Chapter 21 EPE to V to Millikan.notebook

V+_

OIL

Millikan's Oil Drop Experiment

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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?

sourceof variable voltage

flatmetal plates

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spray

Chapter 21 EPE to V to Millikan.notebook

The Electronvolt (eV)

It's a small unit of energy

1 eV = 1.6 x 10-19 Joules

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

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From last night's HW

Answer in EV's: 500 eVAnswer in Joules:  8.0 x 1017 Joules

Chapter 21 EPE to V to Millikan.notebook

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+ + + + + + + + + + + + + + + + + + + + + + + +12 volts

0 volts

3 volts

6 volts

9 volts

ep+

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gains3 electronVolts

gains6 eV's

loses3 eV's

p+

gains6 eV

2 elementary charges x 10 V = 20 eV's

9.128 x 1017 J ____________x =

1.60 x 1019 J____________electron volt

 1 electron volt

x(1.60 x 1019 J) = (9.128 x 1017 J) eV1.60 x 1019 J 1.60 x 1019 J

____________ ____________

x = 570.5 eV

 500 eV

1.60 x 1019 J

=  8 x 1017 J

1400 eV

1.60 x 1019 C )

= 2.24 x 1016 J

31.5 eV

x = 5.04 x 1018 J

to find Joules: use

solve for W = VqW = 10 V x( 2 x 1.6 x1019 C)

= 3.2 x1018 Joules

9.128 x 1017 J ____________x =

1.60 x 1019 J____________electron volt

Set up a proportion this is on page of of RT

crossmultiply and solve for x

 1 x 500 =

Use again

Solve for W = Vq = ( 500 V) (1.6 x 1019 C)

in eV:  +2 x 700 V =

in Joules:                    Use

W = Vq = (700 V)( 2 x

x=

Yep  another proportion

solve for x

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