Complete Atomic Physics

Early Atomic Physics

Core ideas

Atomic physics began when spectra and scattering showed that atoms have internal structure. Rutherford scattering revealed a small charged nucleus, while Balmer lines and the Bohr model showed that bound energies are discrete. Old quantum theory was incomplete, but it introduced angular momentum quantization, correspondence ideas, and the link between frequency and energy differences.

For review, be able to explain Rutherford scattering, use Bohr energies and radii, connect spectral lines to transitions, and state why full wave mechanics was needed. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$E_n=-\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2}\frac{1}{n^2}=-\frac{13.6\,\mathrm{eV}}{n^2}\quad(\mathrm{H})$$

Worked example

A beam of 5.0 MeV alpha particles is directed at a thin gold foil ($Z=79$). For a head-on collision, the distance of closest approach $r_{\min}$ is found by equating the initial kinetic energy to the Coulomb potential energy at the turning point:

$$E_k=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{r_{\min}}.$$

Using $e^2/(4\pi\varepsilon_0)=1.44\;\mathrm{MeV\,fm}$,

$$r_{\min}=\frac{2\times79\times1.44\;\mathrm{MeV\,fm}}{5.0\;\mathrm{MeV}}\approx 45.5\;\mathrm{fm}.$$

This confirms Rutherford’s conclusion that the nucleus is smaller than $10^{-14}$ m, while the atomic radius is $\sim10^{-10}$ m.

Problems with Solutions

Problem 1. Use the Bohr model to compute the radius and energy of the ground state ($n=1$) of hydrogen. Give numerical values in pm and eV. Solution. The Bohr radius is

$$a_0=\frac{4\pi\varepsilon_0\hbar^2}{m_e e^2}\approx 52.9\;\mathrm{pm}.$$

The ground-state energy is

$$E_1=-\frac{m_e e^4}{2(4\pi\varepsilon_0)^2\hbar^2}\approx -13.6\;\mathrm{eV}.$$

Problem 2. In the Geiger—Marsden experiment, about 1 in 8000 alpha particles were scattered by more than $90^\circ$ by a gold foil. Explain why this single-scattering result is incompatible with the Thomson plum-pudding model. Solution. In Thomson’s model the positive charge is spread over the whole atom ($\sim 10^{-10}$ m), so the maximum Coulomb repulsion is far too weak to reverse the momentum of a heavy, energetic alpha particle. The observed large-angle scattering requires a compact, massive nucleus with $r\lesssim 10^{-14}$ m so that the Coulomb force can be enormous at short range.

Problem 3. Calculate the wavelength of the photon emitted in the Balmer $n=3\to 2$ transition in hydrogen. Solution. Using $E_n=-13.6\;\mathrm{eV}/n^2$,

$$\Delta E = 13.6\left(\frac{1}{4}-\frac{1}{9}\right)=13.6\times\frac{5}{36}\approx 1.89\;\mathrm{eV}.$$

Then

$$\lambda=\frac{hc}{\Delta E}=\frac{1240\;\mathrm{eV\,nm}}{1.89\;\mathrm{eV}}\approx 656\;\mathrm{nm}.$$

Section summary. Spectroscopy and scattering forced the quantum view of atomic structure.

The Hydrogen Atom

Core ideas

Hydrogen is the central exactly soluble atom. Separating the Schrodinger equation in spherical coordinates gives quantum numbers $n$, $\ell$, and $m$, radial wave functions, spherical harmonics, degeneracies, and selection rules. Reduced mass, spin-orbit coupling, relativistic corrections, and the Lamb shift refine the simple Coulomb spectrum.

For review, be able to write the Coulomb Hamiltonian, identify quantum numbers and degeneracy, use parity and angular momentum selection rules, and estimate fine-structure scales. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\hat H=-\frac{\hbar^2}{2\mu}\nabla^2-\frac{e^2}{4\pi\epsilon_0 r},\qquad E_n=-\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2n^2}$$

Worked example

The Ly-$\alpha$ transition in hydrogen corresponds to $n=2\to 1$. The wavelength is obtained from the Rydberg formula

$$\frac{1}{\lambda}=R_{\!H}\!\left(1-\frac{1}{4}\right)=\frac{3}{4}R_{\!H},$$

with $R_{\!H}\approx 1.097\times 10^7\;\mathrm{m^{-1}}$. Thus

$$\lambda=\frac{4}{3R_{\!H}}\approx 121.6\;\mathrm{nm}.$$

This ultraviolet line is a cornerstone of astrophysical spectroscopy.

Problems with Solutions

Problem 1. What is the most probable radius of the hydrogen $1s$ state, and what is the probability of finding the electron inside a sphere of radius $a_0$? Solution. The radial probability density is $P(r)=4\pi r^2|\psi_{1s}|^2=(4r^2/a_0^3)e^{-2r/a_0}$. Setting $dP/dr=0$ gives $r=a_0\approx 52.9$ pm. The probability inside $r=a_0$ is

$$\int_0^{a_0}P(r)\,dr=1-5e^{-2}\approx 0.32\;(32\%).$$

Problem 2. Estimate the order of magnitude of the fine-structure splitting for the $n=2$ level of hydrogen. Solution. Fine-structure scales as $\alpha^2$ relative to the binding energy, where $\alpha\approx 1/137$. The $n=2$ binding energy is $3.4$ eV, so

$$\Delta E_{\rm fs}\sim \alpha^2\times 3.4\;\mathrm{eV}\approx \left(\frac{1}{137}\right)^2\times 3.4\;\mathrm{eV}\approx 1.8\times 10^{-4}\;\mathrm{eV},$$

corresponding to $\sim 50$ GHz. (The exact Lamb shift is $4.38\times10^{-6}$ eV, and the Dirac fine structure is $\sim 4.5\times10^{-5}$ eV.)

Problem 3. Calculate the difference in the ionization energy of hydrogen ($^1$H) and deuterium ($^2$H) caused by the reduced-mass correction. Solution. The Rydberg constant for a nucleus of mass $M$ is $R_M=R_\infty/(1+m_e/M)$. For H ($M/m_e\approx 1836$) and D ($M/m_e\approx 3670$),

$$\frac{\Delta R}{R_\infty}\approx \frac{m_e}{M_{\rm H}}-\frac{m_e}{M_{\rm D}}\approx \frac{1}{1836}-\frac{1}{3670}\approx 2.72\times10^{-4}.$$

Since $E_n\propto R$, the ionization energy differs by $\Delta E\approx 13.6\times 2.72\times10^{-4}\;\mathrm{eV}\approx 3.7\times10^{-3}$ eV ($\sim 30$ GHz).

Section summary. Hydrogen supplies the template for quantum numbers, orbitals, and atomic spectra.

Helium

Core ideas

Helium is the simplest atom where electron-electron repulsion matters. The independent-particle picture gives a starting point, but correlation and exchange split singlet and triplet states. Variational methods, perturbation theory, and Hartree—Fock ideas explain why identical-particle symmetry changes energies and spectra.

For review, be able to construct symmetric and antisymmetric two-electron states, distinguish singlet from triplet, estimate screening, and state why exact separation fails. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\hat H=\hat h(1)+\hat h(2)+\frac{e^2}{4\pi\epsilon_0 r_{12}},\qquad \Psi_{\rm total}\ \mathrm{antisymmetric}$$

Worked example

Estimate the ground-state energy of helium using first-order perturbation theory. The unperturbed Hamiltonian uses two hydrogenic electrons with $Z=2$:

$$E^{(0)}=2\times(-Z^2\times 13.6\;\mathrm{eV})=-108.8\;\mathrm{eV}.$$

The first-order correction from electron—electron repulsion is

$$E^{(1)}=\left\langle\frac{e^2}{4\pi\varepsilon_0 r_{12}}\right\rangle\approx \frac{5}{4}Z\times 13.6\;\mathrm{eV}=34.0\;\mathrm{eV}.$$

Thus $E\approx -108.8+34.0=-74.8$ eV, which compares reasonably with the experimental total energy of $-79.0$ eV (first ionization energy 24.6 eV plus He$^+$ ground state $-54.4$ eV).

Problems with Solutions

Problem 1. Explain why there is no $1\,{}^3S_1$ state in helium. Solution. For two electrons both in the $1s$ orbital ($n=1,\ell=0$), the spatial wavefunction must be symmetric. The total wavefunction (spatial $\times$ spin) must be antisymmetric under exchange, so the spin part must be antisymmetric, i.e. the singlet $S=0$. Therefore only the parahelium $1\,{}^1S_0$ state exists; the orthohelium $1\,{}^3S_1$ state is forbidden by the Pauli exclusion principle.

Problem 2. Calculate the ground-state energy of singly ionized helium (He$^+$). Solution. He$^+$ is hydrogenic with $Z=2$, so

$$E_1=-Z^2\times 13.6\;\mathrm{eV}=-54.4\;\mathrm{eV}.$$

Its ionization energy is therefore $54.4$ eV, compared with $13.6$ eV for hydrogen.

Problem 3. In a simple variational treatment of helium, the effective nuclear charge is found to be $Z_{\rm eff}\approx 1.34$ for each electron. Use this to estimate the first ionization energy. Solution. With screening, each electron experiences an effective hydrogenic potential with $Z_{\rm eff}=1.34$. The energy of one electron is

$$E\approx -Z_{\rm eff}^2\times 13.6\;\mathrm{eV}\approx -(1.34)^2\times 13.6\;\mathrm{eV}\approx -24.4\;\mathrm{eV}.$$

The ionization energy is roughly $24.4$ eV, close to the experimental value of $24.6$ eV.

Section summary. Helium introduces correlation, exchange, and approximation methods.

The Alkalis

Core ideas

Alkali atoms have one valence electron outside closed shells. They are hydrogen-like at large radius, but core screening and penetration shift levels by quantum defects. Spin-orbit splitting and optical doublets make alkalis central in spectroscopy, laser cooling, clocks, and quantum control.

For review, be able to use effective one-electron energies, interpret quantum defects, identify fine-structure doublets, and connect spectra to valence-electron wave functions. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$E_{n\ell}\approx-\frac{R_\infty hc}{(n-\delta_\ell)^2},\qquad \Delta E_{\rm so}\propto \langle r^{-3}\rangle\,\bm L\cdot\bm S$$

Worked example

The sodium D-lines ($3p\to 3s$) occur at $589.0$ nm ($D_2$) and $589.6$ nm ($D_1$). The centre-of-gravity energy is about $2.10$ eV above the $3s$ ground state. Using the quantum-defect formula for the $3s$ level,

$$E_{3s}=-\frac{R_\infty hc}{(3-\delta_s)^2}\approx -5.14\;\mathrm{eV},$$

so that $(3-\delta_s)^2=13.6/5.14\approx 2.65$, giving $\delta_s\approx 1.37$. The large $\delta_s$ reflects strong penetration of the $3s$ electron into the Ne core. For the $3p$ level ($E_{3p}\approx -3.04$ eV), one finds $\delta_p\approx 0.86$, showing weaker penetration.

Problems with Solutions

Problem 1. Estimate the quantum defect $\delta_p$ for sodium from the $3p$ level energy ($E_{3p}\approx -3.04$ eV). Solution. Using $E_{n\ell}=-R_\infty hc/(n-\delta_\ell)^2$ with $R_\infty hc=13.6$ eV,

$$(3-\delta_p)^2=\frac{13.6}{3.04}\approx 4.47\;\Rightarrow\;3-\delta_p\approx 2.11\;\Rightarrow\;\delta_p\approx 0.89.$$

Problem 2. The fine-structure splitting of the Na D-lines is $\Delta\lambda=0.6$ nm. Convert this to an energy splitting $\Delta E_{\rm so}$. Solution. For $\lambda\approx 589$ nm,

$$\Delta E_{\rm so}\approx\frac{hc\,\Delta\lambda}{\lambda^2} =\frac{(1240\;\mathrm{eV\,nm})(0.6\;\mathrm{nm})}{(589\;\mathrm{nm})^2} \approx 2.1\times10^{-3}\;\mathrm{eV}\approx 17\;\mathrm{cm^{-1}}.$$

This spin-orbit splitting is much larger than in hydrogen because $\langle r^{-3}\rangle$ is bigger for the penetrating $3p$ valence electron.

Problem 3. Compute the reduced-mass correction to the Rydberg constant for lithium ($M/m_e\approx 12786$). Solution. The correction factor is $1/(1+m_e/M)$. The fractional shift is

$$\frac{\Delta R}{R_\infty}\approx\frac{m_e}{M}\approx\frac{1}{12786}\approx 7.8\times10^{-5}.$$

For a transition at $\lambda=670.8$ nm (Li $2s\to 2p$), the isotope shift between $^6$Li and $^7$Li (mass ratio $\sim 7/6$) is about $0.15$ nm, readily resolved with a grating spectrometer.

Section summary. Alkalis behave like corrected hydrogen atoms with experimentally useful optical lines.

The LS-Coupling Scheme

Core ideas

Many-electron atoms are organized by adding orbital and spin angular momenta. In light atoms, electrostatic interactions usually establish total $L$ and $S$ first, then spin-orbit coupling forms $J$. Term symbols encode $S$, $L$, $J$, parity, and selection rules; Hund’s rules give useful energy ordering.

For review, be able to read and build term symbols, apply angular momentum addition, use Hund’s rules qualitatively, and know when $jj$ coupling replaces LS coupling. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$${}^{2S+1}L_J,\qquad \bm L=\sum_i\bm \ell_i,\quad \bm S=\sum_i\bm s_i,\quad \bm J=\bm L+\bm S$$

Worked example

The ground configuration of carbon is $1s^22s^22p^2$. For the two equivalent $p$ electrons the possible terms are $^1S$, $^1D$ and $^3P$. By Hund’s rules:

1. Maximize $S$: the triplet $^3P$ ($S=1$) lies lowest.
2. Maximize $L$: for the triplet, $L=1$ is fixed.
3. Less than half-filled subshell ($2p^2$): the lowest $J$ is $|L-S|=0$. Hence the ground term is $^3P_0$.

Problems with Solutions

Problem 1. Determine the ground-state term symbol for nitrogen ($1s^22s^22p^3$). Solution. Three equivalent $p$ electrons. Hund’s first rule gives maximum $S=3/2$ (all spins parallel). To satisfy the Pauli principle the spatial state must then have $L=0$ (one electron in each $m_\ell=+1,0,-1$). The subshell is exactly half-filled, so $J=S=3/2$. The ground term is $^4S_{3/2}$.

Problem 2. How many microstates (distinct $|M_L,M_S\rangle$ states) belong to a $^3D$ term? Solution. The statistical weight of a term is $(2S+1)(2L+1)$. For $^3D$, $S=1$ and $L=2$, giving $3\times 5=15$ microstates.

Problem 3. For a $^2P$ term, list the possible $J$ values and the degeneracy of each level. Solution. With $S=1/2$ and $L=1$, the allowed $J$ are $|L-S|=1/2$ and $L+S=3/2$. The levels are $^2P_{1/2}$ and $^2P_{3/2}$. Their degeneracies are $2J+1$, i.e. $2$ and $4$ respectively. The sum $2+4=6$ equals $(2S+1)(2L+1)$, as required.

Section summary. Coupling schemes turn complicated spectra into angular-momentum bookkeeping.

Hyperfine Structure and Isotope Shift

Core ideas

Hyperfine structure comes from nuclear spin interacting with electronic magnetic fields and electric field gradients. Isotope shifts come from nuclear mass and charge-radius differences. These small splittings are essential in clocks, precision tests, and laser spectroscopy.

For review, be able to combine $I$ and $J$ into $F$, compute allowed $F$ values, identify magnetic dipole hyperfine splitting, and separate normal mass, specific mass, and field shifts. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\bm F=\bm I+\bm J,\qquad E_F=\frac{A}{2}\left[F(F+1)-I(I+1)-J(J+1)\right]$$

Worked example

The 21-cm line of interstellar hydrogen arises from hyperfine splitting of the $1s$ ground state ($I=1/2$, $J=1/2$). The total angular momentum is $\bm F=\bm I+\bm J$, giving $F=0$ and $F=1$. The energy formula

$$E_F=\frac{A}{2}\bigl[F(F+1)-I(I+1)-J(J+1)\bigr]$$

yields $E_1=A/4$ and $E_0=-3A/4$, so $\Delta E=A$. The observed transition frequency is $1420.4$ MHz, hence

$$A=h\times 1.4204\;\mathrm{GHz}\approx 5.87\times10^{-6}\;\mathrm{eV},$$

corresponding to $\lambda=21.1$ cm. This radio line maps the cold hydrogen in our Galaxy.

Problems with Solutions

Problem 1. An atom has nuclear spin $I=3/2$ and electronic angular momentum $J=1/2$. Find the allowed values of $F$ and the hyperfine energy shifts relative to the centroid. Solution. The allowed $F$ are $|I-J|,\dots,I+J$, i.e. $F=1$ and $F=2$. Using the same formula:

• $F=1$: $\Delta E_1=(A/2)[2-15/4]=-7A/8$.
• $F=2$: $\Delta E_2=(A/2)[6-15/4]=+9A/8$. The splitting between them is $\Delta E_2-\Delta E_1=2A$.

Problem 2. The nuclear magnetic moment of $^{87}$Rb ($I=3/2$) is $\mu_I=+2.75\,\mu_N$. Calculate the magnetic-dipole hyperfine constant $A$ for the $5s\,{}^2S_{1/2}$ ground state if the electron spin density at the nucleus creates an effective field $B_{\rm eff}\approx 13.7$ T. Solution. The hyperfine interaction is $H_D=A\,\bm I\cdot\bm J/\hbar^2$, with $A=\mu_I B_{\rm eff}/(IJ)$ in order-of-magnitude form for $F$ states. More directly, the splitting between $F=2$ and $F=1$ is $\Delta E=(A/2)[F_2(F_2+1)-F_1(F_1+1)]=2A$. The measured splitting is $h\times 6.835$ GHz, so $A=h\times 3.417$ GHz $\approx 1.41\times10^{-5}$ eV.

Problem 3. Estimate the normal mass shift of the Balmer-$\alpha$ line between hydrogen and deuterium. Solution. The wavelength scales inversely with the reduced-mass Rydberg constant. The fractional shift is

$$\frac{\Delta\lambda}{\lambda}\approx\frac{\Delta R}{R}\approx\frac{m_e}{m_p}-\frac{m_e}{m_d}\approx 2.7\times10^{-4}.$$

For $\lambda=656.3$ nm this gives $\Delta\lambda\approx 0.18$ nm, easily resolved with a medium-resolution spectrometer.

Section summary. Nuclear properties leave precise fingerprints in atomic spectra.

Interaction of Atoms with Radiation

Core ideas

Light drives transitions through the electric dipole interaction, with rates controlled by matrix elements, density of states, detuning, and polarization. Einstein coefficients, Rabi oscillations, optical Bloch equations, selection rules, and spontaneous emission connect quantum amplitudes to observed absorption and fluorescence.

For review, be able to derive dipole selection rules, define Rabi frequency and detuning, distinguish absorption, stimulated emission, and spontaneous emission, and interpret saturation. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$H_{\rm int}=-\bm d\cdot\bm E,\qquad \Omega=\frac{\bm d_{eg}\cdot\bm E_0}{\hbar},\qquad \Delta=\omega-\omega_0$$

Worked example

The spontaneous emission rate of the hydrogen $2p\to 1s$ transition (Ly-$\alpha$) is $A_{21}\approx 6.27\times10^8$ s$^{-1}$. The natural lifetime of the $2p$ state is therefore

$$\tau=\frac{1}{A_{21}}\approx 1.6\;\mathrm{ns}.$$

The corresponding natural linewidth (FWHM) is

$$\Gamma=\frac{\hbar}{\tau}\approx 4.1\times10^{-7}\;\mathrm{eV}\approx 2\pi\times 100\;\mathrm{MHz}.$$

This sets the fundamental limit on the resolution of spectroscopy with this transition.

Problems with Solutions

Problem 1. A two-level atom is driven resonantly with Rabi frequency $\Omega=2\pi\times 10$ MHz. How long does a $\pi$-pulse take to invert the population? Solution. On resonance the population oscillates as $P_2(t)=\sin^2(\Omega t/2)$. A $\pi$-pulse requires $\Omega t_\pi=\pi$, so

$$t_\pi=\frac{\pi}{\Omega}=\frac{1}{2\times10^7}\;\mathrm{s}=50\;\mathrm{ns}.$$

Problem 2. Estimate the saturation intensity $I_{\rm sat}$ for the Na D$_2$ line ($\lambda=589$ nm, $\tau=16$ ns). Solution. For a two-level atom,

$$I_{\rm sat}=\frac{\pi hc}{3\lambda^3\tau} =\frac{\pi(6.63\times10^{-34})(3\times10^8)}{3(5.89\times10^{-7})^3(16\times10^{-9})} \approx 6.4\;\mathrm{mW\,cm^{-2}}.$$

At this intensity the stimulated emission rate equals the spontaneous rate.

Problem 3. A laser with intensity $I=1.0$ mW cm$^{-2}$ at $\lambda=780$ nm is resonant with the Rb D$_2$ transition whose dipole matrix element is $d_{eg}\approx 2.5\times10^{-29}$ C m. Calculate the Rabi frequency. Solution. The electric field amplitude is $E_0=\sqrt{2I/(\varepsilon_0 c)}$. With $I=10$ W m$^{-2}$,

$$E_0=\sqrt{\frac{2\times10}{8.85\times10^{-12}\times3\times10^8}}\approx 86.8\;\mathrm{V\,m^{-1}}.$$

Then

$$\Omega=\frac{d_{eg}E_0}{\hbar} =\frac{2.5\times10^{-29}\times86.8}{1.055\times10^{-34}} \approx 2.1\times10^7\;\mathrm{rad\,s^{-1}},$$

i.e. $\Omega/(2\pi)\approx 3.3$ MHz.

Section summary. Atom-light interaction is controlled by dipole matrix elements and resonance.

Laser Cooling, Trapping, and Modern Experiments

Core ideas

Laser cooling uses momentum exchange between photons and atoms. Doppler cooling, optical molasses, magneto-optical traps, dipole traps, evaporative cooling, and ion traps exploit scattering forces, light shifts, and magnetic gradients. The key idea is to engineer dissipation without losing quantum control.

For review, be able to estimate recoil momentum and Doppler limit, explain MOT restoring forces, distinguish scattering and dipole forces, and name common routes to ultracold gases. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$k_BT_D=\frac{\hbar\Gamma}{2},\qquad \bm F_{\rm sc}\approx \hbar\bm k\,\Gamma\,\frac{s/2}{1+s+(2\Delta/\Gamma)^2}$$

Worked example

The Doppler cooling limit for $^{87}$Rb on the D$_2$ line ($\Gamma=2\pi\times 6.07$ MHz) is

$$T_D=\frac{\hbar\Gamma}{2k_B} =\frac{(1.055\times10^{-34})(3.82\times10^7)}{2(1.381\times10^{-23})} \approx 1.46\times10^{-4}\;\mathrm{K}=146\;\mu\mathrm{K}.$$

In a magneto-optical trap (MOT), six laser beams tuned $\Gamma/2$ below resonance together with a quadrupole magnetic field provide a velocity-damping force and a spatial restoring force, routinely producing $\sim10^7$ Rb atoms at $100\;\mu$K.

Problems with Solutions

Problem 1. Calculate the recoil velocity of a sodium atom after absorbing one D-line photon ($\lambda=589$ nm). Solution. The photon momentum is $p=h/\lambda=6.63\times10^{-34}/5.89\times10^{-7}\approx1.13\times10^{-27}$ kg m s$^{-1}$. For a Na atom ($m\approx 3.82\times10^{-26}$ kg),

$$v_{\rm rec}=\frac{p}{m}\approx\frac{1.13\times10^{-27}}{3.82\times10^{-26}}\approx 2.9\times10^{-2}\;\mathrm{m\,s^{-1}}\approx 3\;\mathrm{cm\,s^{-1}}.$$

Problem 2. How many resonant photons must a Na atom scatter to bring its speed from $v=300$ m s$^{-1}$ to rest? Solution. Each photon removes on average one recoil momentum $mv_{\rm rec}$ along the propagation axis. The number required is

$$N\approx\frac{v}{v_{\rm rec}}\approx\frac{300}{0.029}\approx 1.0\times10^4\;\mathrm{photons}.$$

At a scattering rate of $10^7$ s$^{-1}$ this takes only $\sim 1$ ms.

Problem 3. A MOT uses a magnetic field gradient of $10$ G cm$^{-1}$ ($1.0$ mT cm$^{-1}$). Estimate the Zeeman shift at a displacement $r=5$ mm from the trap centre and compare it with the natural linewidth of Rb. Solution. The field at $r=5$ mm is $B=(10\;\mathrm{G\,cm^{-1}})(0.5\;\mathrm{cm})=5$ G $=5\times10^{-4}$ T. The Zeeman shift for a weak-field $m_F$ sublevel is $\Delta E=\mu_B B$, giving

$$\Delta E=(9.27\times10^{-24})(5\times10^{-4})\approx 4.6\times10^{-27}\;\mathrm{J} \approx h\times 7.0\;\mathrm{MHz}.$$

The natural linewidth of Rb is $\Gamma/2\pi\approx 6.1$ MHz, so the Zeeman shift at $5$ mm is comparable to the line width; this ensures that the spatially dependent restoring force is resonant only on one side of the trap centre.

Section summary. Modern atomic physics uses light and fields to cool, trap, and control single quantum systems.