Javion Henry

Answered

2022-07-27

Solve the equation.
${T}_{2}=$
${T}_{1}=300K,k={1.38066.10}^{-23}$
${T}_{1}^{2}{e}^{\frac{-0.8eV}{k\ast {T}_{1}}}={T}_{2}^{2}\ast {e}^{\frac{-0.4eV}{k\ast {T}_{2}}}$

Answer & Explanation

Rihanna Robles

Expert

2022-07-28Added 18 answers

The LHS of this eqn consists of all known constants, so youcan evaluate it and get a constant ${C}_{1}$ .
The RHS can be written in the form ${x}^{2}\mathrm{exp}\left({C}_{2}/x\right)$ , where x is the temp ${T}_{2}$ and ${C}_{2}=-0.4/k$
Form the eqn ${C}_{1}-{x}^{2}\mathrm{exp}\left({C}_{2}/x\right)=0$ and put it intothe "solver" mode on a TI-83 plus or TI-84 plus to solve forx.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?