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

Transcription

1 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.

2 30-1 Blackbody Radiation and Planck s Hypothesis of Quantized Energy 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). An example of how to construct a blackbody: Figure 30-1 An Ideal Blackbody

3 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.

4 Figure 30-2 Blackbody Radiation at different temperature. Note the radiation peak shifts as temperature increases!

5 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. For blackbody, the peak shift is: Wien s Displacement Law f peak = ( s 1 K 1 ) T 30 1 SI unit: Hz = s -1

6 Exercise 30-1 Find the surface temperature of a blackbody, given that its radiation peak occurs at a frequency of 1.17x10 15 Hz. Solution: With equation 30-1, we have f peak = ( s K ) T Hz = ( s K ) T 15 So, T = 19,900 K

7 Plank s Quantum Hypothesis 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: Quantized Energy E n = nhf, n = 0,1,2, Unit: J (Joule) Planck s Constant, h h = 6.63 x J s 30-3 Si unit: J s

8 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 J s

9 Figure 30-4 The Photon Model of Light: a beam of light consists of many individual photons, each with energy E=h f.

10 Exercise 30-2 Calculate the energy of a photon of yellow light with a frequency of 5.25 x Hz. Given the energy in both Joule and electron volt. Solution: 1 ev = 1.60x10-19 J E = hf = ( J. s)( s ) = J 1eV = = J J( ) ev

11 Active Example 30-1 Find the number of Photons Assume 4.00x10-11 W/m 2 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?

12 Solution: 1. Calculate the area of the pupil of the eye: 2.83 x 10-5 m 2 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 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

13 The 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, W 0 (is constant for a specific material). Figure 30-5 The Photoelectric Effect

14 If an electron is given an energy E by the light and E is greater than W 0, extra energy goes into kinetic energy of the ejected electron. Considering a photon, the maximum kinetic energy K max that a photoelectron can have is K max = E W 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 f 0 ; If the frequency of light is less than f 0, no electron ejected. 2) If the light frequency is greater than the cutoff frequency f 0, increasing the light intensity increases the number of electrons ejected. 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.

15 For the first question, since E = hf 0 = W 0, the cutoff frequency is defined as: Cutoff Frequency f 0, W0 f0 = 30 h 6 SI unit: Hz = s -1 For the second question, since E = hf, the kinetic energy of a photoelectron is K max = hf W K max depends linearly on the frequency, but is independent of the intensity.

16 Figure 30-6 The Kinetic Energy of Photoelectrons and frequency and different metals.

17 Exercise 30-3 The work function for a gold surface is 4.58 ev. Find the cut off frequency f 0, for the gold metal. Solution: 19 W0 (4.58eV )( J / ev ) f0 = = = h J s 15 Hz

18 Example 30-3 White Light on Sodium A beam of white light containing frequencies between 4.00x10 14 Hz and 7.90x10 14 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.

19 Solution Part (a) Cutoff frequency is 19 W0 (2.28eV )( J / ev ) f0 = = = h J s 14 Hz So, the ejected frequency range is 5.50X10 14 Hz to 7.90x10 14 Hz Part (b) With f= 7.90x10 14 Hz (maximum frequency), K max = hf W 0 = ( J s)( Hz) (2.28eV )( J / ev ) = J

20

21

22 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 E 1 v c 2 2 = m 0 c Otherwise its energy and momentum will be infinite, see Equation 29-7, Rest Mass of a Photon M 0 =

23 2) Photons have a finite momentum even though they have no mass. From Equation 29-5, we have p v c 2 1 = m 2 0 v Dividing Equation 30-8 by Equation 30-10,we have p v =, with v = c 2 E c E so p =, with E = c Momentum of a Photon (SI Unit: kg m/s) p = hf c = h, λ hf

24 Exercise 30-4 Calculate the momentum of a photon of yellow light with a frequency of 5.25x10 14 Hz. Solution p = hf c = 34 ( J. s)( = m / s kg m / s 14 Hz) SI unit!

25