27 May 2025

Atomic Spectra

Atomic models, hydrogen wavefunctions, angular momentum, coupling schemes, field splittings, periodic structure, and characteristic X-rays.

bsc semester-vi modern-physics atomic-spectra hydrogen-atom

Atomic spectra are discrete because bound atomic states have discrete energies. The route from Rutherford scattering to the hydrogen Schrödinger equation also shows why an orbit is not a quantum state.

Rutherford scattering and the nuclear atom

For a projectile of charge $ze$ scattered by a nucleus of charge $Ze$, the repulsive Coulomb potential is

\[V(r)=\frac{Zze^2}{4\pi\epsilon_0r}\equiv\frac{K}{r}.\]

Conservation of angular momentum gives $L=\mu vb$, where $b$ is the impact parameter and $\mu$ is the projectile-nucleus reduced mass. With $u=1/r$, Binet’s equation for the outward force $K/r^2$ is

\[\frac{d^2u}{d\phi^2}+u=-\frac{\mu K}{L^2}.\]

Its hyperbolic solution can be written

\[r=\frac{p}{e\cos\phi-1},\qquad p=\frac{L^2}{\mu K},\qquad e^2=1+\frac{2EL^2}{\mu K^2}.\]

The asymptotes satisfy $\cos\phi_\infty=1/e$. Their geometry gives $\theta=\pi-2\phi_\infty$, so

\[\cot\frac{\theta}{2}=\tan\phi_\infty =\sqrt{e^2-1} =\frac{2Eb}{K},\]

where $E=\mu v^2/2$ and $L=\mu vb$ were used in the last step. Therefore

\[b=\frac{K}{2E}\cot\frac{\theta}{2},\qquad E=\frac12\mu v^2.\]

Particles incident in the annulus $2\pi b\,db$ enter the solid angle $d\Omega=2\pi\sin\theta\,d\theta$. Hence

\[\frac{d\sigma}{d\Omega} =\frac{b}{\sin\theta}\left\lvert\frac{db}{d\theta}\right\rvert =\left(\frac{K}{4E}\right)^2\csc^4\frac{\theta}{2}.\]

The observed large-angle events require charge and most of the mass to occupy a region far smaller than the atom; this is the nuclear model inferred from Rutherford’s experiment.

Bohr hydrogen atom and reduced mass

For a circular orbit, Coulomb attraction supplies centripetal force:

\[\frac{\mu v^2}{r}=\frac{e^2}{4\pi\epsilon_0r^2},\]

where nuclear motion is included through the reduced mass

\[\mu=\frac{m_eM}{m_e+M}.\]

Bohr quantization, or equivalently the standing-wave condition $2\pi r=n\lambda$ with $\lambda=h/(\mu v)$, gives

\[\mu vr=n\hbar.\]

Eliminating $v$ produces

\[r_n=\frac{4\pi\epsilon_0\hbar^2}{\mu e^2}n^2,\]

and the total energy is

\[E_n=\frac12\mu v^2-\frac{e^2}{4\pi\epsilon_0r_n} =-\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2}\frac1{n^2}.\]

The same quantization condition is the de Broglie standing-wave relation because $\lambda_{dB}=h/(\mu v)$ and $2\pi r=n\lambda_{dB}$. For emission $n_i\to n_f<n_i$,

\[\frac1\lambda=R_M\left(\frac1{n_f^2}-\frac1{n_i^2}\right),\qquad R_M=R_\infty\frac{\mu}{m_e}.\]

Thus isotope spectra differ slightly because $\mu$ depends on nuclear mass. The Ritz combination principle follows immediately: if $\tilde\nu_{ab}=T_a-T_b$, then $\tilde\nu_{ac}=\tilde\nu_{ab}+\tilde\nu_{bc}$. Bohr’s model explains hydrogenic energies, radii, and the Rydberg law, but not line intensities, fine structure, multielectron atoms, or why an accelerating charge in a stationary orbit does not radiate.

Hydrogen Schrödinger equation

With $V(r)=-e^2/(4\pi\epsilon_0r)$, the time-independent equation is

\[\left[-\frac{\hbar^2}{2\mu}\nabla^2-\frac{e^2}{4\pi\epsilon_0r}\right]\psi=E\psi.\]

In spherical coordinates,

\[\nabla^2=\frac1{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) +\frac1{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) +\frac1{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}.\]

Put $\psi=R(r)\Theta(\theta)\Phi(\phi)$. Separation of $\phi$ gives

\[\frac{d^2\Phi}{d\phi^2}+m_l^2\Phi=0,\qquad \Phi(\phi+2\pi)=\Phi(\phi),\]

so $m_l$ is an integer. The polar equation is

\[\frac1{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right) +\left[l(l+1)-\frac{m_l^2}{\sin^2\theta}\right]\Theta=0,\]

whose regular solutions require $l=0,1,2,\ldots$ and $m_l=-l,\ldots,l$. The angular functions combine into $Y_l^{m_l}$. The radial equation is

\[\frac1{r^2}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right) +\left[\frac{2\mu}{\hbar^2}\left(E+\frac{e^2}{4\pi\epsilon_0r}\right)-\frac{l(l+1)}{r^2}\right]R=0.\]

Writing $u=rR$ removes the first derivative:

\[-\frac{\hbar^2}{2\mu}\frac{d^2u}{dr^2} +\left[-\frac{e^2}{4\pi\epsilon_0r}+\frac{\hbar^2l(l+1)}{2\mu r^2}\right]u=Eu.\]

For a bound state define

\[\kappa=\frac{\sqrt{-2\mu E}}{\hbar},\qquad \rho=2\kappa r,\]

and set $u(\rho)=\rho^{l+1}e^{-\rho/2}v(\rho)$. The factor $\rho^{l+1}$ enforces $u(0)=0$, while $e^{-\rho/2}$ enforces decay as $r\to\infty$. Substitution gives

\[\rho v^{\prime\prime}+[2(l+1)-\rho]v^{\prime} +\left[\frac{\mu e^2}{4\pi\epsilon_0\hbar^2\kappa}-(l+1)\right]v=0.\]

The series for $v$ must terminate; otherwise its $e^\rho$ growth defeats the decaying prefactor. Therefore

\[\frac{\mu e^2}{4\pi\epsilon_0\hbar^2\kappa}=n_r+l+1\equiv n,\]

giving $n=1,2,\ldots$, $l=0,\ldots,n-1$, and the same $E_n$ as the Bohr result. The normalized ground state is

\[\psi_{100}(r)=\frac{e^{-r/a_\mu}}{\sqrt{\pi a_\mu^3}},\qquad a_\mu=\frac{4\pi\epsilon_0\hbar^2}{\mu e^2}.\]

The electric-dipole matrix element $\langle n^{\prime}l^{\prime}m_l^{\prime}\rvert\mathbf r\lvert nlm_l\rangle$ and the angular properties of $\mathbf r$ give the selection rules

\[\boxed{\Delta l=\pm1,\qquad \Delta m_l=0,\pm1},\]

with parity changing between the two states.

Orbital and spin angular momentum

The vector model represents

\[L^2=l(l+1)\hbar^2,\quad L_z=m_l\hbar, \qquad S^2=s(s+1)\hbar^2,\quad S_z=m_s\hbar.\]

For an electron $s=1/2$ and $m_s=\pm1/2$. Its magnetic moments are

\[\boldsymbol\mu_L=-\frac{e}{2m_e}\mathbf L=-\frac{\mu_B}{\hbar}\mathbf L,\] \[\boldsymbol\mu_S=-g_s\frac{\mu_B}{\hbar}\mathbf S,\qquad \mu_B=\frac{e\hbar}{2m_e},\quad g_s\simeq2.\]

The constant $\mu_B=e\hbar/(2m_e)$ is the Bohr magneton. The Stern-Gerlach force $F_z=\mu_z\,dB_z/dz$ separates an atomic beam into discrete components, demonstrating space quantization. In a uniform field, $d\mathbf L/dt=\boldsymbol\mu_L\times\mathbf B$ gives Larmor precession with

\[\omega_L=\frac{eB}{2m_e}.\]

Adding angular momenta gives $\mathbf J=\mathbf L+\mathbf S$ and

\[j=\lvert l-s\rvert,\lvert l-s\rvert+1,\ldots,l+s.\]

Since $2\mathbf L\cdot\mathbf S=J^2-L^2-S^2$, a spin-orbit Hamiltonian $H_{so}=\xi(r)\mathbf L\cdot\mathbf S$ shifts a level by

\[\Delta E_{so}=\frac{\xi}{2}\hbar^2[j(j+1)-l(l+1)-s(s+1)].\]

For weak residual interactions in light atoms, individual orbital momenta form $\mathbf L=\sum_i\mathbf l_i$, spins form $\mathbf S=\sum_i\mathbf s_i$, and then $\mathbf J=\mathbf L+\mathbf S$: this is LS coupling. In heavy atoms, each $\mathbf j_i=\mathbf l_i+\mathbf s_i$ forms first and $\mathbf J=\sum_i\mathbf j_i$: this is jj coupling.

An LS-coupled level is written $^{2S+1}L_J$, where $2S+1$ is the spin multiplicity and $L=0,1,2,\ldots$ is denoted by $S,P,D,\ldots$. For an electric-dipole line, the many-electron rules include $\Delta S=0$, $\Delta L=0,\pm1$ (but not $L=0\leftrightarrow0$), and $\Delta J=0,\pm1$ (but not $J=0\leftrightarrow0$), together with a parity change. These are consequences of angular-momentum and parity matrix elements, not independent empirical restrictions.

Equation-generated level positions: $E_n=-13.6/n^2$ eV, spin-orbit offsets proportional to $[j(j+1)-l(l+1)-s(s+1)]/2$, and linear Zeeman shifts $\Delta E=\mu_Bg_Jm_JB$.

Exchange symmetry, Pauli principle, and periodic structure

For identical particles, exchange cannot produce a physically distinguishable state:

\[\Psi(1,2)=\pm\Psi(2,1).\]

Integer-spin particles use the symmetric sign; electrons are fermions and require the antisymmetric sign. If two electrons occupied the same one-particle state $a$,

\[\Psi=\frac1{\sqrt2}[\psi_a(1)\psi_a(2)-\psi_a(2)\psi_a(1)]=0.\]

This is Pauli exclusion: no two electrons in an atom share all four quantum numbers. Shell capacity follows by counting $m_l$ and spin states:

\[N_n=2\sum_{l=0}^{n-1}(2l+1)=2n^2.\]

Successive filling of subshells, modified by their energies and exchange, produces the periodic table.

Zeeman, Paschen-Back, and Stark effects

In a weak magnetic field, $J$ remains coupled and

\[\Delta E_Z=\mu_Bg_Jm_JB,\]

where

\[g_J=1+\frac{j(j+1)+s(s+1)-l(l+1)}{2j(j+1)}.\]

In a strong field the spin-orbit coupling is overcome. The Paschen-Back shift approaches

\[\Delta E_{PB}=\mu_BB(m_l+g_sm_s).\]

An electric field $\mathbf E=E\hat z$ adds $H^{\prime}=eEz$ for an electron in the nuclear potential. The first-order shift is $\Delta E^{(1)}=eE\langle z\rangle$. It vanishes for a nondegenerate parity eigenstate, but hydrogen states with the same $n$ are degenerate and mix, producing a linear Stark effect. Nondegenerate states begin with the quadratic shift

\[\Delta E^{(2)}=-\frac12\alpha E^2,\]

where $\alpha$ is the static polarizability.

Characteristic X-ray spectra

An incident electron can remove an inner-shell electron. When an outer electron fills the vacancy, a photon carries the energy difference:

\[h\nu=E_{n_i}-E_{n_f}.\]

Screening replaces $Z$ by $Z-\sigma$, so the hydrogenic approximation gives Moseley’s form

\[\sqrt\nu=C(Z-\sigma).\]

The $K_\alpha$ line is predominantly a $2p\to1s$ transition and the $K_\beta$ line a $3p\to1s$ transition, consistent with $\Delta l=\pm1$. These discrete characteristic lines lie on the continuous bremsstrahlung spectrum.

Solved Problems

1. Balmer-$\alpha$ wavelength and photon energy

Find the wavelength and photon energy for the hydrogen transition $n_i=3\to n_f=2$. Use $R_H=1.0967758\times10^7\ \mathrm{m^{-1}}$.

The wavenumber is positive for emission when the lower level is written first:

\[\frac1\lambda=R_H\left(\frac1{2^2}-\frac1{3^2}\right) =R_H\frac5{36}.\]

Hence

\[\lambda=\frac{36}{5R_H}=6.56470\times10^{-7}\ \mathrm m =656.470\ \mathrm{nm}.\]

The photon carries the decrease in atomic energy,

\[E_\gamma=\frac{hc}{\lambda}=3.02595\times10^{-19}\ \mathrm J =1.88865\ \mathrm{eV}.\]

The sign of the atomic energy change is $\Delta E_{atom}=-E_\gamma$; the emitted photon energy is positive. The dimensions follow from $R_H^{-1}$ for length and $hc/\lambda$ for energy.

2. Landé factor and weak-field Zeeman shift

For a $2p_{3/2}$ level in a field $B=0.800\ \mathrm T$, find the shift of the $m_J=3/2$ component.

Here $l=1$, $s=1/2$, and $j=3/2$, so

\[g_J=1+\frac{j(j+1)+s(s+1)-l(l+1)}{2j(j+1)} =1+\frac{15/4+3/4-2}{2(15/4)}=\frac43.\]

Taking the zero-field level as the energy origin,

\[\Delta E_Z=\mu_Bg_Jm_JB =(5.7883818\times10^{-5}\ \mathrm{eV\,T^{-1}}) \left(\frac43\right)\left(\frac32\right)(0.800\ \mathrm T) =9.26141\times10^{-5}\ \mathrm{eV}.\]

The shift is upward because $g_Jm_JB>0$; the $m_J=-3/2$ component shifts downward by the same magnitude. The linear result requires $\mu_BB$ to remain small compared with the spin-orbit splitting.

Descriptive Questions

1. Starting from the separated hydrogen radial equation, explain why regularity at the origin, decay at infinity, and termination of the confluent-hypergeometric series quantize the energy.
2. Derive the shell capacity $2n^2$ from the allowed orbital and spin quantum numbers, and connect the counting to Pauli exclusion.
3. Contrast LS and jj coupling by stating the order in which angular momenta combine and the interaction regime in which each scheme is appropriate.
4. Explain why a nondegenerate parity eigenstate has no first-order Stark shift, whereas degenerate hydrogen levels can show a linear Stark effect.

Numerical Problems

1. A $5.00\ \mathrm{MeV}$ alpha particle approaches a gold nucleus ($Z=79$) head-on. Neglect recoil and use $e^2/(4\pi\epsilon_0)=1.43996448\ \mathrm{MeV\,fm}$. Find the distance of closest approach.

   Answer: $r_{min}=45.5029\ \mathrm{fm}$.

2. For a $2p$ electron, suppose $\xi\hbar^2=2.00\times10^{-5}\ \mathrm{eV}$. Calculate the $j=3/2$ and $j=1/2$ spin-orbit shifts and their separation.

   Answer: $\Delta E_{3/2}=+1.00\times10^{-5}\ \mathrm{eV}$, $\Delta E_{1/2}=-2.00\times10^{-5}\ \mathrm{eV}$; separation $=3.00\times10^{-5}\ \mathrm{eV}$.

3. A spin component with magnetic-moment projection $\mu_z=\mu_B$ traverses a Stern-Gerlach field gradient $dB_z/dz=20.0\ \mathrm{T\,m^{-1}}$. Find the force magnitude.

   Answer: $\lvert F_z\rvert=1.85480\times10^{-22}\ \mathrm N$.

4. Estimate the copper $K_\alpha$ photon energy with $Z=29$, screening constant $\sigma=1$, and the hydrogenic value $13.6\ \mathrm{eV}$. Use the $2p\to1s$ energy difference.

   Answer: $E_{K_\alpha}=7.9968\ \mathrm{keV}$.

Maxima verification worksheet

References

1. Atomic spectroscopy — Wikipedia
2. NIST Atomic Spectra Database
3. B. H. Bransden and C. J. Joachain, Physics of Atoms and Molecules, 2nd ed., chapters on one-electron atoms, angular momentum, and external-field effects.
4. C. J. Foot, Atomic Physics, 2nd ed., chapters 2–5 on hydrogen, helium, fine structure, and atoms in external fields.