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Display information for equation id:math.2555.54 on revision:2555

* Page found: Das ideale Fermigas (eq math.2555.54)

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\begin{align}

& \bar{N}=\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2mkT \right)}^{\frac{3}{2}}}\int_{0}^{\infty }{{}}dy\frac{{{y}^{\frac{1}{2}}}}{\left( {{e}^{y-\eta }}+1 \right)}=\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2mkT \right)}^{\frac{3}{2}}}\left[ \frac{2}{3}{{\left( \frac{\mu }{kT} \right)}^{\frac{3}{2}}}+\frac{{{\pi }^{2}}}{12}{{\left( \frac{\mu }{kT} \right)}^{-\frac{1}{2}}} \right] \\

& \Rightarrow \bar{N}=\frac{2}{3}\frac{\left( 2s+1 \right)}{2}\left( \frac{V}{{{h}^{3}}} \right)4\pi {{\left( 2m\mu  \right)}^{\frac{3}{2}}}\left[ 1+\frac{{{\pi }^{2}}}{8}{{\left( \frac{kT}{\mu } \right)}^{2}} \right] \\

\end{align}

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N ¯ = ( 2 ⁢ s + 1 ) 2 ⁢ ( V h 3 ) ⁢ 4 ⁢ π ⁢ ( 2 ⁢ m ⁢ k ⁢ T ) 3 2 ⁢ ∫ 0 ∞ 𝑑 y ⁢ y 1 2 ( e y - η + 1 ) = ( 2 ⁢ s + 1 ) 2 ⁢ ( V h 3 ) ⁢ 4 ⁢ π ⁢ ( 2 ⁢ m ⁢ k ⁢ T ) 3 2 ⁢ [ 2 3 ⁢ ( μ k ⁢ T ) 3 2 + π 2 12 ⁢ ( μ k ⁢ T ) - 1 2 ] ⇒ N ¯ = 2 3 ⁢ ( 2 ⁢ s + 1 ) 2 ⁢ ( V h 3 ) ⁢ 4 ⁢ π ⁢ ( 2 ⁢ m ⁢ μ ) 3 2 ⁢ [ 1 + π 2 8 ⁢ ( k ⁢ T μ ) 2 ] absent ¯ 𝑁 2 𝑠 1 2 𝑉 superscript ℎ 3 4 𝜋 superscript 2 𝑚 𝑘 𝑇 3 2 superscript subscript 0 differential-d 𝑦 superscript 𝑦 1 2 superscript 𝑒 𝑦 𝜂 1 2 𝑠 1 2 𝑉 superscript ℎ 3 4 𝜋 superscript 2 𝑚 𝑘 𝑇 3 2 delimited-[] 2 3 superscript 𝜇 𝑘 𝑇 3 2 superscript 𝜋 2 12 superscript 𝜇 𝑘 𝑇 1 2 absent ⇒ absent ¯ 𝑁 2 3 2 𝑠 1 2 𝑉 superscript ℎ 3 4 𝜋 superscript 2 𝑚 𝜇 3 2 delimited-[] 1 superscript 𝜋 2 8 superscript 𝑘 𝑇 𝜇 2 {\displaystyle{\displaystyle\begin{aligned} \par&\displaystyle\bar{N}=\frac{% \left(2s+1\right)}{2}\left(\frac{V}{{{h}^{3}}}\right)4\pi{{\left(2mkT\right)}^% {\frac{3}{2}}}\int_{0}^{\infty}{{}}dy\frac{{{y}^{\frac{1}{2}}}}{\left({{e}^{y-% \eta}}+1\right)}=\frac{\left(2s+1\right)}{2}\left(\frac{V}{{{h}^{3}}}\right)4% \pi{{\left(2mkT\right)}^{\frac{3}{2}}}\left[\frac{2}{3}{{\left(\frac{\mu}{kT}% \right)}^{\frac{3}{2}}}+\frac{{{\pi}^{2}}}{12}{{\left(\frac{\mu}{kT}\right)}^{% -\frac{1}{2}}}\right]\\ \par&\displaystyle\Rightarrow\bar{N}=\frac{2}{3}\frac{\left(2s+1\right)}{2}% \left(\frac{V}{{{h}^{3}}}\right)4\pi{{\left(2m\mu\right)}^{\frac{3}{2}}}\left[% 1+\frac{{{\pi}^{2}}}{8}{{\left(\frac{kT}{\mu}\right)}^{2}}\right]\\ \par\end{aligned}}}

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N¯=(2s+1)2(Vh3)4π(2mkT)32∫0∞dyy12(ey−η+1)=(2s+1)2(Vh3)4π(2mkT)32[23(μkT)32+π212(μkT)−12]⇒N¯=23(2s+1)2(Vh3)4π(2mμ)32[1+π28(kTμ)2]

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