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The Bohr model assumes circular electron orbits with quantized orbital angular momentum $N\hslash$. It reproduces the hydrogen energy levels, but wavefunctions in quantum mechanics describe spatial probability distributions rather than definite trajectories. In particular, spherically symmetric Coulomb bound states have zero orbital angular momentum.

The puzzle is the coexistence of reproducible spectral lines and stable chemical atoms. A classical circular orbit in a Coulomb potential can have any negative energy: Coulomb's law alone selects no special radius or emitted frequency. Moreover, its Electron has centripetal acceleration. The Larmor formula then predicts positive radiated power, so conservation of energy makes the orbit shrink. Simply drawing concentric circles explains neither stability nor a discrete spectrum.

Niels Bohr postulated nonradiating stationary states and emission with $h\nu =E_i-E_f$. His circular-orbit condition is $L=n\hslash$. These assumptions override classical radiation; the emitted frequency need not equal an orbital frequency.^[ref]

In On the Constitution of Atoms and Molecules, Part I (1913), pp. 8–9, Niels Bohr interprets existing hydrogen discharge and astronomical spectra. Electrical discharge in a vacuum tube removes Electrons; recombination supplies spectral lines. This is a theoretical interpretation, without a reported spectrograph of his own.^[ref]

An actual optical obstacle appears in Theodore Lyman's measurements: hydrogen discharge light entered an evacuated spectrograph directly through a slit, avoiding a transmitting window. A concave reflection diffraction grating dispersed the light onto photographic film. Pumping preserved the vacuum; a second slit supplied a known spectral displacement for wavelength calibration. Gas contamination and vacuum leaks limited the measurement. This apparatus resolved ultraviolet spectral lines; it did not image orbits.^[ref]

For a separate original calibration exercise, suppose an idealized spectrograph returns synthetic vacuum line positions $x_n={\lambda }_{n\to 2}+b$ for $n=3,4,5,6$, with chosen additive offset $b=0.200\mathrm{nm}$. Assume its dispersion scale is already known and only this offset is uncertain. Compare the finite-mass prediction with the fixed-nucleus prediction ${\lambda }_n^{(\infty )}=(\mu /m_e){\lambda }_{n\to 2}$. Calibrate each model using only the $3\to 2$ line: $\hat{b}=x_3-{\lambda }_3^{\mathrm{model}}$. Then hold that offset fixed and test the other lines with residualsFor the finite-mass model every $d_n=0$. The fixed-nucleus model has $\hat{b}=0.55733\mathrm{nm}$ and residuals $d_4=-0.09264\mathrm{nm}$, $d_5=-0.12100\mathrm{nm}$, $d_6=-0.13400\mathrm{nm}$. Algebraically $d_n=(1-\mu /m_e)({\lambda }_n-{\lambda }_3)$: one additive calibration cannot absorb a multiplicative change of the wavelengths. Allowing an unknown multiplicative dispersion factor as well would make the two models degenerate in this exercise, so independent scale calibration matters.

Awarded to Niels Bohr for atomic structure and radiation from chemical atoms. The Bohr model explains the connection to discrete spectral lines.^[ref]