Review of Quantum Physics !(Rohlf Chapter 3.)

Homework 7!Cavity radiation, etc: Ch. 3, problems 3, 8, 19, 21!Due, Fri. Oct. 2.

Cavity Radiation!Rohlf P63-66

Number of oscillators per unit volume per λ : dNdλ ∝ 1λ4

Average energy per oscillator:

E =EP(E)

0

∞

∫ dE

P(E)0

∞

∫ dEwhere P = Ce−E /kT

E =Ee−E /kT

0

∞

∫ dE

e−E /kT0

∞

∫ dE= kT

Energy per unit volume per unit wavelength:

dudλ ∝ 1

λ4 E ∝ 1λ4 kT

Classical Rayleigh-Jeans law.

Cavity Radiation!(continued)

Energy per unit volume per unit wavelength: dudλ ∝ 1λ4 kT

Energy radiated per unit area per unit wavelength: dRdλ ∝ dudλ

Integrate over allλ : dudλ0

∞

∫ dλ ∝1λ4

0

∞

∫ dλ →∞

Ultraviolet catastrophe!

Classical Rayleigh-Jeans law.

Max Planck 1901

In order to avoid the UV "catastrophe" and make the result agree with the experimental data, turn the integral in E into a sum by quantizing E = nhf for each oscillator frequency f .

Integral → sum.

E =EnPn

n=0

∞

∑

Pnn=0

∞

∑where Pn = Ce−En /kT E =

Ene−En /kT

n=0

∞

∑

e−En /kT

n=0

∞

∑=

nhfe−nhf /kT

n=0

∞

∑

e−nhfn /kT

n=0

∞

∑

E =hf

e−hf /kT − 1⇒

dudλ

∝dRdλ

∝1

λ 5 e−hc/kTλ − 1( )

Roughly agrees with experimental data.Planck did not believe this made any physical sense, and spent the next several years trying to work around it.

Some Useful Constants.

h = 6.63×10-34 J-s = 4.14 ×10-15 eV-s

hc = 1240 eV-nm

≡ h / 2π

c = 197.3 eV-nm = 197.3 MeV-fm = 0.1973 GeV-fm

mec2 = 0.511×106 eV = 0.511 MeV

mpc2 = 938 MeV = 0.938 GeV

mnc2 = 940 MeV = 0.940 GeV

Photoelectric Effect: Einstein 1905!Rohlf, sec. 3.3, P76-80

(This was already covered in Honors Physics II.)

Eemax = hf −φ ⇒ Ephoton = hf

The energy of the electromagnetic field is quantized in little packets of energy (photons)

e k

Compton Effect: A. Compton 1922 !Rohlf, Sec. 4.5. P125-127

Photons carry energy E=hf and also carry momentum as if they are particles of mass=0.

E2 = p2c2 +m2c4

E = hf = hcλ

m = 0

hcλ= pc p = h

λ

E = hf p = hλ

e λ1

λ2

e’

θk

φe

λ2 − λ1 =hmc

1− cosθ( )

Louis de Broglie - 1924!Rohlf, 5.1, p136-140

Photons:particle properties Electrons:wave properties.

p = hλ

E = hf

E2 = p2c2 +m2c4

h2 f 2 = h2c2

λ2+m2c4 f = c

λ⎛⎝⎜

⎞⎠⎟2

+ mc2

h⎛⎝⎜

⎞⎠⎟

2

(This was covered in Honors Physics II.)

Louis de Broglie - 1924!(continued)

p = hλ

E = hf

Non-relativistic limit:

K = p2

2m= p2c2

2mc2 = h2c2

2mc2λ2

λ2 = h2c2

2mc2Kλ = hc

2mc2K

Summary

Wave - Particle Duality

p = hλ

E = hf E2 = p2c2 +m2c4

h2 f 2 = h2c2

λ2 +m2c4 f = cλ

⎛⎝⎜

⎞⎠⎟

2

+ mc2

h⎛⎝⎜

⎞⎠⎟

2

Electron at rest: p = 0, λ = ∞,      f = mc2

hHz

Non-relativistic limit: λ = hc2mc2K

Exercise.

1.Find the quantum frequency of an electron at rest.

2.Find the wavelength of 50 keV kinetic energy electrons which are produced by an x-ray machine operating at a potential of 50 kV. Assume that E0=mc2=0.5 MeV

3.Find the wavelength of the maximum energy photons, which also have an energy of 50 KeV.

4.For the electrons, compare the result of the calculation for the fully relativistic expression and for the non-relativistic limit.

Exercise.

1. f = Eh

= mc2

h= 0.51 ×106 eV

4.14 ×10-15 eV-s≈1.2 ×1020 Hz

2. λ = hc2mc2K

= 1240

2 0.5 ×106( ) 5 ×104( )= .5545 ×10−2 nm

3.   λ = hcpc

= hcE

= 12405 ×104 = 2.5 ×10−2 nm = 0.25 A

o

4.   λ = hcpc

= hcE2 −m2c4

= hc

K +mc2( )2 −m2c4= hc

K 2 + 2Kmc2( )= hc

K 1+ 2mc2

K⎛⎝⎜

⎞⎠⎟

         = 1240

5 ×104 1+ 2(0.5 ×106 )5 ×104

= 12405 ×104 (4.582)

= 0.541×10−2 nm

1.Find the quantum frequency of an electron at rest.

2.Find the wavelength of 50 keV kinetic energy electrons which are produced by an x-ray machine operating at a potential of 50 kV. Assume that E0=mc2=0.5 MeV

3.Find the wavelength of the maximum energy photons, which also have an energy of 50 KeV.

4.For the electrons, compare the result of the calculation for the fully relativistic expression and for the non-relativistic limit.