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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 hydrogenic donor state, replace the bare electron mass by $m^{\ast }=0.1m_e$ and the Coulomb permittivity by ${ϵ}_r{ϵ}_0$ with ${ϵ}_r=16$. Rescaling the Bohr model gives the positive ground-state binding magnitudeThe radius isusing the Bohr radius $a_0=5.29177\times 10^{-11}\mathrm{m}$. The supplied positive hydrogen “energy” formula should be interpreted as a binding magnitude; the bound level's energy relative to the conduction-band edge is $-E_B$.

At $300\mathrm{K}$, $k_{B}T=25.9\mathrm{meV}$, about five times the binding energy. Thus a substantial fraction, ordinarily most of the shallow donors in the dilute nondegenerate regime, will be ionised. An exact fraction also depends on the Fermi level, donor concentration and degeneracies; comparing $E_B$ with $k_{B}T$ alone does not determine it exactly. The large radius explains why a weakly bound donor extends across many unit cells.

Use the nonrelativistic Coulomb potential energy $V=-Z{e}^2/(4\pi {ϵ}_{0}r)$, with the nucleus treated as fixed. Since the wavefunction has zero orbital angular momentum, its Laplacian isSubstituting into the time-independent Schrödinger equation givesFor a constant energy eigenvalue, the coefficient of $1/r$ must vanish. Therefore $\boxed{\beta =Z/a_0}$, where $a_0$ is the Bohr radius. Wavefunction normalization then requiresChoosing the irrelevant overall phase so that $\alpha$ is real and positive gives $\boxed{{\alpha }^2={\beta }^3/\pi }$. The remaining constant term is the hydrogen ground-state energyReplacing $m_e$ by the electron-nucleus reduced mass gives the usual finite-nuclear-mass refinement.

The attractive Coulomb potential energy is $V(r)=-e^2/(4\pi {ϵ}_{0}r)$, soUse precisely the radial normalization stated in the PDF: the radial probability measure is $4\pi r^2|\psi {|}^{2}dr$. The supplied function hasIts integral is one, since ${\int }_0^{\infty }{r}^4{e}^{-r/a_0}dr=24a_0^5$. In particular,This calculation uses the supplied convention, rather than multiplying a separately normalized standard radial function by an extra $4\pi$.

The addition of angular momentum bracket is $j(j+1)-l(l+1)-s(s+1)=3/4-2-3/4=-2$. Substitution in the spin-orbit energy shift in a central potential therefore givesHere $a_0=5.29\times 10^{-11}\mathrm{m}$ is the supplied Bohr radius. The negative sign means the $j=1/2$ member of the $l=1$ pair is shifted downward by this spin-orbit interaction; it is not the total hydrogen binding energy.

The general time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\psi =E\psi$. For a spherically symmetric wave function, ${\nabla }^2\psi ={\psi }^{\prime \prime }+2{\psi }^{\prime }/r$; the Coulomb attraction between the proton and electron is $V=-q^2/(4\pi {ϵ}_{0}r)$. These supply respectively the kinetic and potential terms in the displayed radial equation. Rearranging identifiesas the Bohr radius.

Substitute $\psi =(A+Br)e^{-\lambda r}$ into the rearranged equation and put $ϵ=2mE/{\hslash }^2$. Equating the coefficients of $r,1,r^{-1}$ after cancelling the exponential givesA nonzero state has $A\ne 0$. If $B=0$, these identities give $\lambda =1/a$ and $ϵ=-1/a^2$. If $B\ne 0$, they give $\lambda =1/(2a)$, $B=-A/(2a)$, and $ϵ=-1/(4a^2)$. Thus the two independent eigenfunctions and their corresponding energies areThey are eigenfunctions at different energies, rather than two independent solutions at one fixed energy.

For the higher-energy state, radial probabilities include the spherical volume factor $4\pi r^2$. The normalization constant and $4\pi$ cancel in the ratio. Using the supplied integral at $b=a$ and at infinity,HenceThis calculation fixes the denominator as $8e$ in the source's unparenthesized expression; interpreting it as a product with $e$ would give a negative probability.

The hydrogenic donor state radius is the Bohr radius rescaled by screening and effective mass:The donor-state overlap estimate compares the mean spacing $n_D^{-1/3}$ to this radius. Taking the spacing as $a_B^{\ast }$ gives $n_D\sim (a_B^{\ast }{)}^{-3}=1.32\times 10^{25}{\mathrm{m}}^{-3}=1.32\times 10^{19}\mathrm{c}{\mathrm{m}}^{-3}$. Taking it as the orbital diameter gives a factor eight less, $1.65\times 10^{18}\mathrm{c}{\mathrm{m}}^{-3}$. Thus a geometric estimate is roughly $10^{18}$–$10^{19}\mathrm{c}{\mathrm{m}}^{-3}$, with its contact convention stated. The exponential tails do not have a sharp overlap boundary, and this estimate should not be identified with a precise metal-insulator transition threshold.

Separation of variables in the central Coulomb potential gives ${\psi }_{nlm}=R_{nl}(r)Y_l^m(\theta ,\varphi )$. The radial wavefunction depends on $n$ and $l$, independently of $m$. For the hydrogen radial wavefunction in the first excited shell, put $\rho =r/a_0$. The Bohr radius is $a_0=4\pi {ϵ}_0{\hslash }^2/(\mu e^2)$, where $\mu$ is the electron–proton reduced mass. Choose the real phasesNear the origin, $R_{20}\to 1/(\sqrt{2}{a}_0^{3/2})$, whereas $R_{21}\sim r/(\sqrt{24}{a}_0^{5/2})$. More generally, regular hydrogen radial wavefunctions behave as $r^l$. At large $r$, both displayed functions have a polynomial factor proportional to $r$ multiplying $e^{-r/(2a_0)}$; $R_{20}$ approaches zero from below and $R_{21}$ from above with these phase choices.

The $2s$ radial wavefunction crosses zero once, at $r=2a_0$, and has its negative minimum at $4a_0$. The $2p$ radial wavefunction has no radial node at positive $r$, and reaches its maximum at $2a_0$. The count is $n-l-1$; neither the origin nor the limit at infinity is counted as a radial node. These are sketches of $R$, rather than of $rR$ or the radial probability density $r^2|R{|}^2$.

The radial probability density includes the spherical-shell volume factor:Consequently the radial expectation value isKeeping $m$ and $K$ fixed, the hydrogenic ground-state radius scaling is therefore $⟨r⟩\propto 1/Z$: the $Z=2$ state has half the mean radius of the $Z=1$ state. Equivalently, with $a_{\ast }={\hslash }^2/(mK)$, the mean radii are $3a_{\ast }/2$ and $3a_{\ast }/4$. For the physical electron-nucleus Coulomb potential, $a_{\ast }$ is the Bohr radius of the fixed-nucleus hydrogen atom model.

The exact $Z=2$ comparison in this one-electron model describes the hydrogen-like ion ${\mathrm{He}}^{+}$. A neutral helium atom has two electrons; its exact size also depends on their mutual Coulomb interaction and is not determined by this one-electron calculation. The stated half-radius conclusion is the prediction of the model supplied in the paper.

Let $r$ be the electron-nucleus separation and use the reduced mass $\mu =m_{e}M/(m_e+M)$. Coulomb attraction supplies the relative centripetal force, and Bohr quantization fixes the relative angular momentum:Eliminating $v$ givesThe total energy is $E=\mu v^2/2-Z{e}^2/(4\pi {ϵ}_{0}r)=-Z{e}^2/(8\pi {ϵ}_{0}r)$, henceHere $a_0$ is the ordinary Bohr radius and $R_{\infty }$ the infinite-nuclear-mass Rydberg constant. The electron's radius about the centre of mass is $M{r}_n/(M+m_e)$; for a heavy nucleus this distinction and the reduced-mass correction are small, but the supplied nuclear mass permits them to be stated precisely.

A Wannier-Mott exciton has an internal wave function extending over many unit cells. Its simplest model assumes a three-dimensional bulk crystal, approximately parabolic valence band and conduction band, positive isotropic scalar effective masses $m_e^{\ast },m_h^{\ast }$, a constant scalar relative permittivity ${ϵ}_r$, and a dilute pair population. Replace microscopic crystal forces by those effective masses and dielectric screening, and neglect exchange corrections and coupling to phonons. These assumptions must hold on the spatial and energy scales of the particular exciton.

With relative coordinate $\mathbf{r}={\mathbf{r}}_e-{\mathbf{r}}_h$, the internal Hamiltonian operator is the three-dimensional hydrogen atom problem with a reduced mass and a screened Coulomb potential:Here $e$ is the elementary charge, ${ϵ}_0$ the vacuum permittivity, and $\hslash$ the reduced Planck constant. The internal energy eigenvalues, radius scale, and exciton binding energies areThe principal quantum number is $n=1,2,\dots$. This reduction and these scales are derived in Excitons in Semiconductor Quantum Dots by Matjaž Humar (§3, printed pp. 5–6, equations (13)–(26)).

To check the scales by dimensional analysis, write $C=e^2/(4\pi {ϵ}_0{ϵ}_r)$, which has units of energy times length. Then $a_B^{\ast }={\hslash }^2/(\mu C)$ is a length and $R^{\ast }=\mu C^2/(2{\hslash }^2)$ is an energy. Rescaling $r=a_B^{\ast }\rho$ makes $H_{\mathrm{rel}}/R^{\ast }=-{\nabla }_{\rho }^2-2/\rho$, the dimensionless hydrogen atom operator. The Bohr radius $a_B^{\ast }$ is a wave function scale, not the radius of a classical orbit; for this model's lowest internal bound state, the mean separation is $3a_B^{\ast }/2$.

For the diagrams' hypothetical inputs $m_e^{\ast }=0.10m_0$, $m_h^{\ast }=0.40m_0$, ${ϵ}_r=10$ and $E_g=1.500\mathrm{eV}$, where $m_0$ is the free electron mass, independent substitution gives $\mu =0.080m_0$, $a_B^{\ast }\simeq 6.615\mathrm{nm}$ and $R^{\ast }\simeq 10.885\mathrm{meV}$. Here electronvolts express energy. The exciton binding energies for $n=1,2,3$ are approximately $10.885$, $2.721$ and $1.209\mathrm{meV}$. These are model calculations, not measured material parameters. The large-radius approximation additionally requires $a_B^{\ast }$ to be much larger than the crystal's unit cell length scale. In a thin layer, with anisotropic effective masses, or with spatially varying dielectric screening, this three-dimensional formula is not generally valid.