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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]

For a bound state take $b>0$ and set $\psi (r)=e^{-br}v(r)$. Substitution into the radial equation and cancellation of the common exponential giveSeek a power series $v(r)={\sum }_{j\ge 0}{c}_j{r}^j$ with $c_0\ne 0$. Equating coefficients of $r^j$ yields the polynomial construction of hydrogen S states recurrenceFor any integer $N\ge 1$, choose $b=a/(2N)$. The numerator vanishes at $j=N-1$, while it is nonzero for $j<N-1$, so $v$ is a nonzero polynomial of degree $N-1$. Therefore $\psi$ is continuous at $r=0$ and decays exponentially as $r\to \infty$. The three-dimensional normalization integral isIt is finite near zero because $\psi$ is bounded, and finite at infinity because an exponential dominates every polynomial. Multiplying by its positive reciprocal square root gives a normalizable, nonzero wavefunction. The recurrence ensures the differential equation for $r>0$; at the singular endpoint the regular continuous extension has $2{\psi }^{\prime }(0)+a\psi (0)=0$, so the potentially singular coefficient cancels. No nonexistent value of $1/r$ at zero is substituted.

Since $E=-{\hslash }^2{b}^2/(2m)$, the choice of $b$ givesFor $N=1$, the normalized example is $\psi (r)=\sqrt{b^3/\pi }{e}^{-br}$. For $N=2$, $v(r)$ is proportional to $1-br$. These illustrate the polynomial-times-exponential Coulomb bound states produced for every required energy. The radial derivative at the origin is generally nonzero, a Coulomb cusp consistent with the requested continuity.

Use the physical Coulomb bound state boundary condition: the spherically symmetric wavefunction is regular at the origin and square integrable with measure $4\pi r^{2}dr$. For $E<0$, setAt large $r$, the decaying branch behaves exponentially as $e^{-Kr}$. Substituting $\psi =f{e}^{-Kr}$ into the radial Schrödinger equation givesThe regular solution has a power series $f(r)={\sum }_{j\ge 0}{a}_j{r}^j$, with recurrenceA nonzero regular solution has $a_0\ne 0$. If the series does not terminate, its coefficients eventually have one sign after choosing an overall real phase. For any fixed $0<\delta <1/2$ and sufficiently large $j$, the ratio of consecutive magnitudes is at least $2K(1-\delta )/(j+2)$. The tail therefore grows at least as a positive constant times $e^{2K(1-\delta )r}$ divided by a power of $r$; a finite polynomial of earlier terms cannot cancel it. Multiplication by $e^{-Kr}$ leaves exponential growth, contradicting normalizability.

Thus the series must terminate at degree $N-1$, where $KN=\lambda$ and $N$ is a positive integer. Conversely, termination gives a regular polynomial times $e^{-\lambda r/N}$, which is normalizable. ThereforeThese are the $\ell =0$ members of the hydrogen spectrum. Regularity at the origin is essential: square integrability alone on the punctured radial interval would also admit a singular $1/r$ branch. Such a branch is not a physical eigenfunction of the ordinary three-dimensional Coulomb Hamiltonian; its singularity introduces an inadmissible point-source contribution. In the reduced radial variable $u=r\psi$, the physical condition is $u(0)=0$.

The agreement is in the energy levels, not in electron trajectories. The Bohr model assumes a particle on a circular orbit, with orbital angular momentum quantized as $N\hslash$. The Coulomb bound states here are spatially distributed wavefunctions with no definite classical orbit. Their probability density is spherically symmetric, and their orbital angular momentum is zero because $\ell =0$. Their radial distributions also depend on $N$ and can have nodes. A real radial wavefunction times a global stationary phase has zero probability current, unlike the circulating point charge in the Bohr picture.

Introduce the reduced radial wavefunction $u(r)=r\psi (r)$. Multiplying the radial Schrödinger equation for a central potential by $-2mr/{\hslash }^2$ givesFor the decaying bound state, take $\alpha >0$ and put $u=f{e}^{-\alpha r}$. Differentiation yieldsSince $2mE/{\hslash }^2=-{\alpha }^2$, the two constant multiples of $f$ cancel. Dividing by the nonzero exponential givesThis separates the large-radius exponential decay from the remaining Coulomb bound state equation.

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.