Problem — From lambda to the initial level (Balmer)

Problem

A transition of the hydrogen atom emits a photon with $\lambda = 486$ nm. Knowing it ends on $n_f = 2$, from which initial level does it originate? Identify the Balmer-series line.

Solution

Step 1 — Photon energy.

$\Delta E = \frac{1240}{\lambda(\text{nm})} = \frac{1240}{486} \approx 2{,}55\;\text{eV}$

Step 2 — Energy of the initial level. The photon carries away $\Delta E = E_i - E_f$, with $E_f = E_2 = -3{,}40$ eV:

$E_i = E_f + \Delta E = -3{,}40 + 2{,}55 = -0{,}85\;\text{eV}$

Step 3 — Initial quantum number. From $E_i = -13{,}6/n_i^2$:

$n_i^2 = \frac{13{,}6}{0{,}85} = 16 \quad\Longrightarrow\quad n_i = 4$

Step 4 — Identification. The $4\to 2$ transition is the second Balmer line, $H_\beta$ (cyan).

$\ev{n_i = 4 \quad (4\to 2,\ H_\beta\ \text{line})}$

Collegamenti

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