Question

stuff

Answer 1

Work 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.

Answer 2

if your notation is correct, i dont see way that e^2 cant be factored out of the top

Answer 3

other than that, i got no idea what the setup is doing :)

Answer 4

its all a bit above my reading abilities :/ srry

Answer 5

@abb0t