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Consider a single electron of mass $m_e$ and a point atomic nucleus of mass $M$ and positive electric charge $Ze$, with $e>0$ the elementary charge. Use nonrelativistic classical mechanics, omit external fields and interactions with other electrons, and writeHere $\mu$ is the Coulomb two-body reduced mass, ${ϵ}_0$ the vacuum permittivity, and $h$ the Planck constant. The separation $r$ and relative speed $v$ describe the two-particle relative coordinate; neither body is silently pinned in place. The electron's own radius about the centre of mass is $Mr/(M+m_e)$.

For the assumed circular orbit, Coulomb's law provides the required centripetal acceleration, and the imposed orbital angular momentum condition isSubstitute $v=n\hslash /(\mu r)$ into the first equation and cancel $r^{-2}$ to obtainThe kinetic energy is $K=\mu v^2/2=\kappa /(2r)$ and the Coulomb potential energy is $U=-\kappa /r$, with zero energy at infinite separation. HenceThis derives the energy sequence from the stated circular model; the modern Schrödinger equation derivation is the distinct series-termination quantization of the Coulomb radial equation. The radius above scales as $n^2/Z$, whereas the binding energy scales as $\mu Z^2/n^2$. The Bohr radius article supplies the corresponding Coulomb potential length scale. The nonrelativistic assumption requires $v_n\ll c$, with $c$ the speed of light.

For an original numerical example choose $Z=1$, $M=1.67262192369\times 10^{-27}\mathrm{kg}$, $m_e=9.1093837015\times 10^{-31}\mathrm{kg}$, $e=1.602176634\times 10^{-19}\mathrm{C}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, ${ϵ}_0=8.8541878128\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. Treat these as stipulated constants for the calculation. They give $\mu /m_e=0.99945568$, $r_1=0.0529465\mathrm{nm}$, $E_1=-13.598287\mathrm{eV}$ and $v_1/c\simeq 0.00730$, where $\mathrm{eV}$ denotes an electronvolt. At $n=2$, $r_2=4r_1$ and $E_2=E_1/4$; the ground-state ionization energy is $-E_1$.

Taking $M\to \infty$ gives $\mu \to m_e$. A finite nuclear mass reduces every binding energy by $\mu /m_e$ and increases each separation by $m_e/\mu$. This follows from relative motion, not a fitted correction to individual spectral lines. Replacing the atomic nucleus by a heavier isotope changes the same factor, providing another check of the model.

Model a proton as a uniformly charged thin spherical shell of radius $b$. Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem gives an electron potential energy $-C/b$ inside the shell and $-C/r$ outside. Relative to a point Coulomb potential, the perturbation is consequently $C(1/r-1/b)$ for $r<b$ and zero otherwise.

For the hydrogen atom ground state, time-independent perturbation theory gives the first-order energy correctionSince $E_{1s}=-C/(2a_0)$, the signed fractional change is $\Delta E/E_{1s}=-4b^2/(3a_0^2)+O((b/a_0{)}^3)$. The correction $\Delta E$ is positive: distributing the positive charge weakens the attraction near the origin and reduces the magnitude of the binding energy. Here $a_0$ is the Bohr radius and $b\ll a_0$ is the small-radius approximation.

With Bohr radius $a$, define $A_l=a[i{p}_r/\hslash -(l+1)/r+1/((l+1)a)]/\sqrt{2}$. For regular radial bound states the radial inner product and compatible domains give the displayed factorization. Expanding the first-order factors produces the centrifugal term $l(l+1){\hslash }^2/(2m{r}^2)$ and Coulomb term $-{\hslash }^2/(mar)$. factorization converts a second-order spectral problem into a positivity bound and first-order terminating-state equation.

Consider a fixed proton and one electron, omitting spin, fine structure and the Lamb shift. Let $e>0$ be the elementary charge, $a_0$ the Bohr radius in this fixed-nucleus model, and $\mathbf{E}=\mathcal{E}\hat{\mathbf{z}}$. Because the electron's electric dipole moment is $\mathbf{d}=-e\mathbf{r}$, its electric-dipole interaction isThe first excited hydrogen eigenspace has a common unperturbed energy eigenvalue $E_2$. Write $|s⟩=|2,0,0⟩$ and $|p_m⟩=|2,1,m⟩$ in the hydrogen energy eigenbasis. The electric-dipole selection rules for hydrogen leave only the $s,p_0$ coupling within this eigenspace. Each diagonal matrix element vanishes by parity, but that does not make the entire perturbation vanish. Degenerate perturbation theory requires its eigenvalues in the whole eigenspace. With the usual real wavefunction phases the off-diagonal position matrix element is $⟨s|z|p_0⟩=-3a_0$.^[ref]

The factor three can be checked directly. Set $\rho =r/a_0$ and use normalized radial wavefunctions and spherical harmonics:The angular integral is $\int Y_{00}^{\ast }\cos \theta Y_{10}d\Omega =1/\sqrt{3}$. Including $z=r\cos \theta$ and the radial volume factor $r^2$ then givesReversing the quantum phase of $|p_0⟩$ reverses this sign, with no change to the predicted energy eigenvalues.

Thus the nontrivial two-dimensional block and its determinant equation areConsequently, for the displayed sign convention,The two remaining energy eigenstates $|p_{+1}⟩,|p_{-1}⟩$ have zero first-order shifts. The slopes correspond to opposite electric dipole moments, $⟨d_z⟩=+3e{a}_0$ and $-3e{a}_0$. The electric field selects combinations of opposite parity that already have these dipole expectations at leading order. This is why applying first-order nondegenerate perturbation theory separately to $2s$ and $2p_0$, and obtaining two zero diagonal entries, misses the splitting.

The approximation requires $e{a}_0|\mathcal{E}|$ small compared with the separation to other principal quantum numbers. For real hydrogen atoms, it also requires the relevant electric-dipole interaction to dominate the small splittings omitted here, including the Lamb shift. At sufficiently small electric fields a pre-existing nonzero $2s$–$2p$ separation must be retained before taking the limit. A simple two-state illustration with unperturbed separation $\delta >0$ has shifts relative to its meanFor $3e{a}_0|\mathcal{E}|\ll \delta$, the lower branch moves by $-(3e{a}_0\mathcal{E}{)}^2/\delta +O({\mathcal{E}}^4)$ relative to its zero-field value. At $\delta =0$ the splitting is linear. This interpolation explains the importance of the hydrogen energy-level degeneracy; it is not a complete model of the fine structure multiplet.

A hydrogen atom state with principal quantum number $n=l+1$ has no radial nodes and radial factor $R(r)=A{r}^l{e}^{-r/[a(l+1)]}$, where $a$ is the Bohr radius. Its energy is $-{\hslash }^2/[2m{a}^2(l+1{)}^2]$. With unit-normalized spherical harmonics, the radial normalization gives $|A{|}^2=2^{2l+3}/[a^{2l+3}(l+1{)}^{2l+3}(2l+2)!]$.

The Stark effect changes atomic energy levels in a static electric field. For $\mathbf{E}=\mathcal{E}\hat{\mathbf{z}}$, the electric-dipole interaction is $H^{\prime }=-d_z\mathcal{E}=e\mathcal{E}z$ for an electron, with $e>0$. A nondegenerate state of definite parity has $⟨z⟩=0$, so its leading quadratic Stark effect isThe ground state shift is negative, conventionally $-\alpha {\mathcal{E}}^2/2$ with positive electric polarizability.^[ref]

Degenerate opposite-parity states require degenerate perturbation theory. In ideal Coulomb hydrogen, the $2s,2p_0$ subspace has an off-diagonal matrix element of magnitude $3e{a}_{\mathrm{B}}\mathcal{E}$, where $a_{\mathrm{B}}$ is the Bohr radius. Its mixed states shift byThis Linear Stark effect in the first excited hydrogen eigenspace assumes fine-structure and Lamb splittings are negligible compared with the mixing. In an isolated parity eigenstate with such degeneracies resolved, sufficiently weak fields restore quadratic behaviour. Strong fields require further mixing and can ionize the atom.^[ref]

Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem makes the potential energy of the electron constant inside the charged spherical shell. Relative to the point-proton Coulomb potential, the perturbation isIt is positive: the finite charge distribution weakens the attraction close to the origin.

The first-order energy correction is the expectation value of this perturbation in the unperturbed hydrogen atom ground state. Integrating its probability density over spherical volume givesFor a proton radius much smaller than the Bohr radius, expand the exponential uniformly over this short interval. The leading integral isSince $E_{1s}=-C/(2a_0)$, the signed fractional change isEquivalently, $\Delta E/|E_{1s}|=+4b^2/(3a_0^2)$ to leading order: the energy increases and the binding-energy magnitude decreases. This is the finite proton-shell correction to hydrogen ground energy.

If no small-radius expansion is made, the same time-independent perturbation theory integral gives, with $\beta =b/a_0$,Its expansion starts with $4{\beta }^2/3$, in agreement with the leading result. The integral is exact for the first-order calculation; it is not an exact solution of the perturbed Schrödinger equation.

Use the supplied effective electrostatic term and enforce charge neutrality while varying the chain length: $R=Z/(2n_e)$, with $N$ and $Z$ fixed. The total energy per positive ion becomesThe positive quadratic term supplies degeneracy pressure against compression. Differentiating gives the neutral one-dimensional metal equilibrium density,where $a_B=4\pi {ϵ}_0{\hslash }^2/(m_e{e}^2)$ is the Bohr radius. The second derivative $Z{\pi }^2{\hslash }^2/(12m_e)>0$ verifies a minimum. Within this specified toy energy, the minimizing electron density is independent of $Z$; instead the equilibrium ion spacing is $2R^{\ast }=Z/n_e^{\ast }$.

At the minimum the electrostatic contribution is minus twice the kinetic contribution, so ${\epsilon }_{\ast }=-Z{E}_F^{\ast }/3$. These follow from the given electrostatic approximation, not from a full self-consistent calculation of a real one-dimensional metal. In particular, additional Coulomb terms cannot be appended without the corresponding neutral-background subtraction that defines the supplied finite energy.

Add the electronic diamagnetism of each helium atom to the nuclear magnetic susceptibility, using the same $n_3$ for both contributions in the helium-3 sample:The density cancels when the two terms balance. The sign-change temperature in the independent-atom approximation is thereforeNo numerical atomic radius was supplied, so the exact numerical value remains conditional on that electronic length scale. Taking $⟨r^2⟩\sim a_0^2$, where $a_0$ is the Bohr radius, gives $T_0\simeq 0.32\mathrm{K}$. A slightly more explicit estimate uses the helium variational effective charge $\zeta =27/16$, for which a hydrogenic $1s$ orbital has $⟨r^2⟩=3a_0^2/{\zeta }^2\simeq 1.05a_0^2$, givingAbove this scale the electronic contribution is negative and dominates; below it the positive nuclear contribution dominates within the independent-moment approximation. The estimate is not a precision prediction of a real interacting helium phase.

The number of formula units per volume follows from the mass density:Each neutral formula unit contains $31+33=64$ electrons, so ${\rho }_e=64n_f=1.5726\times 10^{30}{\mathrm{m}}^{-3}$. Let $R^2$ be the electron-weighted mean-square orbital radius, averaged over all these electrons. The supplied Larmor-Langevin susceptibility then gives the dimensionless SI volume magnetic susceptibilitywhere $a_0=0.0529177\mathrm{nm}$ is the Bohr radius. An explicit rough atomic-scale assumption $R\simeq a_0$ gives ${\chi }_0\simeq -2.6\times 10^{-5}$.

The quoted bulk data do not determine $R$: a nuclear charge does not specify an orbital radius in a screened many-electron chemical atom. Consequently the numerical estimate just made has atomic-model uncertainty; it is not a deduction with guaranteed 30 percent accuracy. For example, the alternative assumed $R=2a_0$ gives four times the magnitude while leaving every supplied bulk number unchanged. Atomic wavefunctions, an explicit screened-orbital model, or measured atomic magnetic susceptibilities of materials exhibiting diamagnetism would be needed to support that accuracy. Keeping $R$ explicit makes the later comparison independent of an unadvertised radius assumption. The bare electron mass belongs in this atomic formula; the conduction band effective mass is relevant instead to the large donor envelope in part (c).

The Boltzmann distribution for two atomic levels at thermodynamic equilibrium iswhere $\omega$ is the statistical weight of an atomic level. For a resolved level of total angular momentum $J$, $\omega =2J+1$ when magnetic substates are unresolved.

To extend this counting to the continuum, assume an ideal, nondegenerate plasma with classical free electrons and neglect the small difference between the translational masses of the two heavy ionic species. Let $z_e=e^{{\mu }_e/kT}$ be the electron fugacity in the Saha equation, with the electron chemical potential measured relative to its rest energy. The two electron spin states and the phase-space cell volume $h^3$ giveChemical equilibrium for capture equates the bound species' chemical potential to the sum for its parent ion and free electron. Thus applying Boltzmann factors to a bound level at energy $-I_n$ gives $N_n/N_{+}=({\omega }_n/{\omega }_{+})z_e{e}^{I_n/kT}$. Eliminating the electron fugacity proves the level-specific Saha ionization equation:The factor two is the electron spin multiplicity; the factor with power $3/2$ is the inverse free-electron translational state density. A total-ionization Saha equation sums the bound states into an internal partition function, whereas this expression refers to one specified level.

For dielectronic recombination, write the entrance ion as $X^{(z+1)+}$ in ground state $i$, and the intermediate state of $X^{z+}$ as $d=(j,n\ell )$. Capture excites the core from $i$ to $j$ while binding the incident electron. The resonance energy relative to the entrance continuum is $\bar{E}=E_{ij}-I_{n\ell }>0$. The doubly excited state can eject the electron by autoionization, or undergo radiative stabilization to a bound state.

Let $A_{di}^a$ be the partial autoionization rate back to the entrance ion, $A_a$ the total autoionization rate, $A_r$ the total radiative decay rate, and $A_s$ the rate of radiative decays that stabilize into bound levels. For an isolated narrow resonance, the Saha-Boltzmann equilibrium ratio has the bound-state energy replaced by the positive resonance energy:At equilibrium, inverse dielectronic capture and autoionization fluxes balance: $N_e{N}_i{C}_d=N_d^{\mathrm{eq}}{A}_{di}^a$. This establishes the capture coefficient even when the actual plasma is not in local thermodynamic equilibrium. In the low-density, weak-radiation recombining plasma, the intermediate level instead has the stationary balanceMultiplying its population by the successful stabilization rate gives the actual recombination coefficientHere the equilibrium population was used only to determine the capture rate by detailed balance; it was not assumed to equal the intermediate population when photons escape. This is precisely why the competing decay probabilities appear.

Use the supplied core-transition relation between the Einstein coefficients and atomic oscillator strength,Here $\alpha$ is the fine-structure constant and $a_0$ the Bohr radius, not a recombination coefficient. Factor out $A_{ji}$ and defineSubstitution yieldsThis definition accommodates several entrance and decay channels. In the common spectator-electron approximation with one autoionization channel and stabilization by the core transition, $A_s=A_r=A_{ji}$ and $A_{di}^a=A_a$. If all coupled substates of the captured $n\ell$ electron are summed, $g_d=2(2\ell +1)g_j$, givingFor an intermediate state resolved by fine structure, retain its actual $g_d$ instead of that summed weight. The stabilization probability itself is $A_s/(A_a+A_r)$; the requested ${\beta }_{j,n\ell }$ is not just this branching probability, because the prefactor has already factored out the core radiative rate and statistical weights of atomic levels. Radiative transitions to another still-autoionizing state require that state's eventual survival probability, rather than automatically counting every emitted photon as stabilization.

The total dielectronic recombination coefficient sums all accessible positive-energy resonances. In this isolated-resonance approximation it has the formwith atomic factors included in $c_d$. The resonance-temperature dependence of dielectronic recombination follows by differentiating a single term's logarithm:Thus low temperatures suppress a positive-energy resonance, although resonances very near threshold can still matter; at sufficiently high temperature a fixed set of resonances falls as $T^{-3/2}$. Several core-excitation series can produce several peaks. When autoionization is much faster than radiation, the capture-and-survival product is limited by the stabilizing radiative rate; when capture is weak, autoionization is the limiting inverse-capture rate. For a fixed Rydberg angular channel, autoionization typically decreases at high principal quantum number while the core radiative rate changes little. Capture into very high levels is vulnerable to subsequent collisional or field ionization, so finite-density survival can suppress the total rate. dielectronic recombination is resonant and strongly temperature-selective; its magnitude can be important in coronal ionization balance.

The time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\psi =E\psi$, with attractive Coulomb potential $V(r)=-\kappa /r$, where $\kappa =e^2/(4\pi {\epsilon }_0)$. Separating $\psi =R(r)Y_{\ell }^m(\theta ,\phi )$ uses the spherical harmonic angular eigenvalue $-\ell (\ell +1)$ of the unit-sphere Laplacian. The radial kinetic contribution is $-{\hslash }^2(R^{\prime \prime }+2R^{\prime }/r)/(2m)$, the angular kinetic contribution is the centrifugal term ${\hslash }^2\ell (\ell +1)R/(2m{r}^2)$, and the remaining term is the attractive potential energy. The right side is the energy eigenvalue times the radial wavefunction. Here the stated $m$ neglects nuclear recoil; including it replaces the electron mass by the reduced mass.

For $\ell =0$, substitution of the ground-state exponential givesIts $1/r$ coefficient must vanish and its constant coefficient determines the energy, soHere $a_0={\hslash }^2/(m\kappa )$ is the Bohr radius.

For $R_2=N_2(r+b)e^{-\beta r}$, direct differentiation yieldsCancel the nonzero exponential in the radial Schrödinger equation and compare coefficients of $r$, the constant term, and $1/r$. They give, respectively,ConsequentlyThe positive decay constants ensure normalizability; the second radial wavefunction has one positive-radius node at $2a_0$, as appropriate to the first radial excitation.

If the entire energy gap is carried by one photon, neglecting atomic recoil, conservation of energy gives the requested frequencyThere is a physical qualification: these are the $2s$ and $1s$ states, so the usual single-photon electric-dipole transition is forbidden by the electric-dipole selection rules for hydrogen. Explicitly, both angular factors are $Y_0^0$, and $\int |Y_0^0{|}^2\hat{\mathbf{r}}d\Omega =0$, so every component of $⟨1s|\mathbf{r}|2s⟩$ vanishes. The leading isolated-atom decay is instead two-photon emission, for which ${\nu }_1+{\nu }_2={\nu }_{\mathrm{gap}}$ and individual frequencies are not fixed. This metastability and the two-photon channel are documented in NIST's hydrogen spectral-data compilation. Thus the boxed answer is the energy-gap frequency under the printed single-photon assumption, rather than an assertion of an allowed electric-dipole line.

With the proton treated as fixed and electron mass $m$, the classical energy and angular momentum areFor a circular orbit, $L=mrv$ and Coulomb's law determines its centripetal acceleration:Eliminating $v$ gives $r=4\pi {ϵ}_0{n}^2{\hslash }^2/(m{e}^2)$. Thus the Bohr radius and orbit radii areThe kinetic energy is half the magnitude of the attractive potential energy. HenceThis is the circular-orbit Bohr model; including proton recoil replaces $m$ by the reduced mass.

The time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\Psi =E\Psi$. For the hydrogen atom, separate variables as $\Psi =g(r)Y_{\ell m}(\theta ,\phi )$. The spherical-coordinate Laplacian has radial part $r^{-2}{\partial }_r(r^2{\partial }_r)$ and angular part $r^{-2}{\Delta }_{S^2}$. A spherical harmonic obeys ${\Delta }_{S^2}{Y}_{\ell m}=-\ell (\ell +1)Y_{\ell m}$. The first displayed term is therefore radial kinetic energy; the second is the attractive Coulomb's law potential energy $-e^2/(4\pi {\epsilon }_{0}r)$ of electron and proton; the third is the angular kinetic energy, also called the centrifugal potential. Regular single-valued angular states have $\boxed{\ell =0,1,2,\dots }$, with orbital angular momentum squared ${\hslash }^2\ell (\ell +1)$.

For $g=r^{\alpha }{e}^{-\beta r}$, direct differentiation givesSubstitution into the Radial Schrodinger equation for the hydrogen atom and equality of the coefficients of $r^{-2}$, $r^{-1}$ and the constant term giveThe regular radial solution selects $\alpha =\ell$, rather than the singular branch $-\ell -1$. At $\ell =0$, $r^{-1}$ is locally square integrable but is still inadmissible as a regular Coulomb eigenfunction: its Laplacian has a point-source contribution at the origin. Bound-state decay requires $\beta >0$. Thus, with the Bohr radius $a_0=4\pi {\epsilon }_0{\hslash }^2/(m{e}^2)$,These are the nodeless states with principal quantum number $n=\ell +1$. Increasing $\ell$ raises the energy towards zero, so emission is from the state labelled $\ell +1$ to the one labelled $\ell$. Using the Planck constant $h=2\pi \hslash$ and photon energy $h\nu$, the frequency is

The normalized spherical harmonics separate the angular variables. The Radial Schrodinger equation for the hydrogen atom isWrite $R=A{r}^l{e}^{-\kappa r}$. ThenSubstitution separates a constant term and a $1/r$ term. The latter vanishes precisely when ${\hslash }^2\kappa (l+1)/m=e^2/(4\pi {ϵ}_0)$. With $\kappa =1/[a(l+1)]$ this determines the Bohr radiusThe constant term determinesThese are the maximal-angular-momentum hydrogen states, with principal quantum number $n=l+1$.

Because the angular integral is one, normalization requiresThereforeThe phase of $A$ remains arbitrary.

Take the spherical harmonic $Y_{00}$ to have unit angular norm. The radial normalization condition is thenThereforeEquivalently, if $R$ denotes the entire spherically symmetric wavefunction rather than the radial factor multiplying normalized $Y_{00}$, then $C=({\gamma }^3/\pi {)}^{1/2}$.

The expected radius isFor the ground state, $\gamma =m{q}^2/{\hslash }^2$. The Bohr radius is $a_0={\hslash }^2/(m{q}^2)=1/\gamma$, soThe mean radius is therefore of the Bohr-radius scale and is one and a half times $a_0$. This is the radial normalization of the hydrogen ground state.

For fixed $l$, the radial equation isSubstituteCancellation of the coefficient of $1/r$ requiresthe reduced-mass Bohr radius. The remaining constant term givesEach such no-radial-node state is a circular Coulomb bound state with principal quantum number $n=l+1$.

The stationary Schrödinger equation isThe Laplacian represents the kinetic-energy operator. For a spherically symmetric function, $\Delta \psi ={\psi }^{\prime \prime }+2{\psi }^{\prime }/r$. The attractive Coulomb potential of the nucleus is $V=-q^2/(4\pi {ϵ}_{0}r)$; $E$ is the energy eigenvalue. Multiplication by $-2m/{\hslash }^2$ produces the displayed radial equation with Bohr radiusFor a normalizable bound state, write ${\kappa }^2=-2mE/{\hslash }^2$ and take $\lambda >0$. Substitution of $\psi =(A+Br)e^{-\lambda r}$ ingives, after division by the exponential, the coefficientsIf $B=0$ and $A\ne 0$, these force $\lambda =1/a$ and $\kappa =\lambda$. If $B\ne 0$, the first coefficient gives ${\kappa }^2={\lambda }^2$, the second gives $\lambda =1/(2a)$, and the third gives $B=-A/(2a)$. Up to nonzero scalar factors, the two radial bound states and their energies are consequentlyThey are linearly independent as functions and correspond to different energy eigenvalues. Strictly, they are not two solutions of one fixed-energy differential equation: each is a solution at its own energy.