## Example from Summer Test #1, Question 4a [ENDORSED]

$H_{\psi }=E_{\psi }$

1-D: $E_{TOTAL}\psi (x)=E_{k}\psi (x)+V(x)\psi(x)=-\frac{h^{2}}{8\pi ^{2}m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$

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### Example from Summer Test #1, Question 4a

In the question, we're told the electron transitions from n=3 to n=1. So, our initial is 3 and final is 1. We should apply the concept, Ef-Ei. I am confused because in the answer, the way it was solved shows ...(1/3^2 - 1/1^2). Why is this?
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Chem_Mod
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### Re: Example from Summer Test #1, Question 4a  [ENDORSED]

Diego Zavala 2I
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### Re: Example from Summer Test #1, Question 4a

Because v(frequency)=R[(1/n(1)^2)-(1/n(2)^2)], so v=R[(1/(1)^2)-(1/(3)^2)] and, if multiplied by -h the equation can be rewritten as E=-hR[(1/3^2)-(1/1^2)].