Clase 004

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Chapter 4: Blackbody radiation and the birth of quantum physics: Problems

Luis M. Molina

Departamento de Fsica Teorica, Atomica y Optica

Quantum Physics

Luis M. Molina (FTAO) Chapter 3 Quantum Physics 1 / 5
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Problem 1.1

QuestionExpress Plancks formula for the density of energy T() in a blackbodycavity in terms of the wavelenght . Then, prove Wiens displacementlaw, calculating Wiens constant.

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Problem 1.1

QuestionExpress Plancks formula for the density of energy T() in a blackbodycavity in terms of the wavelenght . Then, prove Wiens displacementlaw, calculating Wiens constant.

SolutionWe start from:

T() =82

c3

h

eh/kT 1(1)

Then, as = c/, we have:

T() = T()

dd

= 8h c3

c33c

21

ehckT 1=

8hc

51

ehc/kT 1(2)

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Problem 1.1

Solution (cont.)Now, to find the wavelength max for which T() reaches a maximum, we equalthe derivative of the function to 0:

T()

= 8hc

hc

kT7ehc/kT

`ehc/kT 12

1

65

ehc/kT 1! = 0=

hc

kT

ehc/kT

ehc/kT 1 5 = 0 (3)

Now taking x = hc/kT, we need to solve the equationxex

ex 1= 5.

This can only be solved numerically, with the result x = 4,965114. Then, finally:

maxT =hc

xk=

6,626 1034J s 2,9979 108m/s

4,965114 1,3806 1023J/K= 2,897 103m K (4)

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Problem 1.2

QuestionA cavity at 6000K has an energy distribution corresponding to a blackbody. We make asmall hole in it with 1mm diameter. Calculate the power irradiated through the hole in thewavelength interval between 5500 and 5510 A, as well as the total power

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Problem 1.2

QuestionA cavity at 6000K has an energy distribution corresponding to a blackbody. We make asmall hole in it with 1mm diameter. Calculate the power irradiated through the hole in thewavelength interval between 5500 and 5510 A, as well as the total power

Solution

As the wavelength interval is rather small, we can approximate the total backbody radiancywithin that interval as:

RT =

ZBA

c

4

8hc

5d

ehc/kT 1

2hc2

5Mehc/MkT 1

() (5)

with = B A and M = (B + A)/2

Then, the total power will be the radiancy times the area of the hole:

P =2 6,626 1034J s (3 108)m2/s2

(0,5505 106m)51

e4,359 1 109m (106 m2) = 0,301W

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Problem 1.3

Solution (cont.)

For the total power, we use Stefans law:

RT = T4 = 5,6704 108 Js1m2K4 (6000K)4 = 7,3488 107 W/m2

P = RT A = 7,3488 107W/m2 (3,1416 106m2) = 230,87W

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Problem 1.3

Solution (cont.)

For the total power, we use Stefans law:

RT = T4 = 5,6704 108 Js1m2K4 (6000K)4 = 7,3488 107 W/m2

P = RT A = 7,3488 107W/m2 (3,1416 106m2) = 230,87W

QuestionIn a thermonuclear explosion the temperature in the fireball is momentarily 107K. Find thewavelength at which the radiation emitted is a maximum.

SolutionWe apply Wiens law, obtaining:

max =2,897 103m K

107K= 2,897 1010m = 2,897 A

which represent rather energetic X-rays.

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