Suppose a particle of ionizing radiation deposits 1.2 MeV in the gas of a Geiger

Question

Suppose a particle of ionizing radiation deposits 1.2 MeV in the gas of a Geiger tube, all of which goes to creating ion pairs. Each ion pair requires 30.0 eV of energy. Randomized Variables E = 1.2 MeV The applied voltage sweeps the ions out of the gas in 1.00 mu s. What is the current, in amperes? This current is smaller than the actual current since the applied voltage in the Geiger tube accelerates the separated ions, which then create other ion pairs in subsequent collisions. What is the current in microamperes if this last effect multiplies the number of ion pairs by 900?

Explanation / Answer

It takes 30eV to create an ion pair.

We have 1.2 MeV of energy. This is 1.2x10 eV

So the number of ions pairs we can make is
1.2x10 / 30 = 4x10 (rounded to 3 significant figures)

This is the maximum number of ion pairs because it assumes all the energy goes to creating ion pairs and that there are no energy losses.

a) Q = It
   I = 4000*1.6*10^-19/10^-6 = 6.4e-10 ampere

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