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Quantization of Light • Blackbody Radiation • Photoelectric Effect • X-rays • Compton Effect Introduction We begin our discussion of the quantum theory by detailing some of the seminal ideas and experiments that led to its development. The first rung in the ladder of the development of the quantum theory is the discovery that electromagnetic radiation is quantized, i.e., consists of discrete packets called photons. The first clue to the quantum nature of electromagnetic (EM) radiation was obtained from an analysis of blackbody radiation. Blackbody Radiation1 At the turn of the 20th century, an explanation was being sought for the spectral distribution of the radiant energy emitted by a heated body. The spectral distribution refers to the relative intensity of the radiation emitted within each wavelength (or frequency) interval. (See figure below.) We know from experience that when a body is heated (e.g., a burner on an electric stove), it will eventually begin to glow a dull red (at a temperature of approximately 1100 K). As the temperature is increased, the visible light emitted shifts toward the blue end of the visible spectrum. All bodies regardless of material, have the same general spectral distribution properties. A blackbody is an idealized object that absorbs (and subsequently re-emits) all the radiation that falls on it. All blackbodies in equilibrium with their surroundings at the same temperature will have the same spectral distribution curves. [Demonstration: Vertical filament lamp, variac, diffraction grating.] Radiation absorbed by a blackbody can be approximated by making a small hole in an otherwise sealed enclosure, such as a hollow block. When radiation enters the hole, very little of the original radiation will emerge because of the small size of the hole. The radiation inside the cavity will be multiply reflected inside the cavity until it is all absorbed. Thus, the hole functions as an absorbing blackbody. Radiation emerging from the hole will likely be radiation emitted by the walls of the cavity with a spectral distribution characteristic of the temperature of the walls. If the cavity is heated to a high

1 See Anderson, Modern Physics & Quantum Mechanics

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enough temperature, the hole will glow because of the visible radiation being emitted. Note that the wavelength at which the peak occurs shifts to higher values as the temperature decreases. An empirical relationship called Wien’s Displacement Law describes this shift:

λpT = constant, (5.1)

where T is the temperature in kelvins. In order to explain blackbody radiation spectra, Rayleigh and Jeans regarded a blackbody as being composed of elementary oscillators that emit radiation of frequency equal to the oscillation frequency. All frequency modes are excited at all temperatures. Each oscillator has an energy kT according to the classical equipartition theorem. Based upon this model, the following expression was obtained:

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8( , ) ,kTT πρ λλ

= (5.2)

where ( , )Tρ λ is the energy per unit volume per wavelength interval at equilibrium temperature T. 2 Eq. (5.2) is the Rayleigh-Jeans law. While the R-J law gave a good description of the blackbody spectrum at large wavelengths, at small wavelengths ( 0λ → ), the energy density ρ →∞ . This calamitous result was known as the “ultraviolet catastrophe.” The shortcomings of the R-J theory stimulated Max Planck3 to take a fresh look at blackbody radiation. Using statistical thermodynamics, Planck postulated that the total energy of the oscillators did not vary continuously. He divided the total energy into a finite number (integer) of “elements” of size ε . In addition, for his final expression for the energy density to fit the data, the energy element ε had to be proportional to the frequency of the radiation: ε = hf , (5.3) where h is a constant now called Planck’s constant. Planck’s final expression was

( ) 15

8( , ) 1 ,hc kThcT e λπρ λλ

−= − (5.4)

which he used to fit the experimental blackbody data. He obtained excellent agreement using h as a fitting parameter, with his best value for h being quite close to the presently accepted value h = 6.626×10-34 J. s. Einstein later showed Planck’s result could only be derived if the energy of each oscillator is quantized in amounts of hf. Thus, the energy of each oscillator could only assume values that were integer multiples of hf: εn = nhf , n = 1,2,3,... (5.5) [Mention cosmic microwave background (T ≈ 2.7 K). Departures from the blackbody spectrum give information about the structure of the early universe.] Example - Show that Planck’s Law reduces to the R-J law for large λ . 2 Note: ( , ) ( / 4) ( , )I c Tλ ρ λ= , where I is the radiated intensity. 3 See: S. Brush, AJP 70, 119 (2002).

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Planck’s law: ( ) 15

8( , ) 1 .hc kThcT e λπρ λλ

−= −

For largeλ , the exponent of e is small. Now, 2 3

2! 3!1 ....x x xe x= + + + + For small x, i.e., x << 1, ex= 1 + x. Thus, 1hc kTe hc kTλ λ≈ + .

Hence, 5 4

8 8( , ) .hc kT kTThc

π λ πρ λλ λ

≈ × =

Photoelectric Effect Planck himself did not fully realize the implications of his quantum hypothesis. However, Einstein did, and used it to explain, in an exceedingly beautiful way, the theretofore puzzling phenomenon called the photoelectric effect. The photoelectric effect is the emission of electrons from a material whose surface has been irradiated by electromagnetic radiation. (In the most common cases, the material is a metal.) It was found that several aspects of the effect were inconsistent with the classical theory of electromagnetic waves. This represented another significant failure of classical theory. The salient characteristics of the photoelectric effect are given below: (1) Electrons are ejected immediately from the metal regardless of the intensity of the radiation. (2) The maximum kinetic energy (KE) of the ejected electrons is independent of the intensity. (3) The maximum KE of the ejected electrons increases with frequency. (4) Below a certain frequency, no electrons are ejected, regardless of intensity. (5) The number of electrons ejected increased with intensity of the radiation. Only item (5) could be explained satisfactorily by classical theory. Einstein departed from classical theory by extending Planck’s quantum hypothesis to the electromagnetic radiation itself. In other words, the radiation (light) itself is quantized in discrete packets called photons, which behave like particles with energy and momentum. Each photon has an energy E = hf. A photon can collide with an electron near the surface of the metal. The photon gives up all its energy to the electron and ceases to exist. The electron, now endowed with extra kinetic energy, can overcome the energy barrier holding it inside the metal and escape from the metal. This interaction is described very simply by the conservation of energy: max ,hf Kφ= + (5.6) where φ is the work function of the metal, which is the minimum energy needed to eject an electron, and Kmax is the maximum kinetic energy of the ejected electron. Einstein’s quantum theory easily explained all the facets of the photoelectric effect. For example, the threshold frequency fo is that for which Kmax = 0, i.e., 0 .hf φ= If f < fo, the photon will not have enough energy to remove an electron. Einstein’s theory was confirmed by an experiment conducted by Millikan (published in 1916). A schematic of the experimental apparatus is shown below. With a voltage applied as shown, electrons tend to be repelled from the plate. However, if they have sufficient KE, they can overcome the retarding potential and reach the plate. For a large enough retarding voltage, no

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electron will be able to reach the plate since all their KE will be converted to electrical PE before reaching the plate. The current drops to zero at the stopping potential or stopping voltage Vs, given by max .seV K= (5.7) Thus, ,seV hf φ= − (5.8) or,

.shV fe e

φ⎛ ⎞= −⎜ ⎟⎝ ⎠ (5.9)

By measuring the slope (h/e) Millikan confirmed that h had the same valued as found by Planck. Einstein was awarded the Nobel Prize in 1921 for his explanation of the photoelectric effect. [Show Millikan data, pp. 78 & 79, Rohlf.]

Example: What is the maximum wavelength of photons capable of ejecting photoelectrons from Li at u = 0.95c? [φ for Li = 2.13 eV.]

K = (γ −1)mc2 =1

1− (0.95)2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

0.511 MeV( ) = 1.13 MeV.

The maximum wavelength (minimum energy) required occurs when K = Kmax . (Thus, the energy

expended by the electron to escape is equal to the minimum value, i.e., the work function.)hcλmax

= φ + Kmax = 1.13 MeV. (φ is negligible.)

Now, hc = 1240 eV ⋅ nm.

Hence, λmax =1240 eV ⋅ nm1.13×106 eV

= 1.1×10−3 nm.

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X-rays and the Compton Effect4 Another mystery in the early days of modern physics was the fact that a monochromatic (single-wavelength) beam of x-rays that was scattered by a sample of material always contained a longer wavelength. Arthur Compton did careful experiments on x-ray scattering and showed that the appearance of the longer wavelength component could be explained by assuming the x-rays were quantized and that the x-ray photons interacted with electrons in the atoms of the material by “colliding” with them. Using conservation of momentum and energy, Compton obtained an expression which described his experimental results beautifully. Before we analyze the Compton Effect, we shall discuss some characteristics of x-rays. X-rays X rays are at the higher frequency (lower wavelength) electromagnetic spectrum. Frequencies of x-rays are more than 1000 times those of visible light. Hence, X-ray photons are more than 1000 times more energetic than visible photons. [Ex-ray ~ 1-10 keV; Evisible ~ l eV.] When electrons with kinetic energies in the keV range are used to bombard a heavy metal, the electrons are decelerated very quickly and they radiate electromagnetic waves. The resulting spectrum of wavelengths is continuous and is called bremsstrahlung (“braking radiation”). Bremsstrahlung is characterized by a sharp cut-off at the low-wavelength end. This sharp cutoff is due to the fact that the maximum energy that an x-ray photon can have is all the kinetic energy of the bombarding electron. If the electrons are accelerated by a voltage V, then we have

.cutoffcutoff

hceV hf

λ= = (5.10)

Thus,

1240 eV nm 1240 nm. (in eV)cutoff

hceV V V

λ ⋅= = = (5.11)

Note that the ratio e/h can be ascertained from the slope of a fcutoff versus V plot.5 Superimposed on the continuous spectrum are a few sharp, discrete peaks, which are characteristic of the target material. We shall see that these peaks are due to distinct energy-level transitions within atoms of the target material. X-ray Diffraction The wavelength of x-rays are comparable to typical spacings between atoms in a crystalline material. Hence, crystalline materials can be used as a natural diffraction grating. When monochromatic x-rays are scattered by a crystal, an analysis of the diffraction pattern gives valuable information about the orientation of planes of atoms in the crystal and symmetries of those planes.6 Further, the distance between adjacent planes can be obtained. Show picture of Laue diffraction of ZnCoTe crystal] 4 May wish to cover Compton scattering first, so students can complete HW assignment. 5 Bremsstrahlung is the inverse of the photoelectric effect: an electron transfers energy to an atom resulting in a photon being emitted. 6 If x-rays directed along direction of symmetry, the diffraction pattern will have that symmetry.

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Consider x-rays incident on two adjacent atomic planes. (See figure below.) Bright spots will be seen in the diffraction pattern if there is constructive interference between the diffracted beams. For each atomic plane, the incident angle is equal to the angle of diffraction (this is why x-ray diffraction is sometimes called Bragg reflection after W.L. and W.H. Bragg, who did the pioneering work in x-ray diffraction). The condition for constructive interference between the two beams is: path difference = ,nλ i.e., 2 sin , (Bragg equation)d nθ λ= (5.12) where n is the order of the diffraction (usually n = 1), θ is the angle of diffraction, d is the distance between adjacent planes, and λ is the wavelength of the x-ray beam. Thus, if λ is known, d can be found. (Conversely, if d is known, λ can be found.) The intensity of the diffraction spots depends on the arrangement of the atoms and the electron density. An analysis of intensity vs. θ gives important structural information about the crystal. This is the field of x-ray crystallography. Compton Scattering7 As mentioned previously, Compton scattering involves the scattering of x-rays by a sample of material in which the scattered beam contains a longer-wavelength component. Compton studied this phenomenon carefully using carbon. His analysis is based on the conservation of relativistic momentum and energy in collision between an x-ray photon and a “free” electron at rest in the “laboratory” frame. The electron is really bound to an atom, but the energy of the x-ray photon is so much greater than the kinetic energy of the valence electron that the electron is effectively at rest. Based upon the photon picture, the scattered photon should have lower energy than the original photon because the original photon gives up energy to the scattered electron. Since the energy of the scattered photon is lower, it must have a smaller frequency and longer wavelength than the original photon. Let us do this more quantitatively.

7 Picture, p.73 Krane

, i

i

ihc

E pλ

= ,

f

f fhc

E pλ

=

Ee, pe

θ

φ

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Conservation of momentum gives: cos cos .i f ep p pθ φ= + (5.13) 0 sin sin .f ep pθ φ= − (5.14) Eq. (5.13) gives cos cos ,e i fp p pφ θ= − (5.15) and Eq. (5.14) gives sin sin .e fp pφ θ= (5.16) Squaring and adding Eqs. (5.15) and (5.16) gives 2 2 22 cos .e i i f fp p p p pθ= − + (5.17) Multiply Eq. (5.17) by c2 to obtain 2 2 2 2 2 2 22 cos .e i i f fp c p c p p c p cθ= − + (5.18) Now, from relativistic energy conservation 2 , i e f eE m c E E+ = + (5.19) or, 2 . e i e fE E m c E= + − (5.20)

Squaring Eq. (5.20) gives ( )22 2 . e i e fE E m c E= + − (5.21) But 2 2 2 2 2.e e eE p c m c= + (5.22) Using Eqs. (5.18), (5.21), and (5.22), we find that ( )22 2 2 2 2 2 2 42 cos . i e f i i f f eE m c E p c p p c p c m cθ+ − = − + + (5.23) After some algebra, one finds that8

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1 1 1(1 cos ),

f i eE E m cθ− = − (5.24)

or, using /E hc λ= , we obtain

( )1 cos .f ie

hm c

λ λ θ− = − (5.25)

h/mec is called the Compton wavelength of the electron.9 Note that the right hand sides of Eqs. (5.24) and (5.25) are always positive, which means that Ef is always less than Ei and 8 pi

2c2 = Ei2, pf

2c2 = Ef2, pipfc2 = EiEf

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f iλ λ> always. Note also that the greatest change in λ occurs when cos 1,θ = − i.e., the photon is backscattered ( 180θ = ° ). In addition, note that the kinetic energy gained by the electron in the scattering process is just given by Ke = Ei – Ef.10 Finally, note that in the Compton spectra, there is always a spectral component at .iλ λ= This occurs because in some collisions of a photon with an atom, the entire atom recoils, rather than a single electron being knocked out. In this case, the me in the Compton wavelength becomes matom, which is 104 times as large as me. Thus, 0,f iλ λ− ≈ i.e., .f iλ λ≈ Example: (p.72, Krane) X-rays of wavelength 0.2400 nm are Compton scattered and the scattered beam is observed at an angle of 60.0° relative to the incident beam. Find: (a) the wavelength of the scattered x-rays, (b) the energy of the scattered x-ray photons, (c) the kinetic energy of the scattered electrons, and (d) the direction of travel of the scattered electrons. Solution: (a) fλ can be found immediately from Equation (5.25):

( )1 cos

0.2400 nm (0.00243 nm)(1 cos 60 )0.2412 nm.

f ie

hm c

λ λ θ− = −

= + − °=

(b) The energy Ef can be found directly from fλ :

1240 eV nm 5141 eV.0.2412 nmf

f

hcEλ

⋅= = =

(c) Using the conservation of energy, we have 2 2 .i e f e f e eE m c E E E m c K+ = + = + + Thus, .e i fK E E= − The initial photon energy Ei is / ihc λ = 5167 eV, and so

Ke = 5167 eV - 5141 eV = 26 eV. (d) Dividing Eq. (5.16) by Eq. (5.15) yields

sin

tan .cos

f

i f

pp p

θφ

θ=

−

9 h/mec = 2.43× 10

-3 nm

10 Ei + mec2 = Ef + mec2

+Ke

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Multiplying top and bottom by c, and recalling that E = pc for photons, we have

sintan

cos

(5141 eV)(sin 60 )(5167 eV) (5141 eV)(cos 60 )1.715.

f

i f

EE E

θφ

θ=

−

°=− °

=

Thus, 59.7 .φ = °