Hydrogen levels, Balmer and Lyman

Problem

Draw the diagram of the first four energy levels of the hydrogen atom ($n = 1, 2, 3, 4$) to correct scale using the values $E_n = -13{,}6/n^2\;\text{eV}$, and mark with arrows the transitions of the Balmer series ($n \to 2$) and the Lyman series ($n \to 1$).

Solution

Step 1 — Level values. With $E_n = -13{,}6/n^2\;\text{eV}$:

$E_1 = -13{,}6, \quad E_2 = -3{,}40, \quad E_3 = -1{,}51, \quad E_4 = -0{,}85\;\text{eV}$

The levels crowd together as they approach $E=0$ (the ionisation limit): the spacing drops rapidly as $n$ grows, as shown in the diagram.

Step 2 — Lyman series ($n \to 1$). Jumps towards the ground state, the most energetic, in the ultraviolet. The smallest ($2\to 1$):

$\Delta E = E_2 - E_1 = -3{,}40 + 13{,}6 = 10{,}2\;\text{eV} \;\Rightarrow\; \lambda = \frac{1240}{10{,}2} \approx 122\;\text{nm}$

Step 3 — Balmer series ($n \to 2$). Jumps towards $n=2$, in the visible. The $H_\alpha$ line ($3\to 2$):

$\Delta E = E_3 - E_2 = -1{,}51 + 3{,}40 = 1{,}89\;\text{eV} \;\Rightarrow\; \lambda = \frac{1240}{1{,}89} \approx 656\;\text{nm} \;(\text{red})$

The $H_\beta$ line ($4\to 2$): $\Delta E = -0{,}85 + 3{,}40 = 2{,}55\;\text{eV} \Rightarrow \lambda \approx 486\;\text{nm}$ (blue-green).

$\ev{\text{Lyman} \to \text{UV};\quad \text{Balmer} \to \text{visible};\quad H_\alpha \approx 656\;\text{nm}}$

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