infamous Posted November 4, 2006 Report Posted November 4, 2006 [math]\sqrt[4]{\frac6\pi}[/math]=1.17557500734 [math]\sqrt[3]{10}[/math]=2.15443469003 [math]\alpha=\left(\frac{e^2}{\hbar c}\right)[/math] fine structure constant [math]re=\left(\frac{\sqrt[4]{\frac6\pi}\times\sqrt[3]{10}^{25}}{\ c^{2}\right)[/math] radius of the electron in cgs units [math]re=\left(\frac{\sqrt[4]{\frac6\pi}\times\sqrt[3]{10}^{19}}{\ c^{2}\right)[/math] radius of the electron in SI units [math]me=\left(\frac{\left(\2\right)^4\times\sqrt[4]{\frac6\pi}\times\sqrt[3]{10}^{-245}\times\left(\pi\right)^2}{\ h^2}\right)[/math] mass of the electron in cgs units [math]me=\left(\frac{\left(\2\right)^4\times\sqrt[4]{\frac6\pi}\times\sqrt[3]{10}^{-296}\times\left(\pi\right)^2}{\ h^2}\right)[/math] mass of the electron in SI units Quote
ronthepon Posted November 6, 2006 Report Posted November 6, 2006 Nernst equation? Let me try... [math]{\Delta}G = -W[/math] [math]{\Delta}G = -q{E^{o}}[/math] but since[math]q = nF[/math] [math]{\Delta}G = -nFE[/math] [math]{\Delta}{G^{o}} = -nF{E^{o}}[/math] For[math]aA {\,}+{\,} bB {\rightarrow} cC {\,}+{\,} dD[/math] [math]Q = {\frac{{[C]^{c}}{[D]^{d}}}{{[A]^{a}}{^{b}}}}[/math] And because [math]{\Delta}G = {\Delta}{G^{o}} + RT{{Ln}Q}[/math] [math]-nFE = -nF{E^{o}} + RT{{Ln}{\frac{{[C]^{c}}{[D]^{d}}}{{[A]^{a}}{^{b}}}}}[/math] or, [math]E = {E^{o}} - {{\frac{RT}{nF}} {{Ln}{\frac{{[C]^{c}}{[D]^{d}}}{{[A]^{a}}{^{b}}}}}}[/math] Wow! this is some amount of work for something that would take me five seconds on paper! Quote
hallenrm Posted November 7, 2006 Report Posted November 7, 2006 Hey, I am interested in learning Latex, can someone tell me how to begin :) Quote
ronthepon Posted November 7, 2006 Report Posted November 7, 2006 see http://hypography.com/forums/tutorials-how-tos/6457-how-use-latex-equations.html http://www.forkosh.com/mimetextutorial.html http://hypography.com/forums/physics-mathematics/6576-latex-fomulas-math-v2-0-a.html Quote
ronthepon Posted November 11, 2006 Report Posted November 11, 2006 [math]t= {\frac{{-\alpha}{+}{\sqrt{\alpha^2 + 4E\beta}}}{4 \beta}}[/math] [math]t= {\frac{{-\alpha}{-}{\sqrt{\alpha^2 + 4E\beta}}}{4 \beta}}[/math] Quote
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