.C. to collide with another neon atom and again ionize that atom. Basically we are going to have a cascade of free electrons traveling towards the anode. The inner wire, anode, is at a high potential while the outer cylinder (cathode) is at a lower potential.

    Variables we know.
    diameter of cathode: 15mm = 15 x10^-3 m
    diameter of anode: 1mm = 1x10^-3 m
    length of Geiger Counter: 40mm = 40x10^-3 m
    radius of neon atom = 154 pm = 154 x10^-10m
    energy to 1st ionize neon = 2080.7 kJ/mol
    pressure in G.C. = 75 torr which equals 8.1x10^-3 kg/m^3 density

    2. Relevant equations
    Gauss Law : E.A = q/εo
    -∫E dr = ΔV
    Mean-Free Path = 1/(pi*r^2*n_v)

    3. The attempt at a solution
    Ok so we need to find how much joules it takes to ionize neon.
    2080.7 kJ/mol = 1.04x10^-2 eV
    1 eV = 1.602x10^-19 J
    2080.7 kJ/mol * 1.04x10^-2 eV * 1.602x10^-19 J = 3.466 x 10^-18 J

    next we find the atom/volume density of neon in the G.C.

    8.1x10^-3 kg/m^3 * 1000g/kg * 1 mole/20.1797g * 6.02x10^23 atoms/mole = 2.416x10^23 atoms/m^3

    this will be out number density n_v.

    next we will find the distance the electron travels before it becomes in contact with a neon atom:

    Mean-Free PAth = 1/(pi*r^2*n_v) r is the radius of the neon atom

    1/(pi* (154 x10^-10m)^2 * 2.416x10^23 atoms/m^3) = 5.55x10^-9 m
    so an electron will on average travel that distance before it becomes in contact with another neon atom.

    Now lets find the electric field inside the G.C.
    Well the cathode is a thin cylindrical shell so it produces no electric field inside the G.C. The wire produces a E though...
    E.A = q/εo
    Surface area of the wire is the derivative of the volume of a cylinder... 2*pi*r*L

    E (2*pi*r*L) = q/εo
    E = Q/(2*pi*r*L*εo)

    now lets find the potential from the anode to the cathode (r --> R)

    -∫Q/(2*pi*r*L*εo) dr
    -Q/(2*pi*L*εo)∫ 1/r dr
    -Q/(2*pi*L*εo)( ln(R) - ln(r) )


    ok so we pretty much have everything. Distance, the potential equation, electric field equation, energy required to ionize now how do we put it all together? This is where I am lost. E is not uniform so I'm not sure if I can solve for the electric force then solve for acceleration and use Vf^2 = Vi^2 + 2*a*Δx or use the energy to find the velocity right before the electron collides with a neon atom. Then what do I do with the velocity. I am pretty sure in some way I am suppose to find the acceleration or force and then i can solve for Q ( the charge on the anode) then I can find everything else...right?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 26, 2012 #2
    Ok i think i got it. We want to create a cascade of electrons from anywhere in the tube. This means i can find the acceleration of the electron thats needed to accelerate it to the velocity needed to ionize a neon atom. And the least likely place of that to happen is at the cathode (furthest from anode) so we have to make sure the acceleration atleast agrees with the total charge (Q) and electric field at hat point. So I find the acceleration needed then use F=ma and find the Force acted on it by the anode. Then using E = Fe/q I can find the total charge on the anode. Then plug the Q into my potential equation, find the potential and now I can find the capacitance which is C = q/V.

    So this is what i did.

    E = 1/2 m v^2
    3.466x10^-18J = 1/2* 9.11x10^-31kg * v^2
    v = 2.76x10^6 m/s
    Vf^2 = Vi^2 + 2*a*x (assume initial v = 0 m/s)
    2.76x10^6 ^2 = 2 *a * 5.55x10^-9m
    a = 6.857x10^16 m/s^2

    E = F/q
    E = ma/q E = Q/(2*pi*R*L*εo) R = radius of cathode

    Q = (2*pi*R*L*εo)*(ma/q)
    Q = (2*pi*7.5x10^-3m*40x10^-3 * εo) * (9.11x10^-31kg*6.857x10^16 m/s^2)/1.602x10^-19C
    Q = +6.506x10^-9C

    Now plug Q into the ΔV integration equation from r --> R
    you get 7919.68V...

    C = Q/V
    C = 6.506x10^-9C/7919.68V = 8.21x10^-13 C.

    Finished? I hope so, I think that's right... lol
    Last edited: Mar 26, 2012
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