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Alisha Vira edited Appendix.tex
over 8 years ago
Commit id: d33dd81e7d47a8320f97882acf18bb7daf572e55
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$$\Delta E_{n}=n(E_{a}+n\frac{\lambda}{L} E_{a})-(n-1)(E_{a}+(n-1)\frac{\lambda}{L} E_{a})$$
$$\Delta E_{n}=E_{a}\frac{\lambda}{L} (2n-1)+E_{a}$$
Therefore, when n=$\frac{1}{2}$:
$$\Delta_{E_{n}}=E_{a}$$ $$\Delta E_{n}=E_{a}$$
In other words:
$$E_{a}=\Delta E_{(0.5)}}$$