hce_nchu
115年
物理
第 47 題
In 1900, the German physicist Max Planck formulated an equation that fit the observed radiance curves for blackbody radiation at all wavelengths: $R(\lambda, T) = \frac{2\pi hc^2}{\lambda^5\left(e^{\frac{hc}{\lambda kT}}-1\right)}$, where $h$ is Planck’s constant, $c$ is the speed of light, $\lambda$ is wavelength, $k$ is Boltzmann’s constant, and $T$ is the absolute temperature. If we integrate the equation over all wavelengths to calculate the total power radiated per unit area, the Stefan-Boltzmann law for blackbody radiation can be derived. Assume $A$ is the area of the radiating surface. Note that $\int_0^\infty \frac{x^n}{e^x - 1} dx = \Gamma(n+1)\zeta(n+1)$, where $\Gamma(n)$ is the gamma function and $\zeta(n)$ is the Riemann zeta function. For references, $\Gamma(n) = (n-1)!$, $\zeta(0) = -\frac{1}{2}$, $\zeta(2) = \frac{\pi^2}{6}$, $\zeta(4) = \frac{\pi^4}{90}$, $\zeta(6) = \frac{\pi^6}{945}$, and $\zeta(8) = \frac{\pi^8}{9450}$. What is the Stefan-Boltzmann law for blackbody radiation?
- A $P_{\text{blackbody}} = \frac{2\pi^3 k^4}{15c^2 h^3}AT^4$
- B $P_{\text{blackbody}} = \frac{2\pi^5 k^4}{15c^2 h^3}AT^4$
- C $P_{\text{blackbody}} = \frac{2\pi^7 k^4}{15c^2 h^3}AT^4$
- D $P_{\text{blackbody}} = \frac{2\pi k^4}{5c^2 h^3}AT^3$
- E $P_{\text{blackbody}} = \frac{2\pi^3 k^4}{5c^2 h^3}AT^5$
思路引導 VIP
在處理這個複雜的積分式時，請觀察分母的指數項 $e^{\frac{hc}{\lambda kT}}$。若我們想將積分化簡為題目提供的標準形式 $\int \frac{x^n}{e^x-1} dx$，你認為應該將哪一串物理量定義為新變數 $x$？而在完成變數代換（包含 $d\lambda$ 的轉換）後，原本分散在各處的溫度 $T$ 最後會組合出幾次方呢？
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AI 詳解
AI 專屬家教
太棒了！你能準確選出 (B)，代表你對近代物理的數學推導有著相當紮實的掌握。這道題目要求我們從普朗克輻射定律 (Planck’s Law) 出發，透過積分求得史蒂芬－波茲曼定律 (Stefan-Boltzmann Law)，是物理學史上連結量子假說與經典熱力學的重要里程碑。
從積分推導物理定律
解題的核心在於變數變換。令 $x = \frac{hc}{\lambda kT}$，則微元項 $d\lambda = -\frac{hc}{x^2 kT} dx$。當我們將這些代換進輻射強度公式並對全波長進行積分時，溫度 $T$ 會因為次方的抵消與結合，最終以 $T^4$ 的形式被提取出來。結合題目提供的特殊函數值，其中 $\Gamma(4) = 3! = 6$ 且 $\zeta(4) = \frac{\pi^4}{90}$，積分項的結果恰好為 $\frac{\pi^4}{15}$，代回常數項後便能得出正確的係數：$$\frac{2\pi hc^2 \cdot (kT)^4}{(hc)^4} \cdot \frac{\pi^4}{15} = \frac{2\pi^5 k^4}{15c^2 h^3}$$
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