Photons spectra and quantization

In the Bohr model, the electron in an atom occupies one of discrete orbits, so the atom's energy can take only certain values $E_1, E_2, E_3, \ldots$. Light is emitted or absorbed in portions called photons, whose energy equals the difference between energy levels: $hf = E_i - E_f$. Because the set of differences $\Delta E$ is discrete, an atom produces not a continuous spectrum but separate lines. Each line corresponds to a frequency $f$ and a wavelength $\lambda = c/f$, so the spectrum acts as a chemical element's fingerprint. When the electron moves to a lower level, a photon is emitted; when it moves to a higher level, a photon is absorbed, and in both cases energy conservation holds.

A photon has frequency $f=5.0\times10^{14}\,\mathrm{Hz}$. Calculate its energy in joules and in electronvolts. Use $h=6.63\times10^{-34}\,\mathrm{J\,s}$ and $1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm{J}$.

In the Bohr model of hydrogen, $E_n=-\frac{13.6}{n^2}\,\mathrm{eV}$. An electron transitions from $n=3$ to $n=2$. Calculate the emitted photon energy and its wavelength.

A hydrogen gas spectrum shows bright lines at $\lambda_1=656\,\mathrm{nm}$ and $\lambda_2=486\,\mathrm{nm}$. Using $hc=1240\,\mathrm{eV\,nm}$ and $E_n=-\frac{13.6}{n^2}\,\mathrm{eV}$, identify which transitions produce these lines. Explain why these atoms do not create a continuous background in the spectrum.

Safe self-study activity: open the PhET simulation “Models of the Hydrogen Atom” or “Neon Lights & Discharge Lamps” (virtual mode only). Set the atom to hydrogen or neon, turn on the spectrum display, and record three visible lines. For each line, calculate the photon energy using $E=\frac{hc}{\lambda}$, with $hc=1240\,\mathrm{eV\,nm}$ and the wavelengths from the simulation. Compare the calculated energies with the level differences shown in the simulation, and write which transitions they correspond to. Do not use real discharge tubes, ultraviolet lamps, or high-voltage equipment: observing through the simulation removes the risk of electric shock and UV exposure.

Show that spectral lines arise from discrete transitions between energy levels, not from a continuous range of energies.

Energy-level diagram of a hydrogen atom: horizontal lines $n=1,2,3,4$, downward arrows for emission (for example, $n=3\to2$ and $n=4\to2$) and upward arrows for absorption. Below the diagram, show two spectra: an emission line spectrum, where each downward emission arrow corresponds to a bright line on a wavelength scale, and an absorption spectrum, where upward absorption arrows correspond to dark lines against a continuous background.

A smartphone screen with area about $0.010\ \text{m}^2$ has luminance $400\ \text{cd/m}^2$ and draws about $0.5\ \text{W}$. A red pixel emits light of wavelength about $650\ \text{nm}$, a blue pixel about $450\ \text{nm}$. One red photon has energy $E=\frac{hc}{\lambda}=\frac{6.63\cdot10^{-34}\cdot3.00\cdot10^8}{650\cdot10^{-9}}\approx3.06\cdot10^{-19}\ \text{J}\approx1.91\ \text{eV}$; a blue photon has about $4.42\cdot10^{-19}\ \text{J}\approx2.76\ \text{eV}$. Thus at $0.5\ \text{W}$ the screen emits on the order of $10^{18}$ photons per second.

This uses light quantization: light arrives as individual photons with energy $E=hf=hc/\lambda$, so red and blue colours require different photon energies. In an LED pixel, an electron transition across the semiconductor band gap sets the discrete energy of the emitted photon.

A microwave oven of power $900\ \text{W}$ produces radiation of frequency $2.45\ \text{GHz}$ and wavelength $\lambda=c/f=\frac{3.00\cdot10^8}{2.45\cdot10^9}\approx0.122\ \text{m}=12.2\ \text{cm}$. One such photon has energy $E=hf=6.63\cdot10^{-34}\cdot2.45\cdot10^9\approx1.62\cdot10^{-24}\ \text{J}\approx1.0\cdot10^{-5}\ \text{eV}$. In $60\ \text{s}$ the oven delivers about $54\,000\ \text{J}$ and emits on the order of $3\cdot10^{28}$ such photons.

The scenario shows the link between photon energy and frequency and the discreteness of absorption: a water molecule changes rotational level only when the photon energy matches a transition; microwave radiation resonates with allowed transitions.

In an aurora, atomic oxygen produces a green line of wavelength $557.7\ \text{nm}$ and a red line of wavelength $630\ \text{nm}$. The photon energies are $E=\frac{hc}{\lambda}$: about $3.56\cdot10^{-19}\ \text{J}\approx2.22\ \text{eV}$ for green, and about $3.16\cdot10^{-19}\ \text{J}\approx1.97\ \text{eV}$ for red. The difference $\Delta E\approx0.25\ \text{eV}$ corresponds to different transitions between oxygen atom levels.

This is a direct illustration of a line spectrum: each colour arises from an electron transition between two discrete energy levels, and the photon energy equals the level difference $hf=E_i-E_f$.