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Chapter 18: Problem 5

The current in the electron beam of a computer monitor is \(320 \mu\) A. How many electrons per second hit the screen?

Short Answer

Expert verified
Answer: \(2 \times 10^{15}\) electrons per second.

Step by step solution

01

Convert current to base units

First, we need to convert the given current from microamperes to amperes. To do this, we'll multiply by \(10^{-6}\): $$ I = 320 \mu\text{A} \times 10^{-6} \text{A}/\mu\text{A} = 3.2 \times 10^{-4} \text{A} $$ Now we have the current in amperes.
02

Rearrange the formula to find the charge

We need to find the charge, \(Q\), in one second (\(t = 1\) s). Rearrange the formula: $$ Q = I \cdot t $$ Plug in the values: $$ Q = (3.2 \times 10^{-4} \text{A}) \cdot (1 \text{s}) = 3.2 \times 10^{-4} \text{C} $$ In one second, the charge amount is \(3.2 \times 10^{-4}\) coulombs.
03

Calculate the number of electrons

To find the number of electrons hitting the screen per second, divide the charge in one second by the elementary charge of one electron: $$ \text{Number of electrons} = \frac{Q}{e} $$ $$ \text{Number of electrons} = \frac{3.2 \times 10^{-4} \text{C}}{1.6 \times 10^{-19} \text{C/electron}} $$ $$ \text{Number of electrons} = 2 \times 10^{15} \ \text{electrons} $$ So, \(2 \times 10^{15}\) electrons per second hit the screen.

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Most popular questions from this chapter

Show that \(A^{2} \times \Omega=W\) (amperes squared times ohms \(=\) watts).
A current of \(10.0 \mathrm{A}\) is carried by a copper wire of diameter $1.00 \mathrm{mm} .\( If the density of the conduction electrons is \)8.47 \times 10^{28} \mathrm{m}^{-3},$ how long does it take for a conduction electron to move \(1.00 \mathrm{m}\) along the wire?
What is the resistance of a \(40.0-\mathrm{W}, 120-\mathrm{V}\) lightbulb?
A current of \(2.50 \mathrm{A}\) is carried by a copper wire of radius $1.00 \mathrm{mm} .\( If the density of the conduction electrons is \)8.47 \times 10^{28} \mathrm{m}^{-3},$ what is the drift speed of the conduction electrons?
What is the current in a \(60.0-\mathrm{W}\) bulb when connected to a \(120-\mathrm{V}\) emf?
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