Worked example: a red photon from the hydrogen atom

Problem

An electron in the hydrogen atom drops from level $n=3$ to level $n=2$. What is the energy and colour of the emitted photon?

Solution

Step 1 — Energy levels. From Bohr’s formula $E_n = -13{,}6/n^2\;\text{eV}$:

$E_3 = -\frac{13{,}6}{9} = -1{,}51\;\text{eV}, \qquad E_2 = -\frac{13{,}6}{4} = -3{,}40\;\text{eV}$

Step 2 — Photon energy. The electron drops to a lower level: the energy released is carried away by the photon,

$E_\text{fotone} = E_3 - E_2 = -1{,}51 - (-3{,}40) = 1{,}89\;\text{eV}$

Step 3 — Wavelength and colour.

$\ev{\lambda = \frac{hc}{E_\text{fotone}} = \frac{1240\;\text{eV·nm}}{1{,}89\;\text{eV}} \approx 656\;\text{nm}}$

It’s a red photon: the famous $H_\alpha$ line of the Balmer series, one of the most characteristic lines in the hydrogen spectrum and a key tool in astrophysics for identifying gas clouds.

Erratum in the text $E_\text{fotone} = 2{,}11\;\text{eV}$: this is a printing error. The $n{=}3\to n{=}2$ jump is worth $\Delta E = 1{,}89\;\text{eV}$, and it is precisely $1{,}89\;\text{eV}$ that gives $\lambda \approx 656\;\text{nm}$ (with $2{,}11\;\text{eV}$ you would get $\approx 588$ nm, yellow, which is not $H_\alpha$).

The book states

Links

Topics: Quantum physics Concepts: Bohr model · Photon Objects: Hydrogen atom