Potential Necessary for Ionization by Collision

[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 cm³, 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 into 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 cm³ 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 metre³

0.004766 → 2.87015 x 10²¹ particles per metre cubed →2.87015 x 10¹⁵ per cm³.

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.

 

[Charles Link]

Perhaps I am misreading this problem, but it appears to be a mercury arc lamp with the electrons as the active particle that creates the ions. The mercury arc lamp can operate at a lower voltage than that which ionizes nitrogen, because Hg has a lower ionization potential (10.4 V) than nitrogen (14.5 V). All that is necessary, as far as I can tell, is you need to get electrons to be more energetic than the ionization potential of nitrogen. (The electrons come off the cathode at very low energy and get accelerated by the potential.) Again, maybe I'm misreading it, but it appears they are giving you a lot of extra miscellaneous information. $\\$ Just a little additional info: Usually these arc lamps operate by thermionic emission of electrons from a heated cathode. The lamps are sometimes started with a much higher voltage=thousands of volts, to get the arc started because the cathode starts out at ambient temperature. The cathode gets heated by the positive ions that crash into it, so once the arc gets started, the Hg arc lamp will operate at voltages just above 10.4 Volts.

 

[haruspex]

it appears to be a mercury arc lamp

I don't think there is any mercury involved. It was mentioned only as a way of expressing the pressure.

 

[Charles Link]

I don't think there is any mercury involved. It was mentioned only as a way of expressing the pressure.

Thank you @haruspex. That's helpful=I read it too quickly. Presumably though they are using some kind of heated cathode with thermionic emission=they don't seem to supply these details...One other way I know of starting an arc is to first have the electrodes in contact and separate them slowly upon running a high current (e.g. 1-2 amps or more) through them.

 

[elpoko]

Thanks for the replies guys. Sorry about the Hg pressure notation, I know it's a little confusing.

Charles, I'm not sure they're leaving out much. The chapter describes only simple parallel plate ionisation chambers, no mention of heated filaments etc so I expect the description is thorough enough for my purposes... I think the main thing required is a way to relate the applied voltage over the distance of 3cm to particle energies/temperature. The chapter's material also mentions an average energy of collision for a corresponding mean free path - 0.0025eV for air at atmospheric pressure. I'm shooting in the dark though.

I feel like it's something simple I'm just overlooking.

Tom

 

[Charles Link]

Thanks for the replies guys. Sorry about the Hg pressure notation, I know it's a little confusing.

Charles, I'm not sure they're leaving out much. The chapter describes only simple parallel plate ionisation chambers, no mention of heated filaments etc so I expect the description is thorough enough for my purposes... I think the main thing required is a way to relate the applied voltage over the distance of 3cm to particle energies/temperature. The chapter's material also mentions an average energy of collision for a corresponding mean free path - 0.0025eV for air at atmospheric pressure. I'm shooting in the dark though.

I feel like it's something simple I'm just overlooking.

Tom

Except for a more refined calculation, I don't think the mean free path is part of the calculation, and the answer (although I'm not supposed to give it, it's too simple not to), is 14.5 volts for nitrogen and 13.5 volts for oxygen. $\\$ Editing: Suppose we attempt a more refined calculation: The electron essentially has to accelerate to the necessary voltage inside the distance of the mean free path... This should be workable. Maybe that's what your instructor is looking for...

 

[Charles Link]

@elpoko Please see my edited comment above.

 

[haruspex]

The electron essentially has to accelerate to the necessary voltage inside the distance of the mean free path.

Yes, I think that is the key here. Except, how to find the density? The calculation of temperature in the OP strikes me as inappropriate. We need the temperatureof the gas, not the temperature equivalent of 14.5eV. Maybe assume room temperature.

 

[Charles Link]

Yes, I think that is the key here. Except, how to find the density? The calculation of temperature in the OP strikes me as inappropriate. We need the temperatureof the gas, not the temperature equivalent of 14.5eV. Maybe assume room temperature.

I think the gas can be assumed to be at room temperature in order to find the density of the gas. The gas molecules can be assumed to be stationary targets as the electron will be moving at high speed. The mean free path of the electron through the gas will depend on the density of the gas molecules along with their cross section. $\\$ Editing: The answer that you get when taking into consideration the short mean free path of the electron at the pressure of 50 mm Hg will probably be considerably larger than 14.5 volts. Even though this was labeled as "Introductory Physics Homework" I believe the textbook is at a level much higher than that. It would be helpful to have a copy of the first chapter of the textbook in solving this, as it is really a rather specialized topic.

 

[Charles Link]

The answer that you get when taking into consideration the short mean free path of the electron with the gas at a pressure of 50 mm Hg will probably be considerably larger than 14.5 volts. Even though this was labeled as "Introductory Physics Homework", (which is what I thought it was in my first response or two), I believe the textbook (listed in the OP as "Nuclear Radiation Physics" Chapter 1 by Lapp and Andrews) is at a level much higher than that. It would be helpful to have a copy of the first chapter of the textbook in solving this, as it is really a rather specialized and also somewhat advanced topic.

 

[elpoko]

I guess it is a bit outside "Introductory Physics."

The answers listed in the back of the book are 15.5KV for nitrogen 11.8KV for oxygen. Seems far out, and unlike a typo. Problem is I don't even have an instructor here to clear things up in person, studying it myself. I haven't been able to get any further.

 

[Charles Link]

I guess it is a bit outside "Introductory Physics."

The answers listed in the back of the book are 15.5KV for nitrogen 11.8KV for oxygen. Seems far out, and unlike a typo. Problem is I don't even have an instructor here to clear things up in person, studying it myself. I haven't been able to get any further.

That doesn't surprise me=answers in the kilovolts. Suggestion: Compute the mean free path $\lambda$ for the electron, where $\lambda=\frac{1}{n \sigma}$ where $n$ is the number of molecules per unit volume, and $\sigma$ is the cross-section area of the molecule. You will see that $\lambda$ is quite small, so that the electron needs to acquire 14.5 volts in a very short distance. Having the answers is helpful, because I was a little surprised when I did the calculation and found the answer to be far in excess of 14.5 volts. $\\$ Additional note: For density $n$, start with $PV=NRT$, where $N$ represents the number of moles of molecules, and $R=.08206$ liters-atm/(mole-degree) . Use $T=295 K$ (room temperature), and convert $N$ to number of molecules per cm^3, $n=(N/V)N_A$, using Avagadro's number $N_A=6.02 E+23$ and converting one liter=1000 cm^3 etc. Pressure $P$ in atm is $P=50/760$. (I included this extra info because it really requires a background in chemistry. If you are doing physics, you may or may not have seen all of this in detail. Alternatively, you could use $PV=N' k_b T$ where $N'$ is the number of molecules and $k_b$ is Boltzmann's constant, but this requires more conversions of units.) $\\$ I'm double-checking my arithmetic, but I compute 55 kV for the first answer. I'm curious to know how the book gets 15.5 kV. A calculation like this normally isn't going to have 3 sig-fig accuracy... For the area of the nitrogen molecule, I used $\sigma=(3.2 E-8)^2 \pi/4$, but a somewhat reduced effective area would also be reasonable in a calculation like this... In any case, I believe I computed the answer 55 kV correctly...