Suppose a particle of ionizing radiation deposits 1.10 MeV in the gas of a Geige

Question

Suppose a particle of ionizing radiation deposits 1.10 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.

(a) The applied voltage sweeps the ions out of the gas in 1.50 µs. What is the current?
____________A

(b) This current is smaller than the actual, 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 if this last effect multiplies the number of ion pairs by 960?
___________A

Explanation / Answer

Ionizing radiation are those radiations which can emit electron out of the atom. There are two kind of ionizing radiations are there

1)Direct

Using charged particle like electrons, protons etc

2)Indirect

eg: Photoelectric effect, compton effect etc

Here we are asked to find out the current.

We know that current=charge/time

I=Q/t=2Ne/t

Ne=1.1 MeV

So, Current,I=2(1.10*106eV/30 eV)(1.6*10-19/1.5*10-6)=7.82*10-9 A

b)I=960*7.82*10-9 A=7.5*10-6A

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