quantized energy levels and atomic spectra

The Bohr model replaces the continuous range of electron energies in an atom with a set of discrete stationary energy levels $E_1, E_2, E_3,\dots$. For hydrogen these levels are approximately $E_n=-13.6/n^2$ (in eV), so the atom can absorb or emit energy only in amounts equal to the difference between two levels. When an electron drops from a higher level $n_i$ to a lower level $n_f$, a photon is emitted with energy $hf=E_{n_i}-E_{n_f}=hc/\lambda$. Conversely, a photon is absorbed only if its energy exactly matches an upward transition, which is why a cool gas removes sharp dark lines from a continuous background. Because the allowed energies are discrete, the resulting emission and absorption spectra consist of separate lines rather than a continuous rainbow. The pattern of lines is a fingerprint of the element, and for hydrogen the Balmer-series lines obey the Rydberg formula $1/\lambda=R_H(1/2^2-1/n^2)$ for $n=3,4,5,\dots$.

For a hydrogen atom, calculate the energy (in eV and J) and the wavelength of the photon emitted when the electron makes a transition from $n=3$ to $n=2$. Use $E_n=-13.6/n^2$ (eV), $h=6.626\times10^{-34}$ (J·s), $c=3.00\times10^8$ (m/s), and $1\ \text{eV}=1.602\times10^{-19}\ \text{J}$.

A hypothetical atom has three energy levels: $E_1=-6.0$ eV, $E_2=-3.0$ eV, and $E_3=-1.5$ eV. List all possible emission photon energies and wavelengths, and state which lie in the visible range (380–750 nm). Use $\lambda(\text{nm}) = 1240/E(\text{eV})$.

A nebula's emission spectrum shows bright lines at 656.3 nm, 486.1 nm, and 434.0 nm. Assuming these are hydrogen Balmer lines (transitions ending at $n=2$), determine the principal quantum number $n$ of the upper level for each line and confirm that the levels are quantized. Use $R_H = 1.097\times10^{-2}$ nm$^{-1}$ and $1/\lambda=R_H(1/2^2-1/n^2)$.

Safe self-study activity: use a free online simulation such as PhET "Models of the Hydrogen Atom" or an atomic-spectra applet. Do not handle real discharge tubes or ultraviolet sources. Turn on the display of energy levels and the spectrum in the simulation. Induce the transitions $n=3\to2$, $n=4\to2$, and $n=5\to2$, and record each emitted wavelength in a table. Compare your values with $\lambda = 1240/\Delta E$ calculated from $E_n=-13.6/n^2$ (eV). Then sketch an energy-level diagram with the three transitions and a spectrum strip with the corresponding lines.

To connect discrete energy levels with observable spectral lines: show that each photon energy equals a level spacing, and each spacing produces one line at a definite wavelength.

A vertical hydrogen energy-level diagram with levels $n=1,2,3,4,5$; downward arrows labeled $3\to2$, $4\to2$, $5\to2$ with photon energies; below, a horizontal spectrum strip aligned so each arrow projects to a colored line at 656 nm, 486 nm, and 434 nm on a wavelength axis.

A $150\ \text{W}$ low-pressure sodium lamp lights a road. Through a hand spectroscope it shows almost nothing but a bright yellow line at $589.0\ \text{nm}$ and $589.6\ \text{nm}$. Each photon carries energy $E = hc/\lambda = (6.63\times10^{-34}\ \text{J·s})(3.00\times10^{8}\ \text{m/s})/(589\times10^{-9}\ \text{m}) = 3.38\times10^{-19}\ \text{J} \approx 2.11\ \text{eV}$; the lamp emits about $4\times10^{20}$ such photons per second.

The lamp gives no continuous rainbow: a sodium atom can occupy only a discrete set of states, so an electron dropping from the $3p$ level to the $3s$ level emits a photon whose energy is exactly the level difference $h\nu = E_{3p} - E_{3s}$. The yellow line seen from far away is a direct measurement of a quantized level spacing.

In the lab a hydrogen tube gives the red $H_\alpha$ line at $656\ \text{nm}$ ($E = 1240/656 = 1.89\ \text{eV}$, the $n=3 \to n=2$ transition, matching $13.6(\tfrac{1}{4}-\tfrac{1}{9}) = 1.89\ \text{eV}$). In a galaxy's spectrum the same line sits at $700\ \text{nm}$, i.e. $z = (700-656)/656 = 0.067$, so the galaxy recedes at about $0.067c \approx 2.0\times10^{4}\ \text{km/s}$.

Every hydrogen atom has the same set of levels, so the set of spectral lines is a fingerprint of the element — the same $E_3 - E_2$ difference in any galaxy. Shifting the whole line pattern (Doppler effect) is used as a speed ruler; the line formula $h\nu = E_n - E_m$ here is exactly the quantized-level theory.

A pyrotechnician adds salts of different metals to the charges: strontium gives a red line at $606\ \text{nm}$ ($E = 2.05\ \text{eV}$), barium a green one at $514\ \text{nm}$ ($E = 2.41\ \text{eV}$), copper a blue one at $455\ \text{nm}$ ($E = 2.73\ \text{eV}$). The gunpowder heats the salts to $1500$–$2500\ \text{K}$, the atoms are excited by collisions and then emit these lines.

Heating supplies energy to the atoms, but they emit only photons of allowed transitions, so the colour of each burst is set by the level differences of a specific element rather than by temperature. Check: $E = hc/\lambda$ converts the measured wavelength into a level energy difference.