hw6solv2

# hw6solv2 - Homework#6 Solutions Question 1 Kittel Kroemer...

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Homework #6 Solutions Question 1) Kittel+Kroemer, Chapter 4, Problem 1. The total number of thermal photons is the sum of photons in each mode: N = X n < s n > = X n 1 e ~ ω n - 1 (1) Replace the sum over n with an integral, similar to the calcualtion in the text on pages 93 and 94. X n 2 × 1 8 Z 0 d n 4 πn 2 (2) The factor of 2 takes care of the two polarizations of light. The factor of 1/8 comes from integrating only over one quadrant (in 3 dimensional n-space). In the integrand ω n = nπc/L . To bring the integral to a dimensionless form we make the change of variables x = π ~ c n . Thus the integral becomes: N = π ± ~ πc ² 3 Z 0 d x x 2 e x - 1 (3) We note that the integral is a known one and equals 2 ζ (3) = 2 . 404. So the ﬁnal answer reads N = 2 . 404 V π 2 ± τ ~ c ² 3 (4) Question 2a) Kittel+Kroemer, Chapter 4, Problem 2. The solar constant gives the power per unit area, which is placed a distance R e - s = 1 . 5 × 10 23 cm away from the sun (the distance from the sun to earth). To get the total power output of the sun we have to integrate over the area over which this radiation is spread: 4 πR 2 e - s . P sun = 4 πR 2 e - s × 0 . 136 J s · cm 2 = 3 . 85 × 10 26 J/s (5) b) To use the Stefan-Boltzmann Law we need the power radiated per area. This can be easily obtained using the radius of the sun r s = 7 × 10 10 cm . J sun = P sun 4 πr 2 s = 6 . 2 × 10 3 J s · cm 2 (6) The Stefan-Boltzmann Law reads J sun = σ B T 4 sun . Solving for the temperature we get T sun = ± J sun σ B ² 1 / 4 = 5750 K (7) 1

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Question 3) Kittel+Kroemer, Chapter 4, Problem 5. For this question we could readily use some results from the previous problem. But we prefer to do it from scratch in order to get familiarity with the methods. The Earth is ﬁrst assumed to be a black body absorber in equilibrium with the radiation from the sun. This means that the earth will radiate the same amount of energy it receives from the sun. We will start by calculating the power of the radiation from the sun using Stefan-Boltzmann law P ± = J ± × 4 πR 2 ± = ( σ B T 4 ± ) × 4 πR 2 ± (8) Note that the circle with a dot in it ( ± ) is a symbol for the sun. The ﬂux density at earth’s orbit is J E = P ± 4 πR 2 e - s = J ± ± R ± R e - s ² 2 (9) The ﬂux absorbed by earth is this value multiplied by the cross section of the Earth (which is just πR 2 E ): Φ in = J E × πR 2 E . In equilibrium this has to be equal to the energy radiated by earth, which again is given by the Stefan-Boltzmann
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hw6solv2 - Homework#6 Solutions Question 1 Kittel Kroemer...

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