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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$\mu$ . 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 other ion pairs in subsequent collisions. What is the current if this last effect multiplies the number of ion pairs by 900?

$=2 \cdot 66 \mu \mathrm{A}$

Physics 103

Chapter 31

Radioactivity and Nuclear Physics

Nuclear Physics

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

University of Winnipeg

Lectures

02:51

In physics, wave optics is…

10:02

Interference is a phenomen…

01:06

The energy of 30.0 eV is r…

00:53

A particle of ionizing rad…

02:11

03:49

In a Geiger tube, the volt…

05:28

A sample of hydrogen atoms…

02:56

A $0.100-\mathrm{mA}$ elec…

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A Review problem. A $\math…

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A hypothetical atom has tw…

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X-ray tube. An X-ray tube …

03:42

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08:44

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but a off the problem. Uh, here, um, were asked to find the total current by total I total current in the Giger counter is equal to the cut. And due to the most do pines plus Karan due to their negative lines, negative wines, then we can find the current due to the post behind. I will be simply don't lumber off charges which is number off charges times charge on each particle. Let's take you, um, let's say e e for four poster charges for I p uh, divided by time. And similarly, for the negative, this will be a total number of charges times Navy to rewrite my teeth. So let's find total number off charges here. So the energy deposited guess is e way have a 10 to the power. There are six in a from Walt and so we can find end. There are a number of charges is equal to the total energy deposited, derided by the energy in each energy used in creating each Hein. So which is a 30 trundled and this gives us one or three times 10 to the power and our five five. So the number off finds we have is one over three times in five. So these number of lines are saying for both poster and egged you. Um So and then, uh, the time we have here is ah, um, tend to the bar minus six seconds, which is one microsecond so we can plug those values here. Um, so here we have one or three, uh, times 10 to the power of five. Five times the charge in each article for Wall Street charges 1 26 constant to power 10 to the power minus 19 Coolum. And this will be since we have a to currents. Then we will take this two times, derided by so this whole already ready by the time we have instant in power minus six seconds. So this gives us in the total current. So let me just there. You see this? This might be confusing. So this will be a total on the total current we get here will be a few times off. So here we will. Two turns off Poon Poon six who were a tree time stint with power attempted power here, Um, minus state in years. Read abuse. So this will be over and burnt Barbie. When we have amplification factor off a gamma, that is a call to 900 900. Then I would actually guarantees actual burnt will be I p plus I in close eye in, uh, divided by the factor we have, which isn't 900. And substituting this very rich people when we got here is this will use, um, carry out here this way. You derided by 900 900. And this gives us the actual Brando from 1.1 war. Um, Night times Dental power minus 10. Minus 10 in Pierre. End off the problem. Thank you for watching.

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