Chapter 30 Quantum Physics

Chapter Outline

30-1 Blackbody Radiation and Planck’s Hypothesis of Quantized

Energy

30-2 Photons and the Photoelectric Effect

30-3 The Mass and Momentum of a Photon

30-5 The de Broglie Hypothesis and Wave-Particle Duality

30-6 The Heisenberg Uncertainty Principle.

30-1 Blackbody Radiation and Planck’s Hypothesis of QuantizedEnergy

Blackbody: an ideal source for light (electromagnetic radiation)

absorption and emission.

1) An ideal blackbody absorbs all the light that is incident on it.

2) An ideal blackbody is also an ideal radiator (See Section 16-6: objects that effective at absorbing radiation are also effective at radiation).

Figure 30-1An Ideal Blackbody

An example of how to construct a blackbody:

Why Blackbody:

1) Objects that absorbs much of the incident light (thought not all of it) are reasonable approximations to a blackbody.

2) It makes mathematical modeling simple.

Blackbody property:

The distribution of energy (electromagnetic radiation as a function of wavelength) in blackbody radiation is independent of the material – it depends only on the temperature T.

Figure 30-2Blackbody Radiation at different

temperature.

Note the radiation peak shifts as temperature increases !

Two properties of blackbody curves in the above figures are particularly important:

1) As the temperature is increased, the area under curve increases, Thus the total energy increases.

2) The peak of the curve moves to higher frequency (shot wavelength), as the temperature is increased.

Wien’s Displacement Law

SI unit: Hz = s-1

130)1088.5( 1110 −⋅×= −− TKsf peak

For blackbody, the peak shift is:

Exercise 30-1

Find the surface temperature of a blackbody, given that its radiation peak occurs at a frequency of 1.17x1015 Hz.

Solution:

With equation 30-1, we have

TKsf peak )1088.5( 1110 −− ⋅×=

TKsHz )1088.5(0117.1 111015 −− ⋅×=×

So, T = 19,900 K

Plank’s Quantum Hypothesis

Quantized Energy

Unit: J (Joule)

2303,2,1,0, −⋅⋅⋅== nnhfEn

The German physicist Max Planck found that:

the radiation energy in a blackbody at frequency f is an integral times h and f, and is quantized as:

Planck’s Constant, h

h = 6.63 x 10-34 J⋅s 30-3

Si unit: J⋅s

30-2 Photons and the Photoelectric Effect

Einstein proposed that light comes in bundles of energy, called photons. A photon is like a “particle” (but light is also a wave).

The photons obey Planck’s law (n=1), and the energy at frequency f is

Energy of a Photon of Frequency f

E = h f 30-4

SI unit: J (Joule)

h = 6.63 x 10-34 J⋅s

Figure 30-4The Photon Model of Light: a beam of light consists of many

individual photons, each with energy E=h f.

Exercise 30-2

Calculate the energy of a photon of yellow light with a frequency of 5.25 x 1014 Hz. Given the energy in both Joule and electron volt.

Solution:

34 14 1 19

1919

(6.63 10 . )(5.25 10 ) 3.48 1013.48 10 ( ) 2.18

1.60 10

E hfJ s s J

eVJ eVJ

− − −

−−

=

= × × = ×

= × =×

1 eV = 1.60x10-19 J

Active Example 30-1 Find the number of Photons

Assume 4.00x10-11 W/m2 of 505-nm light enter the eye. If the light of this intensity and wavelength enters the eye through a pupil that is 6 mm in diameter. How many photons enter the eye per second?

Solution:

1. Calculate the area of the pupil of the eye:

2.83 x 10-5 m2

2. Multiply the intensity by area of the pupil to find the energy per second:

1.13x10-15 J/s

3. Calculate the energy of a photon:

hf = 3.94 x 10-19 J, (with f λ = c )

4. To find the number of photons per second: The pupil energy per second is divided by the one-photon energy,

2870 / s

The Photoelectric Effect

Figure 30-5The Photoelectric Effect

In the experiment (see figure), a Beam of light (photons) hit the surface of a metal and thus eject electrons (photoelectrons).

The minimum amount of energy necessary to eject an electron from a particular metal is called Work Function, W0 (is constant for a specific material).

If an electron is given an energy E by the light and E is greater than W0, extra energy goes into kinetic energy of the ejected electron.

Considering a photon, the maximum kinetic energy Kmax that a photoelectron can have is

Kmax = E – W0 30-5

The experiment shows the following behaviors:

1) To eject electrons, the incident light beam must have a frequency greater than a certain value, called Cutoff Frequency f0; If the frequency of light is less than f0, no electron ejected.

2) If the light frequency is greater than the cutoff frequency f0, increasing the light intensity increases the number of electronsejected.

However, the maximum kinetic energy of a photon dose not increase with the light intensity. The kinetic energy only depends on the frequency of the light.

For the first question, since E = hf0 = W0, the cutoff frequency is defined as:

Cutoff Frequency f0,

SI unit: Hz = s-1

63000 −=

hWf

For the second question, since E = hf, the kinetic energy of a photoelectron is

Kmax = hf – W0 30-7

Kmax depends linearly on the frequency, but is independent of the intensity.

Figure 30-6The Kinetic Energy of Photoelectrons and frequency and

different metals.

Exercise 30-3

The work function for a gold surface is 4.58 eV. Find the cut off frequency f0, for the gold metal.

Solution:

HzsJ

eVJeVh

Wf 1534

190

0 1011.11063.6

)/1060.1)(58.4(×=

⋅××

== −

−

Example 30-3 White Light on Sodium

A beam of white light containing frequencies between 4.00x1014 Hz and 7.90x1014 Hz is incident on a sodium surface, which has a work function 2.28 eV. (a) What is the range of frequencies in this beam of light for which electrons are ejected from the sodium surface? (b) Find the maximum kinetic energy of the “photoelectrons” that are ejected from this surface.

Solution

Part (a) Cutoff frequency is

HzsJ

eVJeVh

Wf 1434

190

0 1050.51063.6

)/1060.1)(28.2(×=

⋅××

== −

−

So, the ejected frequency range is 5.50X1014 Hz to 7.90x1014 Hz

Part (b)With f= 7.90x1014 Hz (maximum frequency),

JeVJeVHzsJ

WhfK

19

1914340max

1059.1)/1060.1)(28.2()109.7)(1063.6(

−

−−

×=

×−×⋅×=

−=

30-3 The Mass and Momentum of a Photon

Photon’s properties:

1) Since photons travel at the speed of light. The rest mass of a photon must be zero

Otherwise its energy and momentum will be infinite, see Equation 29-7, 29-5.

Rest Mass of a Photon

M0 = 0 30-9

8301 202

2

−=− cmcvE

2) Photons have a finite momentum even though they have no mass.

From Equation 29-5, we have

10301 02

2

−=− vmcvp

Dividing Equation 30-8 by Equation 30-10,we have

cvwithcv

Ep

== ,2

hfEwithcEpso == ,

Momentum of a Photon (SI Unit: kg ·m/s)

1130, −==λh

chfp

Exercise 30-4

Calculate the momentum of a photon of yellow light with a frequency of 5.25x1014 Hz.

Solution

smkgsm

HzsJc

hfp

/1016.1/1000.3

)1025.5)(.1063.6(

27

8

1434

⋅×=×

××==

−

−

SI unit !