Robert Eisberg-Quantum Physics-Problem solution Chapter 1-6 PDF

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量子物理習題 CH00 量子物理習題 CH00 Quantum Physics（量子物理）習題 Robert Eisberg（Second edition） ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ Solutions Supplement to Accompany Quantum Physics of Atoms, molecules,solids, Quantum Physics of Atoms, molecules,solids, nuclei, and particles...

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量子物理習題 CH00

量子物理習題 CH00

Quantum Physics（量子物理）習題 Robert Eisberg（Second edition） ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ Solutions Supplement to Accompany

Quantum Physics of Atoms, molecules,solids, nuclei, and particles

Quantum Physics of Atoms, molecules,solids, nuclei, and particles Second Edition Robert Eisberg Prepared by Edward Derringh

Second Edition Robert Eisberg

CH01：Thermal radiation and Planck’s postulate CH02：Photons-particlelike properties of radiation CH03：De Broglie’s postulate-wavelike properties of particles CH04：Bohr’s model of the atom CH05：Schroedinger’s theory of quantum mechanics CH06：Solutions of time-independent Schroedinger equations CH07：One-electron atoms CH08：Magnetic dipole moments, spin and transition rates CH09：Multielectron atoms – ground states and x-ray excitations CH10：Multielectron atoms-optical exciations CH11：Quantum statistics CH12：Molecules CH13：Solids-conductors and semiconductors CH14：Solids-superconductors and magnetic properties CH15：Nuclear models CH16：Nuclear decay and nuclear reactions CH17：Introduction to elementary particles CH18：More elementary particles

CH00-page1

CH00-page2

量子物理習題 CH01

量子物理習題 CH01

Quantum Physics（量子物理）習題

ν

Robert Eisberg（Second edition） CH 01：Thermal radiation and Planck’s postulate

ν

2 2 1 1 ANS： P = A ∫ RT (ν )d ν = Ac ∫ ρT (ν )d ν = Ac ρT (ν av )∆ν 4 4 ν1 ν1

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-1、At what wavelength does a cavity at 6000 0K radiated most per unit wavelength?

2.988× 108 ν1 = c = = 5.4509× 1014 Hz λ1 5.50 ×10 −7

（ 60000 K 時，一空腔輻射體的最強波長為何？） ANS：

ν2 =

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~

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

c ρ (ν )dν . 4 T

ANS：(1-4) ρ T (ν ) ∝ RT (ν )

ρ T (ν )dν =

8πν 2 c3

hν hν

e kT − 1

dν ⇒ 令 x =

hν kT

4 4 3 4 4 4 8π k T x 8π k T π （Hint : dx = ⇒ ρ T (ν ) dν = x h 3c 3 e −1 h 3c 3 15 5 4 2π k （Stefan’s law） 又 RT = σT 4 , σ = 15 c 2 h3

x 3dx π 4 ） = x − 1 15

∫e

8π k 4T 4 π 4 4 2π 5k 4 4 4 )T = RT (ν )dν = ( ρ T (ν )dν = 3 3 h c 15 c 15h3 c2 c 所以 RT (ν ) dν =

c ρ (ν ) dν …………## 4 T

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 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 ′ , ρ T ( ν ′) d ν for cavity 1 was greater than ρT (ν ′ )dν for cavity 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 details of their composition. ANS：

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-4、A cavity radiator at 6000 0 K has a hole 10.0 mm in diameter drilled in its wall. Find the power radiated through the hole in the range 5500～5510Å.（Hint : See Problem2）

（一在 60000 K 的輻射體有一直徑為 10.0mm 的小洞，求由此小洞所輻射出的波長在 5500 ～5510Å 之間的功率。） -1-

c

λ2

=

2.988× 108 = 5.4401× 1014 Hz 5.51× 10−7

1 Therefore, ν av = (ν 1 +ν 2) = 5.46 ×1014 Hz 2

∆ν = ν 2 −ν 1 = 9.9 ×1011 Hz Since ρ T (ν av ) =

8π hν 3av c3

1 hν av

e

kT

−1

hν av hν 8π hν3av Numerically = 1.006 ×10− 13 ， av = 4.37 ， e kT − 1 = 78.04 c3 kT 13 1.006× 10 − ρ T (ν av ) = = 1.289 × 10 −15 78.04 The aera of the hole is A = π r2 = π (5× 10 −3 )2 = 7.854× 10−5 m2

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 ............##

Hance, finally, P =

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-5、(a) Assuming the surface temperature of the sun to be 5700 0K ,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 1.4 ×109 m . (b) What fraction of the sun’s rest mass is lost each year from electromagnetic radiation? Take the sun’s rest mass to be 2.0 ×10 30kg .

ANS：(a) L = 4π R2σ T 4 = 4π (7 × 108 ) 2 (5.67 × 10 −8 )(5700)4 = 3.685× 1026 W （ R sun = 7 ×108 m ）

L=

d dm ( mc2 ) = c2 dt dt

dm L 3.685 ×10 26 = = = 4.094 × 109 kg / s (3 ×10 8) 2 dt c2

(b) The mass lost in one year is dm t = (4.094 ×10 9 )(86400 × 365) = 1.292× 1017 kg dt The desired fraction is , then, ∆M =

f=

∆M 1.292 ×1017 = = 6.5 × 10 −14 …………## 30 2.0 ×10 M -2-

量子物理習題 CH01

量子物理習題 CH01

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-6、In a thermonuclear explosion the temperature in the fireball is momentarily 10 7 0K . Find the wavelength at which the radiation emitted is a maximum. （熱核爆炸，火球溫度達到 107 0 K ，求最大輻射強度之波長為何？） ANS：

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-7、At a given temperature, λ max = 6500Å for a blackbody cavity. What will λ max be if the temperature of the cavity wall is increased so that the rest of emission of spectral radiation is double? （一黑體空腔在某一溫度 λmax = 6500 Å，若溫度增加，以致使此光譜輻射能量加倍時，λmax 應為何？）

ANS：Stefan’s law RT = σ T

4

σT ′ 4 R′ = 2= ⇒ T ′ = 4 2T σ T4 R Wien’s law :

P=

L sun π R2 = π R2 S 4π r 2

The average rate, per m2 , of arrival of energy at the earth’s surface is 1 π R2 S 1 P = = S = (1353W / m 2 ) = 338W / m 2 Pav = 4 π R2 4 π R2 4 4 (b) 338 = σT 4 = (5.67 ×10−8 )T 4

T = 277.860 K …………## NOTE：課本解答 Appendix S，S-1 為(b) 2800 K 。 ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-11、Attached to the roof of a house are three solar panels, each 1m × 2m . Assume the equivalent of 4 hrs of normally incident sunlight each day, and that all the incident light is absorbed and converted to heat. How many gallows of water can be heated from 40℃ to 120℃ each day? ANS：

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-12 、 Show that the Rayleugh-Jeans radiation law, (1-17), is not consistent with the Wien displacement law ν max ∝ T , (1-3a), or λmaxT = const , (1-3b).

0

′ ( 4 2 T ) = (6500 A)T ⇒ λmax λmax ′ T ′ = λmaxT ⇒ λmax ′ =

6500

0

0

A = 5466 A ……## 2 ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-8、At what wavelength does the human body emit its maximum temperature radiation? List 4

assumptions you make in arriving at an answer. （在波長為何時，人體所釋放出的的熱輻射為最大？並列出你的假設。）

ANS： ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-9、Assuming that λ max is in the near infrared for red heat and in the near ultraviolet for blue heat, approximately what temperature in Wien’s displacement law corresponds to red heat? To blue

heat? ANS：

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-10、The average rate of solar radiation incident per unit area on the earth is 0.485cal / cm 2 − min (or 338W / m2 ). (a) Explain the consistency of this number with the solar constant (the solar energy falling per unit time at normal incidence on a unit area) whose value is

1.94cal / cm 2 − min (or 1353W / m2 ). (b) Consider the earth to be a blackbody radiating energy into space at this same rate. What surface temperature world the earth have under these circumstances? L sun 2 4π r r = Earth-sun distance（地球-太陽距離） 。 Lsun = rate of energy output of the sun（太陽

ANS：(a) The solar constant S is defined by S =

釋放能量的速率）。令 R = radius of the earth（地球半徑）。The rate P at which energy impinges on the earth is -3-

ANS：

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-13、We obtain ν

max

in the blackbody spectrum by setting

d ρ T (ν ) = 0 and λ max by setting dν

d ρT (λ ) = 0 . Why is it not possible to get from λmaxT = const to νmax = const ×T simply by dλ c ? This is, why is it wrong to assume that ν maxλ max = c , where c is the speed using λmax =

νmax

of light? ANS：

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-14、Consider the following number : 2,3,3,4,1,2,2,1,0 representing the number of hits garnered by each member of the Baltimore Orioles in a recent outing. (a) Calculate directly the average number of hits per man. (b) Let x be a variable signifying the number of hits obtained by a 4

man, and let f (x ) be the number of times the number s appears. x =

∑ xf ( x) 0 4

∑ f ( x)

. (c) Let

0

p (x ) be the probability of the number x being attained. Show that x is given by 4

x = ∑ xp (x ) 0

ANS：

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ -4-

量子物理習題 CH01

量子物理習題 CH01

1 2 (10 − x) 10 f ( x) = 0 f ( x) =

1-15、Consider the function

−

0 ≤ x ≤10

1-16、Using the relation P ( ε)

all other x

xf ( x) dx

∫

f (x )dx

∞

∫ P(ε )dε = 1, evaluate the integral of (1-21) to deduce

0

(1-22), ε = kT .

∞

∫

ε

e kT and kT

(a) From x = −∞∞

find the average value of x. (b) Suppose the variable x were

−

（由 P (ε )

−∞

ε

∞ e kT 和 ∫ P(ε )dε = 1 兩式，求 1-21 之積分，以證明 1-22 式， ε = kT ） kT 0

discrete rather then continuous. Assume ∆x = 1 so that x takes on only integral values 0,1,2,……,10. Compute x and compare to the result of part(a). (Hint : It may be easier to

ANS：

compute the appropriate sum directly rather than working with general summation formulas.) (c) Compute x for ∆x = 5 , i.e. x = 0,5,10 . Compare to the result of part (a). (d) Draw

c 1-17 、 Use the relation RT (ν ) dν = ρ T (ν ) dν between spectral radiancy and energy density, 4 together with Planck’s radiation law, to derive Stefan’s law. That is, show that

analogies between the results obtained in this problem and the discussion of Section 1-4. Be sure you understand the roles played by ε , ∆ε , and P (ε ) . 10

10

1 (10 − x) 2 dx 10

20 3 1 4 10 50 x − ∫ ∫ x + x 0 3 4 ANS：(a) x = 010 = = 010 1 2 1 2 2 100 10 − + x3 10 x x ∫ 10 (10 − x) dx ∫ (100 − 20x + x )dx 3 0 x

x(100 − 20x + x 2 )dx

0

0

=

20 2 1 3 10 1000 x + x 0 3 4 = 12 = 2.5 1 100 100 − 10 x + x2 100 3 3 1

∑ x10 (10 − x ) 0 10

1

∑ 10 (10 − x)

10

2

=

∑100 x − 20x

2

=

n

∑n= 1

2

+ x3

0 10

=

∑100 − 20 x + x

n (n +1) , 2

10 × 11 10 ×11 × 21 10 × 11 2 ) − 20× +( 2 6 2 10 × 11 10× 11× 21 100 × 11− 20× + 2 6

100 ×

2

1

2

=

n(n +1)(2n +1) , 6

1

2

(c) Set x = 5n ⇒ x =

n

∑n

∑5n 10 (10 −5n ) n= 0 2

1

2π5 k4 2π h ν 3 dν . (Hint : = σ T 4 where σ = h 15c 2 h3 c 2 kTν e −1 ∞

∞

ANS： RT = ∫ RT( ν) dν =

∑ 10(10 − 5n)

∞

q3 dq π 4 ) = q − 1 15

∫e

0

∞

2π k 4 T 4 xdx 2π h ν 3 dν = h 2 c 2h 3 ∫0 e x − 1 c ∫0 kTν e −1

2π k 4T 4 π 4 2π 5k 4 4 4 T =σ T = c 2 h3 15 15 c 2 h3 ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~

825 = 2.143 385

(Hint :

0

=

0

0

∞

RT = ∫

0

50 x −

10

(b) x =

2

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~

2

2

=

n

∑n

∑20 n − 20 n

2

n (n + 1) 2 ] ) =[ 2

+ 5 n3

n= 0

2

∑ 4− 4n + n

2

n= 0

3

1

2

n =0

2 ×3 2 ×3 × 5 2 ×3 2 20 × ) − 20 × + 5( 5 2 6 2 = = 1……## = 2 ×3 2 × 3 × 5 5 4×3− 4× + 2 6 ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~

-5-

1-18、Derive the Wien displacement law, λmaxT = 0.2014 （Hint : Set

hc

λkT

hc d ρ T (λ ) , by solving the equation = 0. dλ k

= x and show that the equation quoted leads to e − x +

that x = 4.965 is the solution.） ANS：

x = 1. Then show 5

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-19、To verify experimentally that the 3 0 K universal background radiation accurately fits a blackbody spectrum, it is decided to measure RT (λ ) from a wavelength below λ max where its value is 0.2RT ( λmax ) to a wavelength above λ max where its value is again 0.2 RT ( λmax ) . Over what range of wavelength must the measurements be made? 2 π k5 t5 x5 c ρ ( λ) = 4 3 x 4 T h c e −1 hc With x = . At λ = λmax , x = 4.965 , by problem 18. Thus λkT

ANS： RT ( λ) =

( kt) 5 h 4c 3 Now find x such that RT (λ ) = 0.2RT ( λmax ) :

RT ( λmax ) = 42.403π

-6-

量子物理習題 CH01

量子物理習題 CH01

2 πk t x ( kt) = (0.2)42.403π 4 3 h4 c3 ex − 1 hc

radiates only about 30% of the energy radiated by a blackbody of the same size and temperature. (a) Calculate the temperature of a perfectly black spherical body of the same

x5 = 4.2403 e x −1 x1 = 1.882 ， x2 = 10.136

size that radiates at the same rate as the tungsten sphere. (b) Calculate the diameter of a perfectly black spherical body at the same temperature as the tungsten sphere that radiates at the same rate.

5 5

5

5

hc 1 (6.626 × 10 −34 )(2.998× 108 ) 1 = (1.38× 10− 23 )(3) kT x x

Numerically, λ =

4 4 ANS：(a) 30 0 0 × σ Ttungsten = σ Tblack ⇒ 300 0 × (2000 + 273) = T4

T = 1682 0 K = 14090 C

4.798× 10− 3 x

λ=

4 4 × 4π (2.3 cm)2 = σ Tblack × 4π r 2 (b) 30 0 0 × σ Ttungsten

3

So that λ1 =

4.798 ×10− = 2.55 mm 1.882

Ttungsten = Tblack ⇒ r = 1.26cm ……##

−3

4.798× 10 = 0.473mm ……## 10.136 ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~

λ2 =

(kT )5 1-20、Show that, at he wavelength λ max , where ρT ( λ) has its maximum ρ T (λ max ) = 170π . (hc) 4 x hc x ANS：If x = , then, by Problem 18, e − x + = 1 ⇒ e x −1 = 5 −x λ max kT 5

Hence, ρT ( λ max) = But, x = 4.965 ,

8π kT 4 λmax

1 4 λmax

Upon substitution, there give ρ T (λ max ) = 170π

(kT )5

……##

4

(hc) ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-21、Use the result of the preceding problem to find the two wavelengths at which ρ T (λ ) has a

value one-half the value atλ max . Give answers in terms of λ max . 5

(kT ) 4 . (hc)

hc

λkT

T

(λ ) d λ

0 ∞



∫ ρ T (λ )dλ

hc 1 (Hint : = 4.965 ; hence Wien’s λmax kT 4

underestimate the integrated energydensity? ANS： ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-24、Find the temperature of a cavity having a radiant energy density at 2000Å that is 3.82 times the energy density at 4000Å. ANS：Let λ ′ = 200nm , λ ′′ = 400nm ; then

hc

1 λ ′5

So that the wavelengths sought must satisfy

Again let x =

∫ρ

zero and λ max ; i.e., show that

approximation is fairly accurate in evaluating the integral in the numerator above.) (b) By the percent does Wien’s approximation used over the entire wavelength range overestimate or

kT 4 ) hc

ANS：By Problem 20, ρT ( λmax ) = 170π

λ max

0

(5 − x) .

= (4.965

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-23、(a) Show that about 25% of the radiant energy in a cavity is contained within wavelengths

8π hc

λ5

1 hc

e λ kT − 1

=

5 1 (kT ) . ⋅170π 2 (hc)4

1

= 3.82 ×

hc

e λ ′ kT − 1

Numerically,

hc

λ ′k λ ′′k

In terms of x, the preceding equation becomes

170 x5 . = e x − 1 16

Solutions are x1 = 2.736 ; x2 = 8.090 . Since, for λ max , x = 4.965 , these solutions give λ1 =1.815 λmax , λ 2 = 0.614λmax …….## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-22、A tungsten sphere 2.30cm in diameter is heated to 2000℃. At this temperature tungsten -7-

so that

e e

35867 T 71734 T

Let x = e

35867 T

−1 −1

1

⇒

hc

eλ ′′ kT − 1

e λ ′′kT −1 hc

e λ ′ kT −1

= 3.82 × (

λ′ 5 ) λ ′′

=

(6.626 × 10 −34 )(2.988× 108 ) = 71734K −7 − 23 (2 × 10 )(1.38× 10 )

=

(6.626 × 10− )(2.988× 10 ) = 35867K (4 × 10− 7)(1.38× 10− 23 )

34

hc

.

1 λ ′′ 5

8

1 = 3.82 × ( )5 = 0.1194 2

; then

35867 1 x− 1 ⇒ x = 7.375 = e T = 0.1194 = 2 x −1 x+ 1

-8-

量子物理習題 CH01

量子物理習題 CH01

T=

35867 = 17950K …………## ln 7.375

所以

NOTE：課本解答 Appendix S，S-1 為18020 0K 。 ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~

d ρ (λ ) d 8π = [ 5 dλ dλ λ

hc e

hc λ kT

−5(e

= 8π hc[

] = 8π hc

−1

hc λ kT

d λ −5 [ ] d λ λhckT e −1

−1)λ −6 + λ −5

hc

hc

λ kT

eλ kT ] ……代入(1)式

hc

(eλ kT − 1)2 其他補充題目

hc

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-001、(a) 簡單敘述黑體輻射的性質，並說明空槍的哪一部份可代表黑體。(b) 黑體輻射的輻 c 8π hc 射強度為 R( λ ) = ( 5 ) 4 λ

1 hc
...