WEBVTT
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A cathode-ray tube, CRT, is a device that produces a focused beam of electrons in a vacuum.
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The electrons strike a phosphor-coated glass screen at the end of the tube, which produces a bright spot of light.
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The position of the bright spot of light on the screen can be adjusted by deflecting the electrons with electrical fields, magnetic fields, or both.
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Although the CRT tube was once commonly found in televisions, computer displays, and oscilloscopes, newer appliances use a liquid crystal display, LCD, or plasma screen.
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Consider a CRT with an electron beam average current of 31.00 microamps.
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How many electrons strike the screen every 5.0 minutes?
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So in this example, we have a beam of electrons travelling through our CRT.
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If we draw a sketch of a cathode-ray tube, we have our cathode which emits the beam of electrons towards the phosphor screen at the other end of the tube.
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Since this beam consists of charged particles moving at a certain speed, we know that it creates a current.
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And this current, which we can call capital 𝐼, is given as 31.00 microamps.
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Considering this beam, we want to solve for how many electrons strike this phosphor screen every 5.0 minutes.
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To figure this out, let’s think for a bit about what electrical current is.
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We know that when we have a directed flow of charged particles, like we have here, that consists of some amount of charge, measured in units of coulombs, every unit of time, measured in seconds.
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And in fact, that’s what an ampere is, the more familiar unit for current.
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It’s a coulomb per second.
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And ampere then represents a certain amount of charge passing a point in a certain amount of time.
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When we consider our question, we know that an individual electron has a certain amount of charge.
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We can call it 𝑞 sub 𝑒.
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And we can recall that the magnitude of this charge is equal to 1.6 times 10 to the negative 19th coulombs.
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That then is an amount of charge and we’re given a time value as well, 5.0 minutes.
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Here’s what we can write then using the information we’ve been given.
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First, we can write that 𝐼 is equal to 31.00 times 10 to the negative sixth coulombs per second, since we now recall that a coulomb per second is an ampere.
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Since this is the amount of charge that hits our phosphor screen every second, if we multiply this value by our time, 5.0 minutes, then that will give us the total amount of charge that hits the screen in five minutes of time.
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But notice that we can’t just multiply these two numbers together as is.
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Why not?
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Because the units don’t match up.
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We’ll want to convert our time of 5.0 minutes into that number of seconds.
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Since we know that 60 seconds make up a minute, we can simply multiply 5.0 by 60.
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And that will be our total time interval, now in units of seconds.
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When we consider multiplying these two numbers together now, notice what happens to the units.
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The units of seconds cancel out from numerator and denominator, leaving us with overall units of coulombs.
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But, we don’t want our final answer in units of coulombs.
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We want it in terms of number of electrons.
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And an electron, we recall, has a charge magnitude given in an amount of coulombs.
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So here’s what we can do.
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If we divide this value by the magnitude of the charge on an electron, then notice that the units of coulombs, the unit of charge, cancels out leaving us with a unitless expression.
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Believe it or not, this is a good thing because the number of electrons will be a unitless value.
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It will simply be a pure number.
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And this number, which we can call 𝑁 sub 𝑒, is what we’ve developed an expression for on the right-hand side of this equation.
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All that’s left to do to solve for 𝑁 sub 𝑒, the number of electrons, is to calculate this term.
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It’s equal to 5.8 times 10 to the 16th electrons.
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This number is greater than 1000 trillion.
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That’s how many electrons strike the screen every 5.0 minutes.