Simple Gas Ionization Detector Problem

In summary, the problem involves a gas-filled parallel plate capacitor with a capacitance of 9.1 x 10-9 F and an energy requirement of 20 eV for each ionization. A particle deposits 2 x 106 eV of energy between the plates and the question is to calculate the size of the voltage pulse produced. Using the equation C = Q/V, the solution involves calculating the number of ionizations, which is found to be 100,000. The decrease in voltage is then calculated to be 1.8E-6, which is consistent with the concept of a voltage pulse. Overall, the solution appears to be correct and makes sense.
  • #1
jumbogala
423
4

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?
 
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  • #2
"Pulse" can be either up or down.
Very convincing - looks correct.
 
  • #3




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!
 

FAQ: Simple Gas Ionization Detector Problem

1. What is a Simple Gas Ionization Detector?

A Simple Gas Ionization Detector is a type of scientific instrument used to detect the presence and concentration of certain gases in a given environment. It works by measuring the amount of ionization that occurs when the gas molecules interact with radiation or electric fields.

2. How does a Simple Gas Ionization Detector work?

A Simple Gas Ionization Detector typically consists of a chamber filled with a gas, an electric field, and electrodes. When a gas molecule enters the chamber, it collides with the gas particles already present, causing ionization. The charged particles are then drawn towards the electrodes, creating a measurable electric current.

3. What types of gases can be detected with a Simple Gas Ionization Detector?

A Simple Gas Ionization Detector can detect a wide range of gases, including but not limited to hydrogen, helium, methane, carbon dioxide, and more. The type of gas that can be detected depends on the specific design and composition of the detector.

4. What are the advantages of using a Simple Gas Ionization Detector?

One of the main advantages of using a Simple Gas Ionization Detector is its sensitivity. It can detect very low levels of gas concentration, making it useful in various scientific and industrial applications. It is also relatively simple and cost-effective compared to other gas detection methods.

5. How accurate is a Simple Gas Ionization Detector?

The accuracy of a Simple Gas Ionization Detector depends on various factors, such as the type of gas being detected, the environmental conditions, and the design of the detector itself. In general, it can provide accurate measurements within a certain range, but it may not be as precise as other more specialized gas detection methods.

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