Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Frostbite

  • 3 years ago

stuff

  • This Question is Closed
  1. Frostbite
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  2. amistre64
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. amistre64
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. amistre64
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    its all a bit above my reading abilities :/ srry

  5. Frostbite
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @abb0t

  6. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy