- #1

elpoko

## Homework Statement

A molecule of nitrogen has a diameter of 3.2 x 10

^{-8}cm and can be ionized upon absorbing 14.5 eV. what potential must be applied to a parallel-plate ion chamber operating at a pressure of 50mm of mercury, and with an electrode separation of 3.0 cm, in order to produce ionization by collision? Repeat the calculations for oxygen molecules where the pertinent constants are 2.9 x 10

^{-8}cm and 13.5 eV, respectively.

In the appendix: Nitrogen Chem. atm. wgt. = 14.008. Atm. mass (MeV) = 13043.556

This is problem 1-4, from the first chapter, "Ionization and Ionization Chambers" from Lapp and Andrews 4th Ed "Nuclear Radiation Physics" (1972). The book uses Centimetre gram seconds CGS units.

## Homework Equations

E

_{kinetic}= kT

k is Boltzmann's constant, T is temperature in degrees Kelvin.

Voltage = Joules/Coulomb

Mean free path

## λ = \frac 1 { 4 \sqrt {2} n r^2 } ##

n is concentration of particles per cm

^{3}, r is radius of spheres in medium. Ie radius of nitrogen in this instance.

Ionic mobility = (cm sec

^{-1}/ (volts cm

^{-1})

The above are found earlier in the chapter and are presumed relevant to the equation.

## The Attempt at a Solution

Key figure is the 14.5 eV necessary for ionization. What voltage across 3cm at 50mm Hg Nitrogen will give an average energy of at least 14.5 eV to particles?

## 14.5 eV = 14.5 * (1.6 * 10^{-19}) = 2.32 * 10^{-18} Joules ##

An average energy of 2.32 x 10

^{-18}Joules or 14.5 eV is needed.

From Boltzmann:

## E = kT ##

## 14.5 = (8.62 * 10^{-5}) * T ##

∴ T = 168213 °K.

Average velocity needed, using kinetic energy formula:

## 2.32 * 10^{-18} = 0.5 * (2.32586705 * 10^{-26}) * v^2 ##

Velocity needed is 14124.3 m/sec.

I'm not sure where to go from here. If I could get a value of the ionic mobility for Nitrogen at 50 mm Hg, then I could just plug that in to get the volts per cm and use 3cm to get voltage required. That value does not seem to be available in the book.

Because the nitrogen diameter is included in the question statement, I believe the mean free path is relevant. To find n, concentration of particles per cm

^{3}use the gas law:

## PV = nRT \rightarrow n = \frac {PV} {RT} ##

Converting 50mm Hg to 6666 Pascals

## n =\frac {6666 * 1} {8.314 * 168213} = 0.004766 ## moles per metre

^{3}

0.004766 → 2.87015 x 10

^{21}particles per metre cubed →2.87015 x 10

^{15}per cm

^{3}.

Plugging in, the mean free path is 0.06015 cm.

From here, I don't know how to relate the mean free path to energy, or then voltage needed. Any help would be much appreciated. I'm doing some home study as I have an interest in nuclear physics. There may be some issues with units above as I'm jumping between old CGS and new SI and back again, but I'm far more concerned about the actual strategy used to solve the question than numerical slips.