Problem — The n=3 level of hydrogen and the H-alpha line

Problem

Using $E_n = -13{,}6/n^2$ eV, calculate the energy of the $n = 3$ level of the hydrogen atom. Then find the wavelength of the photon emitted in the $n = 3 \to n = 2$ transition and identify the line.

Solution

Step 1 — Energy of the $n=3$ level.

$E_3 = -\frac{13{,}6}{3^2} = -\frac{13{,}6}{9} \approx -1{,}51\;\text{eV}$

Step 2 — Photon energy for $n=3\to n=2$. We also need $E_2$:

$E_2 = -\frac{13{,}6}{2^2} = -3{,}40\;\text{eV}$

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

Step 3 — Wavelength.

$\lambda = \frac{1240}{\Delta E} = \frac{1240}{1{,}89} \approx 656\;\text{nm}$

Step 4 — Identification. $656$ nm is red light: it is the $H_\alpha$ line, the first line of the Balmer series (transitions to $n_f = 2$).

$\ev{E_3 = -1{,}51\;\text{eV}, \qquad \lambda \approx 656\;\text{nm}\ (H_\alpha,\ \text{red})}$

Collegamenti

Argomenti: Quantum physics Concetti: Bohr model · Photon Competenze: Unit conversion