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CH00





CH00

Quantum Physics 量 理 Robert Eisberg Second edition ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ Solutions Supplement to Accompany Quantum Physics of Atoms, molecules,solids, nuclei, and particles Second Edition Robert Eisberg Prepared by Edward Derringh

Quantum Physics of Atoms, molecules,solids, nuclei, and particles SecondEdition Robert Eisberg

CH01 CH02 CH03 CH04 CH05 CH06 CH07 CH08 CH09 CH10 CH11 CH12 CH13 CH14 CH15 CH16

Thermal radiation and Planck’s postulate Photons-particlelike properties of radiation De Broglie’s postulate-wavelike properties of particles Bohr’s model of the atom Schroedinger’s theory of quantum mechanics Solutions of time-independent Schroedinger equations One-electron atoms Magneticdipole moments, spin and transition rates Multielectron atoms – ground states and x-ray excitations Multielectron atoms-optical exciations Quantum statistics Molecules Solids-conductors and semiconductors Solids-superconductors and magnetic properties Nuclear models Nuclear decay and nuclear reactions

CH17 Introduction to elementary particles CH18 More elementary particles

CH00-page1CH00-page2





CH01
ν2 ν1 ν





CH01

Quantum Physics 量 理 Robert Eisberg Second edition CH 01 Thermal radiation and Planck’s postulate ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-1 ANS ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ At what wavelength does a cavity at 6000 K radiated most per unit wavelength? 60000K 輻
0

ANS

P = A ∫ RT (ν )dν =

2 1 1 Ac ∫ ρT (ν )dν = Ac ρT (ν av )∆ν 4 ν1 4

ν1 = ν2 =

c

λ1
c

= =

2.988 ×108 = 5.4509 × 1014 Hz 5.50 ×10−7 2.988 × 108 = 5.4401× 1014 Hz 5.51× 10−7 1 (ν 1 + ν 2 ) = 5.46 ×1014 Hz 2

λ2

4 1-2 Show that the proportionality constant in (1-4) is . That is, show that the relation between c
c spectral radiancy RT (ν ) and energy density ρT(ν ) is RT (ν )dν = ρT (ν )dν . 4 ANS (1-4) ρT (ν ) ∝ RT (ν )

Therefore, ν av =

∆ν = ν 2 −ν 1 = 9.9 ×1011 Hz Since ρT (ν av ) =
3 8π hν av 3 c

1 e
hν av kT

ρT (ν )dν =

8πν 2 c3

hν e kT − 1


dν ⇒ 令 x =

hν kT

−1

Numerically
Hint :

8π hν c3

3 av

= 1.006 ×10−13

hν av = 4.37 kT

e kT − 1 = 78.04

hν av

⇒ ρT (ν )dν =

8π k 4T 4 x3 8π k 4T 4 π 4 dx =x 3 3 h c e −1 h3c 3 15 2π 5 k 4 15c 2 h3
4

∫e

x 3dx π 4 = x − 1 15

RT = σ T 4 , σ =

Stefan’s law
5 4

1.006 × 10−13 = 1.289 ×10−15 78.04 The aera of the hole is A = π r 2 = π (5 × 10−3 ) 2 = 7.854 × 10−5 m 2

ρT (ν av ) =

ρT (ν )dν =

8π k T π 4 2π k 4 = ( )T 4 = RT (ν )dν h3c 3 15 c 15h3c 2 c
4 4

Hance, finally, P =

1 1 Ac ρT (ν av )∆ν = (7.854 × 10−5 )(2.998 ×108)(1.289 ×10−15 )(9.9 × 1011 ) 4 4

P = 7.51W ............##

RT (ν )dν =

c ρT (ν )dν …………## 4

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-5 (a) Assuming the surface temperature of the sun to be 57000 K ,use Stefan’s law, (1-2), to determine the rest mass lost per second to radiation by the sun. Take the sun’s diameter to be~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-3 Consider two cavities of arbitrary shape and material, each at the same temperature T, connected by a narrow tube in which can be placed color filters (assumed ideal) which will allow only radiation of a specified frequency ν to pass through. (a) Suppose at a certain frequency ν ′ , ρT (ν ′)dν for cavity 1 was greater than ρT (ν ′)dν forcavity 2. A color filter which passes only the frequency ν ′ is placed in the connecting tube. Discuss what will happen in terms of energy flow. (b) What will happened to their respective temperatures? (c) Show that this would violate the second law of thermodynamic; hence prove that all blackbodies at the same temperature must emit thermal radiation with the same spectrum independent of the...
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