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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
Classical Rayleigh-Jeans law.
Number of oscillators per unit volume per λ :
Average energy per oscillator:
∞
E
EP(E)dE
∫
0
= ∞
∫0 P(E)dE
where P = Ce− E/kT
∞
E
Ee
dE
∫
0
= ∞
− E/kT
∫0 e dE
− E/kT
= kT
Energy per unit volume per unit wavelength:
du
1
1
∝ 4 E ∝ 4 kT
dλ λ
λ
dN 1
∝
dλ λ 4
Cavity Radiation!
(continued)
Classical Rayleigh-Jeans law.
Energy per unit volume per unit wavelength:
du
1
∝ 4 kT
dλ λ
Energy radiated per unit area per unit wavelength:
∞
du
Integrate over all λ : ∫ d λ ∝
dλ
0
Ultraviolet catastrophe!
∞
1
∫0 λ 4 dλ → ∞
dR du
∝
dλ dλ
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 P
n
E =
n
n= 0
∞
∑P
∞
∑E e
n
where Pn = Ce− En /kT
E =
n= 0
∞
∑ e− E /kT
n
n
n= 0
E =
n= 0
hf
e− hf /kT − 1
⇒
du
dλ
∝
dR
dλ
∝
∞
− En /kT
λ
5
(e
1
− hc/kT λ
=
∑ nhfe−nhf /kT
n=0
∞
∑ e−nhf /kT
n
n=0
)
−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
me c 2 = 0.511× 10 6 eV = 0.511 MeV
m p c 2 = 938 MeV = 0.938 GeV
mn c 2 = 940 MeV = 0.940 GeV
Photoelectric Effect: Einstein 1905!
Rohlf, sec. 3.3, P76-80
(This was already covered in Honors Physics II.)
k
e
Eemax = hf − φ ⇒
E photon = hf
The energy of the electromagnetic field is
quantized in little packets of energy (photons)
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.
E 2 = p2 c2 + m 2 c4
e’
φe
λ1 e
m=0
θk
λ2 λ2 − λ1 =
hc
E = hf =
λ
h
(1− cosθ )
mc
hc
= pc
λ
p=
h
λ
E = hf
p=
h
λ
Louis de Broglie - 1924!
Rohlf, 5.1, p136-140
(This was covered in Honors Physics II.)
Photons:particle properties  Electrons:wave properties.
h
p=
λ
E = hf
E 2 = p2 c2 + m 2 c4
2 2
h c
h f = 2 + m2 c4
λ
2
2
⎛ c ⎞ ⎛ mc ⎞
f = ⎜ ⎟ +⎜
⎝ λ ⎠ ⎝ h ⎟⎠
2
2
2
Louis de Broglie - 1924!
(continued)
p=
h
λ
E = hf
Non-relativistic limit:
p2
K =
2m
p2 c2
=
2mc 2
h2 c2
λ =
2mc 2 K
2
h2 c2
=
2mc 2 λ 2
λ=
hc
2mc 2 K
Summary
Wave - Particle Duality
p=
h
λ
E = hf
h2 c2
2 2
h f = 2 + m2 c4
λ
Electron at rest: p = 0,
E 2 = p2 c2 + m 2 c4
2
⎛ c ⎞ ⎛ mc ⎞
f = ⎜ ⎟ +⎜
⎝ λ ⎠ ⎝ h ⎟⎠
2
λ = ∞, Non-relativistic limit:
λ=
2
mc 2
f=
Hz
h
hc
2mc 2 K
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.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.
E
mc 2
0.51 × 10 6 eV
1. f =
=
=
≈ 1.2 × 10 20 Hz
-15
h
h
4.14 × 10 eV-s
2. λ =
hc
2mc 2 K
=
1240
(
)(
2 0.5 × 10 6 5 × 10 4
)
= .5545 × 10 −2 nm
o
hc hc
1240
−2
3. λ =
=
=
= 2.5 × 10 nm = 0.25 A
pc E 5 × 10 4
4. λ =
hc
=
pc
hc
E 2 − m2c4
=
hc
( K + mc )2 − m 2 c 4
1240
=
5 × 10
4
2(0.5 × 10 6 )
1+
5 × 10 4
2
=
=
hc
( K 2 + 2Kmc2 )
1240
= 0.541× 10 −2 nm
4
5 × 10 (4.582)
=
hc
⎛ 2mc 2 ⎞
K ⎜ 1+
K ⎟⎠
⎝