- $A\left(n+\frac{1}{2}\right)$
- $A(1-n^2)$
- $A\left(\frac{1}{n^2}-\frac{1}{4}\right)$
- $An^2$
- $-\frac{A}{n^2}$

Solution :

The energy levels of the hydrogen atom are given in terms of the principal quantum number $n$ and a positive constant $A$ by the expression:

**Choice 5 is the answer.**

Recall that the energy levels of the hydrogen atom is given by \[E_n = -\frac{13.6\,\text{eV}}{n^2}.\] It's just one of those things that you've got to remember, as it takes time you don't have to derive it from first principals (I only ever remember that $E_n\propto \frac{1}{n^2}\,$ since the constant of 13.6 eV is usually unimportant).

- $A\left(n+\frac{1}{2}\right)$
- $A(1-n^2)$
- $A\left(\frac{1}{n^2}-\frac{1}{4}\right)$
- $An^2$
- $-\frac{A}{n^2}$

Solution :

Recall that the energy levels of the hydrogen atom is given by \[E_n = -\frac{13.6\,\text{eV}}{n^2}.\] It's just one of those things that you've got to remember, as it takes time you don't have to derive it from first principals (I only ever remember that $E_n\propto \frac{1}{n^2}\,$ since the constant of 13.6 eV is usually unimportant).

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