Geiger counter TPR FL Passage

[549]

FL TPR4, PS, Passage 2, Item 8

This passage talked about a Geiger counter, in which a capacitor, attached to a battery and a resistor, is surrounded with an inert gas. The gas is ionized and sends a current through the circuit, resulting in a voltage pulse, and emitting a click.

The question asked what change to the system will cause a decrease in sensitivity of the Geiger counter.
A. Using a gas with a higher ionization energy
B. Using a gas with a lower dielectric constant

I thought that both of these answers were good, but I picked B (A is correct) because I figured that changing the capacitance was a more direct was to change to voltage pulses, therefore the sensitivity. I understand why A works, but I don't know how to clearly choose that over choice B.

Any help appreciated!

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[GRod18]

I would say that as you lower the dialectric constant, you are lowering capacitance. So if you are lowering capacitance less charge would be stored when the ions goes through the capacitor based on Q = CV if ionized charges stay the same, as we lower capacitance, voltage would go up, therefore your geiger counter should be more sensitive and emit more clicks when you lower capacitance.

This is my reasoning, but perhaps someone else can offer you more input.

 

[vsl5]

A geiger counter works by attracting ionized gas particles to the wire and counts them as "events". If you have more ionized gas particles, you will hear more clicking. If ionization energy is increased, less gas particles are ionized and so less detection will occur. The voltage pulse u are refering to is caused by the ionized gas collected at the wire, not the voltage due to capacitance, thats why B is wrong.

 

[549]

Oh, okay- thanks!