Problem — Ranking transition frequencies

Problem

Order from lowest to highest the frequencies of the photons emitted in the transitions $3\to 2$, $4\to 2$, $4\to 3$, $5\to 2$ of the hydrogen atom, and state which fall in the visible range (Balmer series).

Solution

Step 1 — Levels. $E_2 = -3{,}40$; $E_3 = -1{,}51$; $E_4 = -0{,}85$; $E_5 = -13{,}6/25 = -0{,}544$ (eV).

Step 2 — Photon energies ($\Delta E = E_i - E_f > 0$, and $\nu = \Delta E/h$ is proportional to $\Delta E$):

$\begin{aligned} 4\to 3:\ &\Delta E = -0{,}85-(-1{,}51) = 0{,}66\;\text{eV} \\ 3\to 2:\ &\Delta E = -1{,}51-(-3{,}40) = 1{,}89\;\text{eV} \\ 4\to 2:\ &\Delta E = -0{,}85-(-3{,}40) = 2{,}55\;\text{eV} \\ 5\to 2:\ &\Delta E = -0{,}544-(-3{,}40) = 2{,}86\;\text{eV} \end{aligned}$

Step 3 — Ordering the frequencies. Since $\nu \propto \Delta E$:

$\ev{\ \nu_{4\to 3} < \nu_{3\to 2} < \nu_{4\to 2} < \nu_{5\to 2}\ }$

Step 4 — Which are visible. Transitions to $n_f = 2$ form the Balmer series, which falls in the visible range: $3\to 2$ ($656$ nm, red), $4\to 2$ ($486$ nm, cyan), $5\to 2$ ($434$ nm, violet). The $4\to 3$ transition has $\Delta E = 0{,}66$ eV $\Rightarrow \lambda \approx 1875$ nm: it falls in the infrared (Paschen series), not visible.

Collegamenti

Argomenti: Quantum physics Concetti: Bohr model · Photon Competenze: Ranking reasoning