Simple Gas Ionization Detector Problem

[jumbogala]

Homework Statement

I'm not sure if this is really introductory physics... tell me if you think I'd have more luck in the advanced physics section.

The problem: You have a gas filled counter, in the form of a parallel plate capacitor. It has capacitance 9.1 x 10^-9 F. It takes 20 eV of energy for each ionization in the capacitor to occur.

2 x 10⁶ eV of energy is deposited between the plates by a particle. What is the size of the voltage pulse produced?

Homework Equations

C = Q/V

The Attempt at a Solution

I took a guess, but I have no idea if it's right.

I said that 2E6 / 20 = # of ionizations = 100 000

Each ionization produces an electron and an ion, with charge e. The electrons move to the positive plate and the ions move to the negative plate. Because the charges on the plates will counteract each other, we're lowering Q by (100 000*e) = 1.6E-14.

Then (change in charge) / C = (change in voltage)

(1.6E-14) / 9.1E-9 = 1.8E-6 is the voltage pulse seen (as a decrease).

Is that right? Does it make sense that the voltage decreases - doesn't pulse imply increase?

 

[Delphi51]

"Pulse" can be either up or down.
Very convincing - looks correct.

 

[kerimek]

Hello, thank you for your question. I believe this is a problem that falls under introductory physics as it involves basic principles of electricity and energy.

Your attempt at a solution is on the right track, but there are a few things that can be clarified and corrected.

Firstly, the idea of a voltage pulse can be confusing in this context. A voltage pulse can refer to a change in voltage, either an increase or a decrease. In this case, the voltage pulse would be a decrease, as you correctly calculated.

Next, it is important to note that the energy deposited by the particle is not directly related to the number of ionizations that occur. The energy deposited is the total energy required for all the ionizations to occur. In this case, since it takes 20 eV for each ionization, the 2 x 10^6 eV of energy deposited can produce a maximum of 100,000 ionizations (2 x 10^6 / 20 = 100,000).

Your calculation for the change in charge is correct, but the units should be coulombs (C) instead of electron charge (e). So the change in charge is 1.6 x 10^-14 C.

Finally, your calculation for the voltage pulse is also correct, but the units should be volts (V) instead of microvolts (uV). So the voltage pulse seen is 1.8 x 10^-6 V, which is a decrease in voltage.

I hope this helps clarify the problem and your solution. Keep up the good work in your studies of physics!