Chapter 5: Q30P (page 244)

(a)Use Equation5.113 to determine the energy density in the wavelength range $d λ$ . Hint: set $ρ (ω) dω = \bar{ρ} (λ) dλ$ , and solve for $\bar{ρ (λ)}$
(b)Derive the Wien displacement law for the wavelength at which the blackbody energy density is a maximum
$λ_{\max} = \frac{2.90 \times 10^{- 3} mK}{T}$
You'll need to solve the transcendental equation $(5 \times x) = 5 e^{- x}$ , using a calculator (or a computer); get the numerical answer accurate to three significant digits.

Short Answer

Expert verified

$(a) \bar{ρ} (λ) = \frac{16 π^{2} hc}{λ^{5} (e^{2 hc / k_{e} Tλ} - 1)} (b) λ_{\max} = \frac{2.90 \times 10^{- 3} mK}{T}$

Step by step solution

Given

The energy density in wavelength range is given by:

$ρ (ω) dω = \bar{ρ} (λ) dλ$

Determining the energy density in the wavelength rangedλ

$ω = 2 π c = \frac{2 π c}{λ}, so dω = - \frac{2 πc}{λ^{2}} dλ, and ρ (ω) = \frac{h}{π^{2} c^{3}} \frac{{(2 πc)}^{3}}{λ^{3} (e^{2 πhc / k_{B} Tλ} - 1)} ρ (ω) |dω| = 8 πh \frac{1}{λ^{3} (e^{2 πhc / k_{B} Tλ} - 1) |- \frac{2 πc}{λ^{2}} dλ|} = \bar{ρ} (λ) dλ \bar{ρ (λ) = \frac{16 π^{2} hc}{λ^{3} (e^{2 πhc / k_{B} Tλ} - 1)}}$

For density we want only the size of the interval, not its sign)

Deriving the Wien displacement law for wavelength

To find maximize, we need to calculate:

d \bar{ρ} / dλ = 0; 0 = 16 π^{2} hc [\frac{5}{λ^{6} (e^{2 πhc / k_{B} Tλ} - 1)} - \frac{e^{2 πhc / k_{B} Tλ} (2 πhc / k_{B} T)}{λ^{6} {(e^{2 πhc / k_{B} Tλ} - 1)}^{2}} (- \frac{1}{λ^{2}})] A = \frac{2 πhc}{k_{B} T} 5 = \frac{A}{λ} \frac{e^{A / γ}}{e^{A / γ} - 1} 5 e^{- A / λ} = 5 - \frac{A}{λ}

Using mathematic $\frac{A}{λ} = 4.966$

$λ_{\max} = \frac{A}{4.966} = \frac{2 π hc}{4.966 k_{B} T} λ_{\max} = \frac{2.90 \times 10^{- 3} mK}{T}$

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Short Answer

Step by step solution

Given

Determining the energy density in the wavelength rangedλ

Deriving the Wien displacement law for wavelength

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