Difference between revisions of "Ion Diffusion"

From New IAC Wiki
Jump to navigation Jump to search
Line 52: Line 52:
<math> n(x,y,z,t) = \frac{s}{(4 \pi D_L t)^{1/2} } e^{\left (-\alpha t - \frac{(z-v_d t)^2}{4D_L t} -\frac{r^2}{4D_T t} \right) X  \sum_{m=0}^\infy \frac{(r^2/4D_Tt)^m} {m!} \right[ 1- e^{frac{(r^2/4D_Tt)^m}{m!}}</math>
<math> n(x,y,z,t) = \frac{s}{(4 \pi D_L t)^{1/2} } e^{\left (-\alpha t - \frac{(z-v_d t)^2}{4D_L t} -\frac{r^2}{4D_T t} \right) X  \sum_{m=0}^\infy \frac{(r^2/4D_Tt)^m} {m!} \right[ 1- e^{frac{(r^2/4D_Tt)^m}{m!}} \left]</math>

Revision as of 20:03, 3 September 2013

Importance of Ion Diffusion Study

Ion diffusion and mobility data in a gas theoretically and experimentally are important. First, experimental data of ion mobility provides information to understand the ion-molecule interactions that takes place as medium that has (E/N) > 2 Td (thermal equilibrium limit). Also it is used to calculate the ion-ion recmbination coefficients and the rate of dispersion in the gas due to the ion mutual repulsion.Finally, diffusion and mobility data give information to understand the electrical discharges in gases. Although A lot of data has been collected on the ion transport in gas in the last decade, recent measurements show a lot of incorrect data due to ignore the change in the ion identity through a chemical interaction as the ion travels in a gas.<ref name="Mason"> E.A. Mason & E.W. McDaniel, Transport properties of ions in gases, John Wiley and sons, 1988 </ref>

In case of using GEM detector as gaseous chambers, there are some assumptions to reach an analytical solution for the ions' diffusion in the gas. The solution treats a single type of ions in a uniform density N, moving in an area of a uniform electric field. The ion number density is low enough so the space-charge field is negligible. For a small ionic flux density <ref name="Mason"/>:

[math] J(r,t) = v_d n (r,t)- D.\nabla n(r,t) [/math]

The ionic diffusion equation is written as:

[math] \frac{\partial n (r,t)}{\partial t}+ \nabla .J + \alpha n = 0 [/math]

Substituting the value of J(r,t):

[math] \frac{\partial n (r,t)}{\partial t}- \nabla .D \nabla n + v_d. \nabla n + \alpha n = 0 [/math]

Since the source in circular shape in a plan normal to the the electric field, Adding it to the previous equation in Cartesian coordinate:

[math] \frac{\partial n (x,y,z,t)}{\partial t}= D_T (\frac{\partial^2 n (x,y,z,t)}{\partial^2 x} + \frac{\partial^2 n (x,y,z,t)}{\partial^2 y} ) + D_L \frac{\partial^2 n (x,y,z,t)}{\partial^2 z} - v_d \frac{\partial n}{\partial z} - \alpha n + \beta(x,y,z,t) [/math]

Assuming the origin in the middle of the source, the solution for the equation is :

[math] n(x,y,z,t) = \int_{-\infty}^t dt' \int_{-\infty}^{\infty} dx' \int_{-\infty}^{\infty} dy' \int_{-\infty}^{\infty} dz' \frac{\beta(x',y',z',t')}{4 \pi (t-t')^{2/3} D_T D_L^{1/2}} . \exp \left (-\alpha (t-t') - \frac{(x-x')^2 + (y-y')^2}{4D_T (t-t')} - \frac{[z-z' - v_d (t-t')]^2}{4D_L (t-t')}\right ) [/math]

in cylindrical coordinate:

[math] n(x,y,z,t) = \int_0^\infty r'dr' \int_0^{2 \pi}d\theta' \int_{-\infty}^\infty dz' \int_{-\infty}^t dt' \frac {\beta(x',y',z',t')}{4 \pi (t-t')^{2/3} D_T D_L^{1/2}} . \exp \left (-\alpha (t-t') - \frac{r'^2 +r^2- 2rr'cos(\theta - \theta')}{4D_T (t-t')} - \frac{[z-z' - v_d (t-t')]^2}{4D_L (t-t')}\right ) [/math]

U-233 source is thin coating of radius [math]r_0 [/math] and thickness [math] \delta z [/math], emits a fission fragment in a time [math]\delta t [/math], the source [math] \beta(r',z',t')[/math] is mathematically described :

[math] \beta(r',z',t') = \frac{b}{\pi r ^2} S(r'-r_0)\delta (z') \delta (t') [/math]

Assuming the coat is a uniform surface and its thickness is infinitesimal compared to the fission fragment range in gas. Substituting in intead of [math] \beta [/math] :

[math] n(x,y,z,t) = s e^{-\alpha t - \frac{-(z- v_d t)^2/4D_L t}{(4 \pi t)^{3/2} D_L^{1/2} D_T}} \int_0^{r_0} r'dr' \int_0^{2 \pi} d\theta' e^ {\frac{r'^2 +r^2- 2rr'cos \theta'}{4D_T t}} [/math]

Using intergal tables [math] 1/\pi \int d\theta e^{\pm x cos\theta} = \sum_{m=0}^\infty \frac{(x/2)^{2m}}{(m!)^2} [/math] , then integrating over r' by parts:

[math] \int_0^a x^m e^{-x} dx = m! ( 1- e^{-a} \sum_{i=0}^m \frac {a^i}{i!}) [/math] so the solution looks like:


[math] n(x,y,z,t) = \frac{s}{(4 \pi D_L t)^{1/2} } e^{\left (-\alpha t - \frac{(z-v_d t)^2}{4D_L t} -\frac{r^2}{4D_T t} \right) X \sum_{m=0}^\infy \frac{(r^2/4D_Tt)^m} {m!} \right[ 1- e^{frac{(r^2/4D_Tt)^m}{m!}} \left][/math]