Physics 202 - University of Virginia

Physics 202Professor P. Q. Hung

311B, Physics Building

Physics 202 – p. 1/30

Quantum Physics

Blackbody Radiation

Blackbody: A perfect absorber of radiation ofall wavelengths.

A blackbody reradiates the energy that itabsorbs with a universal spectrum(wavelength or frequency)

The hotter the blackbody is the more energy itemits

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Quantum Physics

Blackbody Radiation

Physics 202 – p. 2/30

Quantum Physics

Blackbody Radiation

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Quantum Physics

Blackbody Radiation

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Quantum Physics

Blackbody Radiation

Radiation with long wavelengths obey theWien’s displacement lawλmax T = 0.2898 × 10−2 m.K

Recall that f = cλ, compare the above

equation with the one in the book.λmax is the value of the wavelength where theradiation intensity versus wavelength reachesits maximum.

Physics 202 – p. 4/30

Quantum Physics

Blackbody Radiation

Radiation with long wavelengths obey theWien’s displacement lawλmax T = 0.2898 × 10−2 m.K

Recall that f = cλ, compare the above

equation with the one in the book.λmax is the value of the wavelength where theradiation intensity versus wavelength reachesits maximum.

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Quantum Physics

Blackbody Radiation

Physics 202 – p. 5/30

Quantum Physics

Blackbody Radiation: Example 1Near what wavelength is the most energyradiated from an object at room temperature?

Room temperature: T ≈ 2930K.

λmax = 0.2898×10−2 m.KT

= 0.2898×10−2 m.K2930K

0.989× 10−5 m. This about in the middle of theinfrared range.

Physics 202 – p. 6/30

Quantum Physics

Blackbody Radiation: Example 1Near what wavelength is the most energyradiated from an object at room temperature?

Room temperature: T ≈ 2930K.

λmax = 0.2898×10−2 m.KT

= 0.2898×10−2 m.K2930K

0.989× 10−5 m. This about in the middle of theinfrared range.

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Quantum Physics

Blackbody Radiation: Example 2The eye is most sensitive to a wavelength of500 nm. At what temperature would the intensityof an object be maximum at this wavelength?

T = 0.2898×10−2 m.Kλmax

= 0.2898×10−2 m.K500×10−9 m

= 5796 0K.

This is about the temperature on the surfaceof the sun.

Physics 202 – p. 7/30

Quantum Physics

Blackbody Radiation: Example 2The eye is most sensitive to a wavelength of500 nm. At what temperature would the intensityof an object be maximum at this wavelength?

T = 0.2898×10−2 m.Kλmax

= 0.2898×10−2 m.K500×10−9 m

= 5796 0K.

This is about the temperature on the surfaceof the sun.

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Quantum Physics

Planck’s hypothesis

Classical model of blackbody radiation.: acavity filled with electromagnetic radiationwith a hole punched into it which then leaksthat radiation.

Rayleigh-Jeans ⇒ The energy density ofradiation (from which one can calculate theintensity) inside the cavity is inverselyproportional to the square of the wavelengthand proportional to the temperature.u(λ, T ) = 8π

c1λ2kT

Only good for long wavelength (fit the data)!

Physics 202 – p. 8/30

Quantum Physics

Classical model of blackbody radiation.: acavity filled with electromagnetic radiationwith a hole punched into it which then leaksthat radiation.

Rayleigh-Jeans ⇒ The energy density ofradiation (from which one can calculate theintensity) inside the cavity is inverselyproportional to the square of the wavelengthand proportional to the temperature.u(λ, T ) = 8π

c1λ2kT

Only good for long wavelength (fit the data)! Physics 202 – p. 8/30

Quantum Physics

The energy density blows up as λ → 0! Thisis called the ultraviolet catastrophe.

Experimentally, one finds u(λ, T ) → 0 asλ → 0.

Physics 202 – p. 9/30

Quantum Physics

The energy density blows up as λ → 0! Thisis called the ultraviolet catastrophe.

Experimentally, one finds u(λ, T ) → 0 asλ → 0.

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Quantum Physics

Planck: In order to fit the data, he proposed tomodify the above formula.u(λ, T ) = (8π

chλ3 )

λkT)−1

For large λ, e( hc

λkT)≈ 1 + hc

λkT⇒

Rayleigh-Jeans formula.

u(λ, T ) → 0 as λ → 0.This was a fit to the data. To derive it fromfirst principles, Planck had to assume that theenergy if blackbody radiation is quantized.

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Quantum Physics

chλ3 )

λkT)−1

For large λ, e( hc

λkT)≈ 1 + hc

λkT⇒

Physics 202 – p. 10/30

Quantum Physics

chλ3 )

λkT)−1

For large λ, e( hc

λkT)≈ 1 + hc

λkT⇒

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Quantum Physics

Physics 202 – p. 11/30

Quantum Physics

Planck’s Hypothesis:En = nh ff = c

Energy comes in bundles of hf and is notcontinuous. The above equation representswhat is known as Energy quantization.Energy comes in increment of fundamentalquanta of energy hf .

Physics 202 – p. 12/30

Quantum Physics

Planck’s Hypothesis:En = nh ff = c

Energy comes in bundles of hf and is notcontinuous. The above equation representswhat is known as Energy quantization.Energy comes in increment of fundamentalquanta of energy hf .

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Quantum Physics

Fit to the data givesh = 6.63 × 10−34 J.s = 4.14 × 10−15eV.s

What is the physical meaning of the quantumhf?

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Quantum Physics

Fit to the data givesh = 6.63 × 10−34 J.s = 4.14 × 10−15eV.s

What is the physical meaning of the quantumhf?

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Quantum Physics

Einstein and the Photoelectric effectWhat is the photoelectric effect?Set Up:

Two parallel plates, one called the cathodeand another called the anode. A potentialdifference can be applied to the 2 plates.

Light shines on the cathode.

If the anode is positive, it is observed thatelectrons that are ejected from the cathode,reach the anode and thus setting up acurrent.

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Quantum Physics

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Quantum Physics

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Quantum Physics

Einstein and the Photoelectric effect

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Quantum Physics

If the cathode is made positive, it is observedthat there are still some electrons that haveenough kinetic energy to reach the anode.

There is a potential difference called the“stopping potential”, V0, at which and beyond,no electron can reach the anode.

The “stopping potential” V0 is found to beindependent of the intensity of light, and isfound to be dependent on the frequency.What is the explanation for this phenomenon?

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Quantum Physics

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Quantum Physics

Physics 202 – p. 16/30

Quantum Physics

Einstein and the Photoelectric effectEinstein:

Light is composed of quanta of energy hfcalled photons.

The photons penetrate the surface of thecathode and their energy is given completelyto the electrons.

φ: Energy necessary to remove an electronfrom the surface of the metal. It is called thework function and is characteristic of themetal.

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Quantum Physics

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Quantum Physics

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Quantum Physics

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Quantum Physics

The maximum kinetic energy isKmax = eV0 = hf − φ

One can plot V0 versus f . The slope is h/e.

Physics 202 – p. 19/30

Quantum Physics

The maximum kinetic energy isKmax = eV0 = hf − φ

One can plot V0 versus f . The slope is h/e.

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Quantum Physics

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Quantum Physics

Threshold frequency and wavelength, f0 andλt:φ = hf0 = hc

Useful number:hc = 1.24 × 104 eV.Å

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Quantum Physics

Threshold frequency and wavelength, f0 andλt:φ = hf0 = hc

Useful number:hc = 1.24 × 104 eV.Å

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Quantum Physics

Einstein and the Photoelectric effect: ExampleThe threshold wavelength for potassium is5, 640 Å. What is the work function for potassium?Solution:φ = hc

= 1.24×104 eV.Å5,640 Å

= 2.2 eV

The work functions for metals are typically in afew of eV’s.

Physics 202 – p. 22/30

Quantum Physics

Mass and momentum of a photon

Recall: vc

and E2 = p2c2 + (mc2)2

Photon travels with a speed of light ⇒ E = pc⇒ m = 0. The photon has zero mass.

E = hf ⇒ p = hf

Physics 202 – p. 23/30

Quantum Physics

Recall: vc

and E2 = p2c2 + (mc2)2

E = hf ⇒ p = hf

Physics 202 – p. 23/30

Quantum Physics

Recall: vc

and E2 = p2c2 + (mc2)2

E = hf ⇒ p = hf

Physics 202 – p. 23/30

Quantum Physics

Compton EffectWhat happens when an electromagnetic wave isincident on a charged particle such as anelectron?

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Quantum Physics

Compton Effect

Classical picture:When the charged particle is hit by an EMwave, it oscillates (due to the oscillatingelectric field). In turns, it gives off radiationbecause of the oscillation (acceleration). Thisradiation, according to the classical picture,would have the same frequency orwavelength as the incident EM wave.

Observation:The radiation has in general a differentwavelength.

Physics 202 – p. 25/30

Quantum Physics

Compton Effect

Classical picture:When the charged particle is hit by an EMwave, it oscillates (due to the oscillatingelectric field). In turns, it gives off radiationbecause of the oscillation (acceleration). Thisradiation, according to the classical picture,would have the same frequency orwavelength as the incident EM wave.

Observation:The radiation has in general a differentwavelength.

Physics 202 – p. 25/30

Quantum Physics

Compton EffectCompton:

Collision is considered to be between aphoton of energy hf and momentumhf/c = h/λ and an electron. Recall that for amassless particle, E = pc = hf ⇒

p = hf/c = h/λ.

The electron would absorb energy energydue to the recoil and the scattered photonwould have less energy and thus a lowerfrequency!

Physics 202 – p. 26/30

Quantum Physics

Compton EffectCompton:

Collision is considered to be between aphoton of energy hf and momentumhf/c = h/λ and an electron. Recall that for amassless particle, E = pc = hf ⇒

p = hf/c = h/λ.

The electron would absorb energy energy dueto the recoil and the scattered photon wouldhave less energy and thus a lower frequency!

Physics 202 – p. 26/30

Quantum Physics

Compton Effect

Let θ be the angle between the direction ofthe initial photon and the direction of thescattered photon. Let the initial wavelengthbe λ and that of the scattered photon be λ

Energy and momentum conservation givesλ

− λ = hmec

(1 − cos θ)h

mec= hc

mec2 = 0.0243 Å is called the Comptonwavelength of the electron.

Physics 202 – p. 27/30

Quantum Physics

Compton Effect

Let θ be the angle between the direction ofthe initial photon and the direction of thescattered photon. Let the initial wavelengthbe λ and that of the scattered photon be λ

Energy and momentum conservation givesλ

− λ = hmec

(1 − cos θ)h

mec= hc

mec2 = 0.0243 Å is called the Comptonwavelength of the electron.

Physics 202 – p. 27/30

Quantum Physics

Compton effect

Physics 202 – p. 28/30

Quantum Physics

Compton effect: ExampleX-rays of wavelength λ = 22 pm = 22 × 10−12 m(photon energy = 56 keV ) are scattered from acarbon target, and the scattered rays aredetected at 850 to the incident beam. What is theCompton shift of the scattered rays? Whatpercentage of the initial X-ray photon energy istransferred to an electron in such a scattering?

Physics 202 – p. 29/30

Quantum Physics

Compton effect: Example

∆λ = hmec

(1 − cos θ) = 0.0243 Å (1 − cos 850) =

0.2218 Å ≈ 2.2 × 10−12m = 2.2pm

Frac = E−E′

E= hf−hf

hf= λ

λ′ = ∆λ

λ+∆λ=

2.2pm22pm+2.2pm

≈ 9%

The photoelectric effect + the Compton effect⇒ a convincing support of the photonconcept of light. Since light also exhibits awave nature ⇒ duality: light has both waveand particle properties.

Physics 202 – p. 30/30

Quantum Physics

∆λ = hmec

(1 − cos θ) = 0.0243 Å (1 − cos 850) =

0.2218 Å ≈ 2.2 × 10−12m = 2.2pm

Frac = E−E′

E= hf−hf

hf= λ

λ′ = ∆λ

λ+∆λ=

2.2pm22pm+2.2pm

≈ 9%

Physics 202 – p. 30/30

Quantum Physics

∆λ = hmec

(1 − cos θ) = 0.0243 Å (1 − cos 850) =

0.2218 Å ≈ 2.2 × 10−12m = 2.2pm

Frac = E−E′

E= hf−hf

hf= λ

λ′ = ∆λ

λ+∆λ=

2.2pm22pm+2.2pm

≈ 9%

Physics 202 – p. 30/30

Physics 202 - University of Virginia

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