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Frostbite
 3 years ago
stuff
Frostbite
 3 years ago
stuff

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Frostbite
 3 years ago
Best ResponseYou've already chosen the best response.0Work so far; \[C _{V,m}=\left( \frac{ \partial U _{m} }{ \partial T } \right)_{V}=\frac{ N _{a} \left( \langle \epsilon ^{2} \rangle  \langle \epsilon \rangle ^{2} \right) }{ kT ^{2} }\] \[\langle \epsilon \rangle=\sum_{i}^{n}p _{i} \epsilon _{i}\] \[\langle \epsilon \rangle=\frac{ 4 }{ q } \epsilon _{0}+\frac{ 2e ^{ \beta \epsilon} }{ q } \epsilon=\frac{ 2e ^{ \beta \epsilon} }{ q } \epsilon\] \[\langle \epsilon ^{2} \rangle=\sum_{i=0}^{1} p _{i} \epsilon ^{2} _{i}=\frac{ 4 }{ q } \epsilon _{0} ^{2}+\frac{ 2e ^{ \beta \epsilon} }{ q } \epsilon ^{2}=\frac{ 2e ^{ \beta \epsilon} }{ q } \epsilon ^{2}\\\] \[C _{V,m}=\frac{ N _{A}\left( \frac{ 2e ^{ \beta \epsilon} }{ q } \epsilon ^{2}\left( \frac{ 2e ^{ \beta \epsilon} }{ q } \epsilon \right)^{2} \right) }{ kT ^{2} }\] Now from here things are going down hill. I know I need to substitute the equation: \[E=\frac{ hc }{ \lambda } \Leftrightarrow \epsilon=hc \bar{v}\] But at the same time I like to make the expression more simple.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0if your notation is correct, i dont see way that e^2 cant be factored out of the top

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0other than that, i got no idea what the setup is doing :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0its all a bit above my reading abilities :/ srry
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