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Chapter 3: Q14E (page 93)

According to Wien's Law, the wavelength $位_{m a x}$ of maximum thermal emission of electromagnetic energy from a body of temperature Tobeys
localid="1660036169367" $位_{m a x} T = 2.898 脳 10^{- 3} m 路 K$
Show that this law follows from the spectral energy density obtained in Exercise 13. Obtain an expression that, when solved, would yield the wavelength at which this function is maximum. The transcendental equation cannot be solved exactly, so it is enough to show thatlocalid="1660036173306" $位 = 2.898 脳 10^{- 3} m 路 \frac{K}{T}$ solves it to a reasonable degree of precision.

Short Answer

Expert verified

The Wien's law, $位_{\max} T = 2.9 脳 10^{- 3} mK$ .

Step by step solution

01

Given data

Wein鈥檚 Law and spectral energy density.

02

Formula used

Planck's formula in terms of wavelength

$\frac{d U}{d 位} = \frac{8 蟺 V h c}{(e^{h c / 位 k_{B} T} - 1)} 脳 \frac{1}{位^{5}}$

Some useful values:

$h = 6.63 脳 10^{- 34} J 路 s, k_{B} = 1.38 脳 10^{- 23} J 路 K^{- 1}, a n d c = 3 脳 10^{8} m 路 s^{- 1}$

03

Calculation using Planck’s formula

To find the peak of the blackbody radiation, the derivative should be equal to ' 0 '. At that point, the value of $位$ will correspond to the maximum. The constant term will not affect to the peak of the radiation

$\frac{d}{d 位} [\frac{8 蟺 V h c}{(e^{h c / 位 k_{B} T} - 1)} 脳 \frac{1}{位^{5}}] = 0$

Solving the above will give,

Therefore,

$\frac{x e^{x}}{(e^{x} - 1)} - 5 = 0 (x - 5) e^{x} + 5 = 0$

Solving this equation, $x = 4.9651$

Therefore,

$\begin{array}{rcl} 位_{\max} T & = & \frac{h c}{x k_{B}} \\ = & \frac{(6.63 脳 10^{- 34} J 路 s) 脳 (3 脳 10^{8} m 路 s^{- 1})}{4.9651 脳 1.38 脳 10^{- 23} J 路 K^{- 1}} \\ = & 2.9 脳 10^{- 3} m 路 K \end{array}$

04

Conclusion

The Wien's law, $位_{\max} T = 2.9 脳 10^{- 3} mK$ .

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Most popular questions from this chapter

A ball rebounds elastically from the floor. What doesthis situation share with the ideas of momentum conservation discussed in connection with pair production?

You are conducting a photoelectric effect experiment by shining a light of 500 nmwavelength at a piece of metal and determining the stopping potential. If, unbeknownst to you, your 500 nm source actually contained a small amount of ultraviolet light, would it throw off your results by a small amount, or by quite a bit? Explain.

Show that the angles of scatter of the photon and electron in the Compton effect are related by the following formula:

$c o t \frac{胃}{2} = (1 + \frac{h}{m c 位}) t a n 蠒$

The electromagnetic intensity of all wavelengths thermally radiated by a body of temperature T is given by

$I = 蟽 T^{4}$ where $蟽 = 5.66 脳 10^{- 8} W 路 m^{2} 路 K^{4}$

This is the Stefan-Boltzmann Law. To derive it. show that the total energy of the radiation in a volume V attemperature T is $U = 8 蟺^{5} k_{B}^{4} V T^{4} / 15 h^{3} c^{3}$ by integrating Planck's spectral energy density over all frequencies. Note that

${鈭�}_{0}^{鈭�} \frac{x^{3}}{e^{x} - 1} d x = \frac{蟺^{4}}{15}$

Intensity, or power per unit area, is then the product of energy per unit volume and distance per unit time. But because the intensity is a flow in a given direction away from the blackbody, c is not the correct speed. For radiation moving uniformly in all directions, the average component of velocity in a given direction is $\frac{1}{4} c$ .

From equations (3-4) to (3-6) obtain equations (3.8). It is easiest to start by eliminating 蠁 between equations (3-4) and (3-5) using cos²蠁+蝉颈苍²蠁 = 1 The electron speed u may then be eliminated between the remaining equations.

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