The expression for I ( f ) , the Planck distribution law in terms of frequency f .

Question

Want to see the full answer?

Answer 1

Question

Chapter 39, Problem 39.61P

(a)

To determine

The expression for I(f) , the Planck distribution law in terms of frequency f .

(b)

To determine

To show: the integral ∫0∞I(λ)=2π5k415c2h3T4 .

(c)

To determine

The constants in part (b) and to show that σ has the value given in Section 39.5 .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

A blackbody is an object with a radiation spectrum that is dependent solely on its tempera- ture. A blackbody spectrum (or spectral radiancy curve) is described by the Planck Radiation Law. (a) i. Sketch the spectral radiancy curves for blackbodies with temperatures of T = 4000 K and T = 6000 K, respectively. Describe the main differences between the two curves in terms of the appropriate physical laws defined as a function of tempera- ture. ii. What is the wavelength at peak intensity for each blackbody? State the part of the electromagnetic spectrum to which each wavelength belongs. (b) Use the Planck Radiation Law to determine the power radiated per unit area between the wavelengths A 500 nanometres and λ = 503 nanometres for the T 6000 K blackbody. What fraction of the blackbody's radiancy lies in this wavelength range? =

Through what potential difference ΔVΔV must electrons be accelerated (from rest) so that they will have the same wavelength as an x-ray of wavelength 0.130 nmnm? Use 6.626×10−34 J⋅sJ⋅s for Planck's constant, 9.109×10−31 kgkg for the mass of an electron, and 1.602×10−19 CC for the charge on an electron. Express your answer using three significant figures. =89.0 V Through what potential difference ΔVΔV must electrons be accelerated so they will have the same energy as the x-ray in Part A? Use 6.626×10−34 J⋅sJ⋅s for Planck's constant, 3.00×108 m/sm/s for the speed of light in a vacuum, and 1.602×10−19 CC for the charge on an electron. Express your answer using three significant figures. Second question is what I need help on! Thanks!

The wavelength λmax at which the Planck distribution is a maximum can be found by solving dρ(λ,T)/dT = 0. Differentiate ρ(λ,T) with respect to T and show that the condition for the maximum can be expressed as xex − 5(ex − 1) = 0, where x = hc/λkT. There are no analytical solutions to this equation, but a numerical approach gives x = 4.965 as a solution. Use this result to confirm Wien’s law, that λmaxT is a constant, deduce an expression for the constant, and compare it to the value quoted in the text.