Chapter 3

THE QUANTUM THEORY OF LIGHT

Chapter Outline

• 3.1 Hertz’s Experiments—Light as an Electromagnetic Wave

• 3.2 Blackbody Radiation Enter Planck The Quantum of Energy

• 3.3 The Rayleigh–Jeans Law and Planck’s Law (Optional) Rayleigh–Jeans Law -Planck’s Law

• 3.4 Light Quantization and the Photoelectric Effect

• 3.5 The Compton Effect and X-Rays X-Rays The Compton Effect

• 3.6 Particle–Wave Complementarity

• 3.7 Does Gravity Affect Light? (Optional)

3.1 HERTZ’S EXPERIMENTS—LIGHT AS AN ELECTROMAGNETIC WAVE

• Light as an Electromagnetic waves

• The Electromagnetic waves behaves like light

• would be reflected by metal mirrors,

• would be refracted by dielectrics like glass.

• would exhibit polarization and interference, and would travel outward from the wire through a vacuum with a speed of 3.0 x 108 m/s.

• These properties led to the unifying and simplifying postulates that light was also a type of Maxwell wave or electromagnetic disturbance.

3.2 BLACKBODY RADIATION

• A blackbody is defined as:

an object that absorbs all

the electromagnetic

radiation falling on it and

consequently appears black

• The opening to the cavity

inside a body is a good

approximation of a blackbody

• A ray entering the cavity can leave it only after multiple reflection

• For a sufficiently small aperture all incoming radiation will be trapped inside the cavity , the surface of the aperture appears as a black body

Blackbody Radiation

• Problem : predict the spectral energy density u(f,T) of blackbody radiation

• According to Kirchhoff, the emitted power is proportional to the power absorbed. Or

where ef is the power emitted per unit area per unit frequency by

a particular heated object, Af is the absorption power and J( f, T)

is a universal function ,f: is the frequency, and T is the absolute

temperature of the body.

A blackbody absorbs all the incident power that is Af =1 for all

frequencies and so Kirchhoff’s law becomes

Equation 3.2 shows that the power emitted per unit area per

unit frequency by a blackbody depends only on temperature

and frequency. So blackbody is “ideal radiator”.

Stefan showed experimentally that the total power per unit area

emitted at all frequencies by a hot solid, e total, was proportional to the

fourth power of its absolute temperature. Therefore, Stefan’s law may

be written as

Stefan Law

• where e total :is the power per unit area emitted at the surface of the blackbody at all frequencies,

• ef :is the power per unit area per unit frequency emitted by the blackbody,

• T : is the absolute temperature of the body,

• : is the Stefan–Boltzmann constant, given by 5.67 x 10-8 Wm-2 .K-4.

• For any body

• a: coefficient less than 1

EXAMPLE 3.1

• Estimate the surface temperature of the Sun from the following informations. The Sun’s radius is given by Rs = 7.0 x108 m. The average Earth–Sun distance is R = 1.5 x 1011 m. The power per unit area (at all frequencies) from the Sun is measured at the Earth to be 1400 W/m2. Assume that the Sun is a blackbody

• Solution: For a black body, we take a =1, so Equation 3.4 gives

where the notation e total(Rs) stands for the total power per unit area at the

surface of the Sun. Because the problem gives the total power per unit

area at the Earth, e total(R), we need the connection between e total(R) and e total(Rs)

or

Using Equation 3.5, we have

or

• As can be seen in Figure 3.3, the wavelength marking the maximum power emission of a blackbody, max, shifts toward shorter wavelengths as the blackbody gets hotter. i.e

Wien put his formula

where max is the wavelength in meters

and T is the absolute temperature of the glowing object

Wein Formula

• From Wein formula we can find the temperature of sun’s surface.

• Assume the peak sensitivity of human eye (about 500nm for blue-green light) is max,

Then using Wein formula , we have

With good agreement as in previous example.

Spectral energy density u(f,T) of a blackbody

• We defined J( f, T), the power radiated per unit area per unit frequency by the blackbody.

• Consider the spectral energy density , u( f, T), which is “ energy per unit volume per unit frequency of the radiation within the blackbody cavity”

• Both J( f, T) and u( f, T) are proportional . i.e

J( f, T) α u( f, T).

• It is found that the constant of proportionality is c/4 i.e

• Wein proposed his exponential law for u(f,T) (1893) as

Where A and β are constants

This formula had been confirmed by experiment of Fridchich Paschen

Plank Radiation Formula(1990)

• Plank found the general formula for u(f,T) as

where h is Planck’s constant =6.626 x 10-34 J . s, and kB is Boltzmann’s constant =1.380 x 10-23 J/K. We can see that Equation 3.9 has the correct

limiting behavior at high and low frequencies with the help of a few

approximations.

•At high frequencies, where hf/kBT >>1, we omit 1 in the denominator

Which is Wein’s exponential law

At low frequencies, where hf / kBT <<1

• Then we have

Which show that the spectral energy density is proportional to T in the low-frequency u( f, T) α T

Derivation of the Stefan –Boltzman law from the blank formula

• Pank formula is given by

• Stefan-boltzman :

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Which is the Stefan-Boltzman constant

The Quantum of Energy

• The energy spectrum of a classical oscillator is continuous

A: amplitude, k spring constant

• Plank assumed that the walls of a glowing cavity were composed of billions of vibrating submicroscopic electric charges, called “resonators” all vibrating at different frequencies

2

2

1kAE

• . Hence, each oscillator should emit radiation with a frequency corresponding to its vibration frequency.

• Plank assume that the energy spectrum of a resonator is discrete:

En =hf(n) ,n=1, 2, 3, ….

h: plank’s constant =6.626 x 10-34 J .s

he also showed that emission of radiation of frequency f occurred when a resonator dropped to the next lowest

energy state. Thus the resonator can change its energy

only by the difference E according to

That is, it cannot lose just any amount of its total energy,

but only a finite amount, hf, the so-called quantum of

energy.

Figure 3.9

Allowed energy levels

according to Planck’s

original hypothesis for an

oscillator with frequency

f. Allowed transitions are indicated by the double-headed arrows.

3.4 PHOTOELECTRIC EFFECT -1905

• When light is incident on a clean metalic surface, electrons are emitted from the metal

with a range of velocities and that the maximum kinetic energy of photoelectrons, Kmax. does not depend on the intensity of the incident light.

• The maximum kinetic energy of, Kmax.

is proportional to the frequency of the incident light.

Figure 3.14 (a)Photoelectric effect apparatus.

(b) A plot of photocurrent versus applied voltage. The graph shows that

Kmax is independent of light intensity I for light of fixed frequency

(c) A graph showing the dependence of Kmax on light frequency.

• Threshold frequency f0 : the frequency of light which has just enough energy to knock an electron to be released with zero kinetic energy, K=0.

• Figure (c) shows that no photoelectrons are emitted below f0 .

• The photoelectric effect takes place only if the frequency of the incident light exceeds a threshold value f0

Measurement of Kmax.

• In figure (a), a retarding voltage is applied to measure the maximum kinetic energy Kmax.

• the retarding voltage is gradually increasing until the most energetic electrons are stopped and the photo current becomes zero.

• At this point

where Vs : stopping volt

• In figure (b), the expected photocurrent by increasing the intensity , which could be explained classically, but the result showed that Kmax. doesn’t depend on the intensity of light.(expected result).

• The other unexpected result is the linear dependence of Kmax. on light frequency shown in figure (c)

:Work function

• The work function of the metal is define as the minimum energy with which an electron is bound in the metal

Einstein’s theory of the photoelectric effect

• The energy of light is not distributed evenly over the classical wave front, but is concentrated in discrete regions (or in “bundles”), called quanta, each containing energy, hf. i.e a light quantum was so localized that it gave all its energy, hf, directly to a single electron in the metal.

• That is

hfKeVs max

Where : work function of the metal

• When electron is released with K=0, the above relation becomes

hffhf

0min0

That is light with f<f0 has insufficient energy to free an electron

and the photocurrent =0 for f<f0

Table 3.1 Work Functions of Selected Metals

Potasium ,20V to eject electron

EXAMPLE 3.5 The Photoelectric Effect in Zinc

• Philip Lenard determined that photoelectrons released from Zinc by ultraviolet light were stopped by a voltage of 4.3 V. Find Kmax. and vmax. for these electrons

• Solution

To find vmax., we set the work done by the electric field equal to the change

in the electron’s kinetic energy, to obtain

X-RAYS • X-rays were discovered in 1895 by the German physicist

Wilhelm Roentgen

• X-rays are produced by bombarding a metal target (copper, tungsten, and molybdenum are common) with energetic electrons having energies of 50 to 100 keV

• X-rays extremely penetrating type of Radiation (electromagnetic)

• The minimum continuous x-ray wavelength, λmin, is found to be independent of target composition and depends only on the tube voltage, V.

X-rays are produced by bombarding a metal with energetic

electrons having energies of 50 to 100 keV

• all of the incident electron’s kinetic energy is converted to electromagnetic energy in the form of a single x-rays photon. For this case we have

or

where V is the x-ray tube voltage

THE COMPTON EFFECT

• Compton Effect is the scattering of x-rays from free electrons:

• Classical theory:

• Classical theory predicted that incident radiation of frequency f0 should accelerate an electron in the direction of propagation of the incident radiation, and that it should cause forced oscillations of the electron and reradiate at frequency f’ , where f’≤ f0 .

• Also, the scattered radiation should depend on the time and the intensity of the incident radiation.

Quantum model:

• Compton showed experimentally that the wavelength shift of x-rays scattered at a given angle is absolutely independent of the intensity of radiation and the length of exposure, and depends only on the scattering angle.

• The lower scattered frequency f’ , because the incident photon gives some of its energy to the recoiling electron.

• Photon in addition to carrying energy hf, cary momentum hf/c and scatter like particles.

• Brag Equation:

• Scattering of x-rays from parallel crystal planes:

• Consider two successive planes of atoms as shown in Figure 3.20.

• For single plane, A, will scatter constructively if the angle of incidence Ѳi equals the angle of reflection Ѳr

• Atoms in successive planes (A and B) will scatter constructively at an angle if the path length difference for rays (1) and (2) is a whole number of wavelengths, n.

Figure 3.20 Bragg scattering of x-rays from successive planes of atoms.

Constructive interference occurs for ABC equal to an integral number of

wavelengths

From the diagram, constructive interference will occur when

And because AB = BC = d sinѲ , it follows that

This is Bragg equation

before

after

The Compton Effect

before

after (َb) Quantum model

Compton scattering – Experimental setup

• Rotating crystal spectrometer

• The wavelength was measured with a rotating crystal spectrometer

using graphite (carbon) as the target. The intensity was determined by

a movable ionization chamber that generated a current proportional to

the x-ray intensity.

•It is found that

Which is called Compton scattering formula, which shows

the increase in wavelength in photon's wavelength when it is scattered through an angle Ѳ, it depends only on Ѳ. • h/mc is called the Compton wavelength = 0.0243 Å =0.00243nm

Conservation of energy

where E is the energy of the incident photon, E’ is the energy of the scattered photon, mec

2 is the rest energy of the electron, and Ee is the total relativistic energy of the electron after the collision.

conservation of momentum

where p is the momentum of the incident photon, p’ is the momentum of the scattered photon, and pe is the recoil momentum of the electron

x-direction

y-direction

• The relation between Ee and Pe is given

•

Equations 3.29 and 3.30 may be solved simultaneously to eliminate , the

electron scattering angle, to give the following expression for pe2

A photon has a mass equals to zero, we have

If the relations E = hf and p = hf/c are substituted into Equations 3.28 and 3.31, these become respectively

and

we eliminate Ee and pe by substituting Equations 3.33 and 3.34 into

the expression for the electron’s relativistic energy,

After some algebra , one obtains Compton’s result for the increase in a

photon’s wavelength when it is scattered through an angle θ.

DOES GRAVITY AFFECT LIGHT? Photon has zero mass, but its effective inertial mass, mi

,may reasonably

be taken to be the mass equivalent of the photon energy, E, or

The same result is obtained if we divide the photon momentum by the photon

speed c:

• The effective inertial mass determines how the photon responds to an

applied force such as that exerted on it during a collision with an electron

• The gravitational mass of an object determines the force of gravitational attraction of that object to another, such as the Earth

• The inertial mass of all material bodies

is equal to the gravitational mass

• Assume that the photon, like other

objects, also has a gravitational mass

equal to its inertial mass. In this case a

photon falling from a height H should

increase in energy by mgH and

therefore increase in frequency,

although its speed cannot increase and

remains at c.

• Experiments, show this increase in

frequency and confirm that the photon

indeed has an effective gravitational

mass of hf/c2

Schematic diagram of the

falling-photon experiment.

Relation between f and f‘ may be derived by applying conservation

of energy to the photon at points A and B

Because the photon’s kinetic energy is E = pc = hf and its potential

energy is mgH, where m = hf/c 2, we have

or

The fractional change in frequency, f/f, is given by

For H=50m

Gravitational redshift from a high-density star

• The increase in frequency for a photon falling inward suggests a decrease in frequency for a photon that escapes outward to infinity against the gravitational pull of a star

• This effect, known as “gravitational red shift,” would cause an emitted photon to be shifted in frequency toward the red end of the spectrum

• The redshift may be derived once again by conserving photon energy

From the conservation of energy

Using hf for the photon’s

kinetic energy and GMm/R for its potential energy, with

m equal to hf/c 2 and Rs equal to the star’s radius,

yields

or

EXAMPLE: The Gravitational Redshift for a White Dwarf

• White dwarf stars are extremely massive, compact stars that have a mass on the order of the Sun’s mass concentrated in a volume similar to that of the Earth. Calculate the gravitational redshift for 300-nm light emitted from such a star.

• Solution

• Black Holes

• for a very massive star in the course of its life cycle to become so dense that the term

GM/Rsc 2 >1 in that case equation

hf

CR

GMhfhf

s

2

'

suggests that the photon cannot escape from the star,

because escape requires more energy than the photon

initially possesses. Such a star is called a black hole because it emits no light and acts like a celestial vacuum

cleaner for all nearby matter and radiation

• How we observe the black hole?

• Through the gravitational attraction.

• Search for x-rays produced by inrushing matter attracted to the black hole

• Black hole: An object (Star) of sufficiently high density can trap light forever

H.W

•2,4,21,28,47