   Chapter 31, Problem 4PE

Chapter
Section
Textbook Problem

Suppose a particle of ionizing radiation deposits 1.0 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 1.00 (s. What is the current? (b) This current is smaller than the actual current since the applied voltage in the Geiger tube accelerates the separated ions, which then create ether ion pairs in subsequent collisions. What is the current if this last effect multiplies the number of ion pairs by 900?

To determine

(a)

The current in the Geiger tube.

Explanation

Given:

Energy deposited by the ionizing radiation

E=1.0 MeV=1.00×106eV

Energy required to create an ion pair

Ei=30.0 eV

Time taken to sweep of the ion pairs

t=1.00 μs=1.00×106s

Formula used:

The number of ion pairs n is given by,

n=EEi.....(1)

Since there are 2 charges in each pair created, therefore, the total charge Q is given by,

Q=2ne......(2)

Where, e is the electronic charge.

Current I is the rate of flow of charge and is given by,

I=Qt.....(3)

Calculation:

Calculate the number of ion pairs created in the gas by substituting the values of E and Eiin equation (1).

n=EEi=1.00×106eV30.0 eV=3.333×104

Calculate the total charge created in the Geiger tube

To determine

(b)

The current in the Geiger tube if the number of ion pairs created is 900 times that calculated in part (a).

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