Checkpoint: Hydrogen and Quantum Numbers

This checkpoint consolidates the key results of Module 15. Before attempting the exercise, review the three interlocking ideas below.

The hydrogen energy formula

Solving the time-independent Schrödinger equation for an electron in a $1/r$ Coulomb potential yields a discrete set of bound-state energies labelled by the principal quantum number $n = 1, 2, 3, \ldots$:

The value $-13.6\,\text{eV}$ is the ground-state energy; it is set by the combination $m_e e^4 / [2(4\pi\epsilon_0)^2\hbar^2]$ evaluated with SI constants and converted to eV. The negative sign signals a bound state. As $n$ grows the levels become less negative and crowd toward the ionisation limit at $E = 0$.

Quantum-number bookkeeping

Three integers label every hydrogen orbital:

| Symbol | Name | Allowed values | |--------|------|----------------| | $n$ | principal | $1, 2, 3, \ldots$ | | $\ell$ | orbital angular momentum | $0, 1, \ldots, n-1$ | | $m_\ell$ | magnetic | $-\ell, -\ell+1, \ldots, \ell$ |

For fixed $n$ the number of distinct $(n, \ell, m_\ell)$ triplets is

The identity $\sum_{k=0}^{n-1}(2k+1) = n^2$ follows because the left side is the sum of the first $n$ odd integers, which equals $n^2$. For example, $d(4) = 1 + 3 + 5 + 7 = 16 = 4^2$.

Photon emission and the Rydberg formula

When an electron falls from level $n_i$ to a lower level $n_f$, the atom emits a photon whose energy equals the energy the electron gives up:

The factor $1/n_f^2 - 1/n_i^2$ is positive because $n_f < n_i$. This is the Rydberg formula, and it reproduces every hydrogen spectral line to high precision.

The Balmer series arises when $n_f = 2$. The first four lines are:

| Transition | $E_\gamma$ (eV) | Colour | |------------|-----------------|--------| | $3 \to 2$ | $1.89$ | red ($656\,\text{nm}$) | | $4 \to 2$ | $2.55$ | blue-green ($486\,\text{nm}$) | | $5 \to 2$ | $2.86$ | blue ($434\,\text{nm}$) | | $6 \to 2$ | $3.02$ | violet ($410\,\text{nm}$) |

Let us verify the $4 \to 2$ entry. With $E_1 = -13.6\,\text{eV}$:

The emitted photon carries

Converting to wavelength via $\lambda = hc/E_\gamma = 1240\,\text{eV·nm} / 2.55\,\text{eV} \approx 486\,\text{nm}$, which falls in the blue-green part of the visible spectrum — matching experimental observation.

Try it

This is a numerical exercise — return a number. Compute the energy (in eV) of the photon emitted when a hydrogen electron falls from $n = 4$ to $n = 2$. Use $E_1 = -13.6\,\text{eV}$ and the formula $E_n = E_1/n^2$.

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