Problem — Ranking the energies of the H levels

Problem

Order, from the most negative (lowest) to the highest, the energies of the levels $n = 1, 2, 3, 4$ of the hydrogen atom and explain why the levels bunch up as they approach zero.

Solution

Step 1 — Computing the levels. With $E_n = -13{,}6/n^2$ eV:

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

Step 2 — Ordering. From the most negative (lowest) to the highest:

$\ev{E_1 < E_2 < E_3 < E_4}$

The ground level $n=1$ is the deepest; as $n$ rises, energy grows (becomes less negative) towards zero.

Step 3 — Why they bunch up. Energy goes as $-1/n^2$. The spacing between successive levels is

$E_{n+1} - E_n = -13{,}6\left(\frac{1}{(n+1)^2} - \frac{1}{n^2}\right)$

which decreases rapidly as $n$ grows: from $n=1$ to $n=2$ the jump is $10{,}2$ eV, from $n=3$ to $n=4$ only $0{,}66$ eV. Since $1/n^2 \to 0$ as $n\to\infty$, infinitely many levels accumulate just below $E = 0$, which is the ionisation threshold: above it the electron is free and the energies become continuous.

Collegamenti

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