Chapter 21 EPE to V to Millikan.notebook

spray ­ ­ ­ ­ contents

15

14

13 5 12 ­ ­ 11 4 ­ 10

OIL 9

8 3 7 ­ 6

2 5

4 ­

3 1

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90° 0 0 flat metal plates ­

+ Millikan's _ V Oil Drop source of variable voltage Experiment

Summary of Millikan's Oil drop experiment: purpose: to Determine the charge of the electron (the elementary charge!) Synopsis: determine the charge on many oil droplets, then find the largest number that is commonly divisible into all the charges found. That would be the charge of one electron.

1. When oil drops are sprayed out the gain some electrons from the nozzle. 2. some droplets fall between the oppositely charged plated and the voltage can be selectively adjusted to "suspend" some of the drops - in essence balancing the downward pull of gravity with an equal but oppositce upward electrical force. 3. When the voltage is removed, the droplet falls at terminal velocity, which can be measured. The terminal veloicity of the droplet can then be used to determine the drop's weight, thereby determining the electric force. 4. Now knowing the electric force and the stregth of the electric field, Millikan could calculated the Amount of chargeon the droplet in Coulombs. 5. Collecting a variety of charges, he could infer the elctron mass by finding the largest # which could divided evenly into all the data. Easy, right? Chapter 21 EPE to V to Millikan.notebook The Electron­volt (eV) contents It's a small unit of energy

The electron volt is the amount of an energy a single elementary charge gains or loses when moved across one volt of potential difference.

1 eV = 1.6 x 10-19 Joules

From last night's HW

Answer in EV's: ­500 eV Answer in Joules: ­ 8.0 x 10­17 Joules Chapter 21 EPE to V to Millikan.notebook

contents 0 volts ------

3 volts p+ e­ p+ 6 volts gains gains loses 3 electronVolts 6 eV's 3 eV's 9 volts gains 6 eV + + + + + + + + + + + + + + + + + + + + + + + + 12 volts 2 elementary charges x 10 V = 20 eV's

to find Joules: use

solve for W = Vq W = 10 V x( 2 x 1.6 x10­19 C) = 3.2 x10­18 Joules

Set up a proportion this is on page of of RT 9.128 x 10­17 J 1.60 x 10­19 J ______= ______x electron volt 9.128 x 10______­17 J 1.60 x 10­19 J = ______x electron volt cross multiply and solve for x ______x(1.60 x 10­19 J) = (9.128 x 10______­17 J) eV 1.60 x 10­19 J 1.60 x 10­19 J

x = 570.5 eV

­ 1 x 500 = ­ 500 eV

Use again

Solve for W = Vq = ( 500 V) (­1.6 x 10­19 C) = ­ 8 x 10­17 J

in eV: +2 x 700 V = 1400 eV

in Joules: Use

W = Vq = (700 V)( 2 x 1.60 x 10­19 C ) = 2.24 x 10­16 J

Yep ­ another proportion 31.5 eV 1 electron volt = x 1.60 x 10­19 J

solve for x x = 5.04 x 10­18 J