/
The photoelectric effect The photoelectric effect

The photoelectric effect - PowerPoint Presentation

alida-meadow
alida-meadow . @alida-meadow
Follow
648 views
Uploaded On 2017-04-16

The photoelectric effect - PPT Presentation

Contents Einsteins proposed experiment Solving photoelectric problems Example 1 Example 2 Whiteboard Photon vs wave theory Waves Photons Color Brightness Energy per photon changes ID: 537953

work energy kinetic function energy work function kinetic photon stopping potential light metal electrons wavelength photons strikes 602 ejected

Share:

Link:

Embed:

Download Presentation from below link

Download Presentation The PPT/PDF document "The photoelectric effect" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.


Presentation Transcript

Slide1

The photoelectric effectContents:Einstein’s proposed experimentSolving photoelectric problemsExample 1 | Example 2WhiteboardPhoton vs wave theorySlide2

Waves Photons

Color

Brightness

Energy per photon changes

E = hf

Radio waves vs X-rays

Wavelength Changes

Amplitude Changes

# of photons changes

many = bright

few = dim

CCD Devices, High speed film

big

 = red

small  = blue

small = dim

big = bright

LightSlide3

Photo Electric EffectHow it worksEinstein’s idea

+

-

V

Reverse the voltage

Photon Energy = Work + Kinetic Energy

hf

=

 +

E

max

 - Work function

(Depends on material)

hf

– photon energy

E

max

– max KE of photoelectrons

(=

eV

s) – where V

s is the stopping potentialSlide4

Millikan’s setupSlide5

Photon Energy = Work + Kinetic Energyhf =  + Emax Example 1: A certain metal has a work function of 3.25 eV. When light of an unknown wavelength strikes it, the electrons have a stopping potential of 7.35 V. What is the wavelength of the light?

117 nmSlide6

Photon Energy = Work + Kinetic Energyhf =  + Emax Example 2: 70.9 nm light strikes a metal with a work function of 5.10 eV. What is the maximum kinetic energy of the ejected photons in eV? What is the stopping potential?

12.4 VSlide7

Example 1: A certain metal has a work function of 3.25 eV. When light of an unknown wavelength strikes it, the electrons have a stopping potential of 7.35 V. What is the wavelength of the light?Photon energy = Work + Kinetic EnergyPhoton Energy = 3.25 eV + 7.35 eV = 10.60 eVConvert to J: (10.60 eV)(1.602E-19 J/eV) = 1.69812E-18 JE = hf = hc/ ,  = hc/E = (6.626E-34 Js)(3.00E8 m/s)/(1.69812E-18 J

)

= 1.17059E-07 m = 117 nm

Photon Energy = Work + Kinetic Energy

hf =

 + E

max

hf =

hf

o + eV

 - Work function (Depends on material)fo

- Lowest frequency that ejectse - Electron chargeV - The uh stopping potentialSlide8

Example 2: 70.9 nm light strikes a metal with a work function of 5.10 eV. What is the maximum kinetic energy of the ejected photons in eV? What is the stopping potential?E = hf = hc/ = (6.626E-34 Js)(3.00E8 m/s)/(70.9E-9) = 2.80367E-18 JConvert to eV: (2.80367E-18 J)/(1.602E-19 J/eV) = 17.5 eVPhoton Energy = Work + Kinetic Energy17.5 eV = 5.10 + Kinetic Energy

Kinetic Energy = 12.4 eV, so 12.4 V would stop the electrons.

Photon Energy = Work + Kinetic Energy

hf =

 + E

max

hf =

hf

o

+ eV

 - Work function (Depends on material)f

o - Lowest frequency that ejectse - Electron chargeV - The uh stopping potentialSlide9

Ag (silver)

4.26

Au (gold)

5.1

Cs (cesium)

2.14

Cu (copper)

4.65

Li (lithium)

2.9

Pb (lead)

4.25

Sn (tin)

4.42

Chromium

4.6

Nickel

4.6

Metal

Work Function

Whiteboards:

Photoelectric effect

1

|

2

|

3

|

4Slide10

6.00 eVPhoton energy = Work function + Kinetic energy of electronPhoton energy = 2.15 eV + 3.85 eV = 6.00 eVPhotons of a certain energy strike a metal with a work function of 2.15 eV. The ejected electrons have a kinetic energy of 3.85 eV. (A stopping potential of 3.85 V) What is the energy of the incoming photons in eV?Slide11

147 nmE = hf = hc/,Photon energy = Work function + Kinetic energy of electronPhoton energy = 3.46 eV + 5.00 eV = 8.46 eVE = (8.46 eV)(1.602 x 10-19 J/eV) = 1.3553 x 10-18 J

E = hf = hc/,

 = hc/E = (6.626 x 10

-34

Js)(3.00 x 10

8

m/s)/(1.3553 x 10

-18

J)

 = 1.4667x 10-07 m = 147 nm

Another metal has a work function of 3.46 eV. What is the wavelength of light that ejects electrons with a stopping potential of 5.00 V? (2)Slide12

6.67 VE = hf = hc/, 1 eV = 1.602 x 10-19 JPhoton energy = Work function + Kinetic energy of electronE = hf = hc/ = (6.626 x 10-34 Js)(3.00 x 108

m/s)/(112 x 10

-9

m)

E = 1.7748 x 10

-18

J

E = (1.7748 x 10

-18

J)/(1.602 x 10-19 J/eV) = 11.079 eV

Photon energy = Work function + Kinetic energy of electron11.079 eV = 4.41 eV + eVs

11.079 eV - 4.41 eV = 6.6688 eV = eVsVs = 6.67 V

112 nm light strikes a metal with a work function of 4.41 eV. What is the stopping potential of the ejected electrons? (2)Slide13

3.70 eVE = hf = hc/, 1 eV = 1.602 x 10-19 JPhoton energy = Work function + Kinetic energy of electronE = hf = hc/ = (6.626 x 10-34 Js)(3.00 x 108

m/s)/(256 x 10

-9

m)

E = 7.7648 x 10

-19

J

E = (7.7648 x 10

-19

J)/(1.602 x 10-19 J/eV) = 4.847 eV

Photon energy = Work function + Kinetic energy of electron4.847 eV = Work function + 1.15 eV11.079 eV - 1.15 eV = 3.70 eV

256 nm light strikes a metal and the ejected electrons have a stopping potential of 1.15 V. What is the work function of the metal in eV? (2)Slide14
Slide15

Einstein’s Photon theory predicts:Photon energy = work function + Kinetic energy of electronhf =  + Ekmax E

kmax

=

hf

-

(slope, y intercept)

Photon Theory predicts:

E

kmax

rises with frequencyIntensity of light ejects

moreWave Theory predicts:

Ekmax rises with Amplitude (Intensity)Frequency should not matterSurvey says

Millikan does the experiment

1915 conclusion1930 conclusion