Question
Find the binding energy of a hydrogen atom in the state n = 2.
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Answer
$\text{n}_1=2,\ \text{n}_2=\infty$
$\text{E}=\frac{-13.6}{\text{n}^2_1}-\frac{-13.6}{\text{n}^2_2}$
$\text{E}=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$
$\text{E}=13.6\Big(\frac{1}{\infty}-\frac{1}{4}\Big)$
$\text{E}=\frac{-13.6}{4}=-3.4\text{eV}$
$\text{E}=\frac{-13.6}{\text{n}^2_1}-\frac{-13.6}{\text{n}^2_2}$
$\text{E}=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$
$\text{E}=13.6\Big(\frac{1}{\infty}-\frac{1}{4}\Big)$
$\text{E}=\frac{-13.6}{4}=-3.4\text{eV}$
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