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The Bohr model combines classical circular motion in the Coulomb potential with explicitly quantum postulates. A one-electron atom of nuclear charge $+Ze$ has electron–nucleus reduced mass $\mu$. With $C=Z{e}^2/(4\pi {\epsilon }_0)$, the relative circular orbit obeysBohr postulates permitted nonradiating stationary orbits with $L=n\hslash$, for $n=1,2,\dots$. These assumptions giveFor hydrogen with the nucleus taken infinitely heavy, the $n=1$ radius is the Bohr radius, $a_0\simeq 5.29\times 10^{-11}\mathrm{m}$. The finite reduced mass gives the small nuclear-motion correction, and increasing $Z$ contracts the one-electron orbit.

The circular-motion equation gives $T=C/(2r)$ while the potential energy is $-C/r$, so the total energy is $-C/(2r)$. Substituting the quantized radius yieldsFor the heavy-nucleus hydrogen approximation, $E_n\simeq -13.6\mathrm{eV}/n^2$. The ionization threshold is zero energy, so removing the ground-state electron requires about $13.6\mathrm{eV}$.

The model further postulates that radiation is emitted or absorbed only in transitions between stationary orbits, with photon energy $h\nu =|E_i-E_f|$. Emission from $n_i>n_f$ givesThus it reproduces the principal hydrogen spectral series, their limiting wavelengths, and the gross hydrogenic energy levels. For example, $n=3\to 2$ in hydrogen has $\Delta E\simeq 1.89\mathrm{eV}$ and $\lambda \simeq 656\mathrm{nm}$, a visible line. Absorption reverses the transition when the photon matches the energy gap.

The achievement is that quantized energies explain discrete spectra, but the model's stationary orbits are postulates rather than consequences of classical electrodynamics: an accelerating classical charge would radiate. It does not provide a general calculation of line intensities or transition selection rules and fails as a quantitative theory of many-electron atoms. Spin, fine structure and the full Zeeman effect require further physics. Modern quantum mechanics replaces definite electron orbits by wavefunctions and quantum probabilities. In particular, the true hydrogen ground state has $\ell =0$, so its orbital angular momentum vanishes; the model's rule $L=n\hslash$ should not be identified with the modern orbital eigenvalue rule $L^2=\ell (\ell +1){\hslash }^2$. Bohr's energy quantization remains a successful hydrogenic approximation, while its literal orbital picture does not.

For a photon, the photon energy formula and the vacuum relation $\nu =c/\lambda$ give $E_{\gamma }=hc/\lambda$. For an Electron, the de Broglie wavelength gives the same momentum magnitude $p=h/\lambda$, but the nonrelativistic kinetic energy is $K_e=p^2/(2m_e)$. Using $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $m_e=9.10938\times 10^{-31}\mathrm{kg}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$,Thus, to the precision of the wavelength,The electron answer is its kinetic energy, as required for the diffraction experiment; its relativistic energy would additionally include the approximately $511\mathrm{keV}$ rest energy. Since $K_e\ll m_e{c}^2$, the nonrelativistic approximation is well justified. Equal wavelengths imply equal momenta, not equal energies for particles with different energy–momentum relations.

A fundamental observer follows the mean cosmic matter flow and is at rest with respect to the homogeneous cosmological frame. Cosmological isotropy permits no preferred spatial acceleration, rotation vector or anisotropic expansion for this flow: the congruence is geodesic, irrotational and shear-free, with isotropic expansion. Its orthogonal homogeneous spatial slices define simultaneity. Choosing fixed spatial labels along these worldlines and their common proper time as $t$ gives synchronous coordinates, $g_{tt}=c^2$ and $g_{ti}=0$.

In the Friedmann-Lemaître-Robertson-Walker metric, $t$ is this cosmic proper time and $\chi$ is radial comoving coordinate. On a constant-$t$ radial slice, the proper distance in cosmology from $0$ to a fixed ${\chi }_0$ is $D(t)=a(t){\chi }_0$. ThereforeThis is the Hubble parameter. The metric describes expansion when $\dot{a}>0$; its form alone also permits static or contracting histories, so expansion additionally requires that sign of the scale factor derivative.

For a radial photon, the metric components relevant to its motion are $g_{tt}=c^2$ and $g_{\chi \chi }=-a^2(t)$, both independent of $\chi$. The angular momentum components vanish. The covariant-momentum form of the geodesic equation therefore givesOnly this radial contraction is needed: angular metric coefficients can depend on $\chi$, but multiply zero angular components. The conserved quantity is the lower-index $p_1=p_{\chi }$, rather than the contravariant component $p^1$.

For outward travel put $P=-p_{\chi }>0$. The null condition $c^2(p^t{)}^2-a^2(p^{\chi }{)}^2=0$ givesHere the time coordinate is $t$, rather than $ct$. A comoving observer has $u^t=1$ and measures photon energy $E=p_{\mu }{u}^{\mu }=p_t$. Thus energy and frequency decrease as $a^{-1}$, giving the cosmological redshiftThe reciprocal relation for wavelengths follows from their common local propagation speed $c$.

Finally center the spatial coordinates on the receiving comoving observer. At the source, the transverse spatial line element is $dl=a(t_E)S_K(\chi )d\theta$. For a short transverse rod and a small observed angle $\Delta \theta$,Thus the angular diameter distance is $d_A=a(t_E)S_K(\chi )$. Writing the reception epoch as $t_0=t_R$ gives the displayed formula in the question. The length is the proper transverse length at emission; $t_0$ cannot be replaced by an unrelated epoch while retaining that expression.

The energy carried by one photon is proportional to its frequency $\nu$, with proportionality constant the Planck constant $h$. The equivalent forms use angular frequency $\omega =2\pi \nu$ and vacuum wavelength $\lambda =c/\nu$. Different observers can measure different photon energies because they measure different frequencies; the photon remains massless. In signature $(+---)$, an observer with four-velocity $u^{\mu }$ measures $E=p_{\mu }{u}^{\mu }$, where $p^{\mu }$ is the photon's four-momentum and $u^{\mu }{u}_{\mu }=c^2$.