Chapter 3: Q13E (page 93)

Equation (3-1) expresses Planck's spectral energy density as an energy per range df of frequencies. Quite of ten, it is more convenient to express it as an energy per range of wavelengths, By differentiating $f = C / λ$ we find that $d f = - C / λ^{2} d λ$ . Ignoring the minus sign (we are interested only in relating the magnitudes of the ranges df and $d λ$ ). show that, in terms of wavelength. Planck's formula is
$\frac{d U}{d λ} = \frac{8 π V h c}{(e^{h c / λ k_{B} T} - 1)} \times \frac{1}{λ^{5}}$

Short Answer

Expert verified

Planck's formula in terms of wavelength,

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

Step by step solution

Given data

On differentiating, $f = \frac{c}{λ}$ , we get $d f = \frac{- c}{λ^{2}} d λ$ .

Formula used

Planck's formula is given by

$\frac{d U}{d f} = \frac{h f}{e^{h f / K_{B} T} - 1} \times \frac{8 π V}{c^{3}} f^{2}$

Calculation using Planck’s formula

Rearranging Planck's formula:

$d U = \frac{h f}{e^{h f / K_{B} T} - 1} \times \frac{8 π V}{c^{3}} f^{2}$

By substituting the given information,

$d U = \frac{h f}{e^{h f / h_{B} T} - 1} \times \frac{8 π V}{c^{3}} \times \frac{c^{2}}{λ^{2}} \times \frac{c}{λ^{2}} d λ \frac{d U}{d λ} = \frac{8 π V h c}{e^{h f / λ k_{B} T} - 1} \times \frac{1}{λ^{5}}$

Conclusion

Planck's formula in terms of wavelength,

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

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Short Answer

Step by step solution

Given data

Formula used

Calculation using Planck’s formula

Conclusion

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