Chapter 3: Q14E (page 93)

According to Wien's Law, the wavelength $λ_{m a x}$ of maximum thermal emission of electromagnetic energy from a body of temperature Tobeys
localid="1660036169367" $λ_{m a x} T = 2.898 \times 10^{- 3} m \cdot K$
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" $λ = 2.898 \times 10^{- 3} m \cdot \frac{K}{T}$ solves it to a reasonable degree of precision.

Short Answer

Expert verified

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

Step by step solution

Given data

Wein’s Law and spectral energy density.

Formula used

Planck's formula in terms of wavelength

$\frac{d U}{d λ} = \frac{8 π V h c}{(e^{h c / λ k_{B} T} - 1)} \times \frac{1}{λ^{5}}$

Some useful values:

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

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 $λ$ will correspond to the maximum. The constant term will not affect to the peak of the radiation

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

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,

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

Conclusion

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

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Short Answer

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Given data

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Calculation using Planck’s formula

Conclusion

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