02 Quantum Physics.pptx

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Green House Effect

Plank’s Constant

Energy

of a photon

is proportional to

the frequency of the photon.

E α

f

E =

h

·

f

Plank’s Constant

In his studies of black-body radiation, Maxwell Planck discovered that electromagnetic energy is emitted or absorbed in discrete quantities.

Planck’s

Equation:

E = hf

(

Apparently, light consists of tiny bundles of energy called photons, each having a well-defined quantum of energy.

Apparently, light consists of tiny bundles of energy called

photons, each having a

Homework:

How many photons come out

of the laser every second?

1.02 x 1016 photons

Energy in Electron-volts

Photon energies are so small that the energy is better expressed in terms of the electron-volt.

One electron-volt (eV) is the energy of an electron when accelerated through a potential difference of one volt.

One electron-volt (eV) is the energy of an electron when accelerated through a potential difference of one volt.

-Energy in Electron-volts

Photon energies are so small that the energy is better expressed in terms of the electron-volt.

One electron-volt (eV) is the energy of an electron when accelerated through a potential difference of one volt.

One electron-volt (eV) is the energy of an electron when accelerated through a potential difference of one volt.

1 eV = 1.60 x 10-19 J 1 keV = 1.6 x 10-16 J

1 MeV = 1.6 x 10-13 J

-+ + + + + + + + + 12 Volts

12 Volts = 12 Joules/Coulomb 12 Volts = 12 eV/e

-U_e = 12 eV K = 12 eV

-= h·

f

h·

c

λ

Example 1: What is the energy of a photon of yellow-green light (l = 555 nm)?

First we find f from wave equation: c = fl

E = 3.58 x 10-19 J

E = 3.58 x 10-19 J _Or E_E = 2.24 eV _{= 2.24 eV}

Since 1 eV = 1.60 x 10-19 J

;

c

hc

f

E hf

l

l







34 8 -9

(6.626 x 10 J s)(3 x 10 m/s)

555 x 10 m

E





Useful Energy Conversion

Since light is often described by its wavelength in

nanometers (nm) and its energy E is given in eV, a conversion formula is useful. (1 nm = 1 x 10-9 m)

If l is in nm, the energy in eV is found from: Verify the answer

in Example 1 . . .

-19

(in Joules)

hc

; 1 eV 1.60 x 10 J

E

l





9

-19

(1 x 10 nm/m)

(in eV)

Photoelectric Effect

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-Photo Electric Effect

2.9 eV

6.5 eV

9.1 eV

3.3 eV

5.9 eV

Find the work function of the metal.

= 3.2 eV

-Find the energy of the photon.

Photo Electric Effect

= 3.2 eV

-+ + + + 6 Volts

Photo Electric Effect

Find the energy of the photon.

Not enough information!

= 3.2 eV

Photo Electric Effect

Find the energy of the photon.

Use a graph with at least five data

points to find

(with three sig figs)

:

• _{Work function} of ? material.

• _{Threshold Frequency.}

Slope of a Straight Line (Review)

The general equation for a straight line is:

y = mx + b

y = mx + b

The x-intercept x_o occurs when line crosses x axis or when y = 0.

The slope of the line is the rise over the run:

x_o x

y

The slope of a line:

Finding Planck’s Constant, h

Using the apparatus on the previous slide, we determine the stopping potential for a number of incident light frequencies, then plot a graph.

Note that the x-intercept f_o

is the threshold frequency.

f_o

Stopping potential

Frequency V

Finding h constant

Example 3: In an experiment to determine Planck’s constant, a plot of stopping potential versus frequency is made. The slope of the curve is 4.13 x 10-15 V/Hz. What is Planck’s

constant? f_o Stopping potential Frequency V y x Slope

h = e(slope) = (1.6 x 10-19C)(4.13 x 10-15 V/Hz) Experimental Planck’s h = 6.61 x 10-34 J/Hz

Experimental Planck’s h = 6.61 x 10-34 J/Hz

0

h

W

V

f

e

e

 



 



 

4.13 x 10 V/Hz

h

Slope

e

Example 4: The threshold frequency for a given surface is

1.09 x 1015 _{Hz. What is the stopping potential for incident}

light whose photon energy is 8.48 x 10-19 _J?

Photoelectric Equation:

W = (6.63 x 10-34 Js)(1.09 x 1015 Hz) =7.20 x 10-19 J

Stopping

potential: Vo = 0.800 V

A

CathodeIncident light_Anode

+

-V

c

0

E hf





W eV



0

;

eV

 

E W W



hf

-19 -19 -19

0

8.48 x 10 J 7.20 x 10 J 1.28 x 10 J

eV







-19

0 _-19

1.28 x 10 J 1.6 x 10 J

E

=

m

∙

c

2

The Gold bar has

a mass of

9 x 10

Joules

E

= 1∙(

3 x 10

)

E

=

m

∙

c

2

m = 5.97 x 1024 kg

V = 29,800 m/s

KE = 2.65 x 1033 J

E = m∙c2

2.65 x 1033 = m∙(3 x 108)2

The earth orbits the sun with a

kinetic energy of

2.95 x 10

kilograms

KE = ½ 5.97 x 1024 · 29,800 2

KE = ½ m·v2

E

= m∙

c

p

= m∙

v

m =

p

/

v

E

= ∙

p

c

v

E

= ∙

p

c

c

E

=

p

∙

c

Momentum of a Photon

h

∙

c

λ

=

p

∙

c

h

p = 17 kg·m/s K = 11 Joules p = 8 kg·m/s

K = 4 Joules p = 15 kg·m/s

K = 7 Joules

p, Vector Sum

K, Algebraic Sum

Conservation

Compton Effect

λ = 5.70 nm

λ = 6

.50 n_m V =

? ? ?

Find velocity using:

• _Momentum

6.13 x 104 m/s

Energy

Momentum

c

1.99x10-25 / 5.70x10-9

1.99x10-25 / 6.50x10-9

-E = 4.297 x 10-18

(6.63x10-34 / 5.70x10-9)2

(6.63x10-34 / 6.50x10-9)2

-3.07 x 106 m/s

p = 4.297 x 10-18

Waves and Particles

We know that light behaves as both a wave and a particle. The rest mass of a photon is zero, and its wavelength can be found from momentum.

Wavelength of a photon:

All objects, not just EM waves, have wavelengths which can be found from their momentum

de Broglie Wavelength:

hc

E

pc

l





h

p

l



h

mv

The circumference of the innermost orbit, according to this picture, is equal to one wavelength.

• The second has a circumference of two electron wavelengths.

Example 5: What is the de Broglie wavelength of a 90-eV electron? (me = 9.1 x 10-31 kg.)

-e- 90 eV

l = 0.122 nm

l = 0.122 nm K = ½ mv2

1.44 x 10-17 = ½ (9.1 x 10-31) v2

v = 5.626 x 106

6.626 x 10-34

(9.1 x 10-31)(5.626 x 106)

λ =

-19

-17

1.6 x 10 J

90 eV 1.44 x 10 J 1 eV

K  _ _ 

 

h

h

p

mv

Electron Diffraction

35.2 cm

85.6 cm

B = 723 mT

Separation = 18.25 cm

Compare the DeBroglie Wavelength of the electron based on its momentum

to the wavelength measured by interference with double slit diffraction. λ

-λ_DeBroglie = 2.079 x 10-6 _m

λ_Interference = 2.084 x 10-6 _m 0.24% E of D

F_E = qE

F_B = qvB

V =

46.1_{7 V}