[CHM 330] Question about the worksheet

[Andrew J. Pounds]

So the eigenvalues for the particle in a box are


En=n2h28mL2E_n=\frac{n^2 h^2}{8 m L^2}

For the 1D particle in a box problem from the worksheet we needed the 
LUMO and HOMO transition energy. The LUMO (n=6) to the HOMO (n=5) 
transition energy is given by the difference of these two energy levels...


E6−E5=62h28mL2−52h28mL2=(62−52)h28mL2E_6-E_5=\frac{6^2h^2}{8mL^2}-\frac{5^2 
h^2}{8mL^2}=(6^2-5^2)\frac{h^2}{8mL^2}

On 12/1/23 00:01,  wrote:
> On number 1 of the worksheet we solved from LUMO to HOMO and our n 
> value was 6^2-5^2. If we are asked to solve from HOMO to LUMO would we 
> then plug in 5^2 -6^2 or would it just be the higher number minus the 
> lower?
>
> Thank you,
>
>
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-- 
*/Andrew J. Pounds, Ph.D./*
/Professor of Chemistry and Computer Science/
/Director of the Computational Science Program/
/Mercer University/
/1501 Mercer University Drive, Macon, GA 31207 /
/(478) 301-5627/
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