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The Faraday constant is the electric charge associated with one mole of elementary charges, $F=N_{\mathrm{A}}e\approx 96485.33212\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$. The product of the Avogadro constant and elementary charge is exact under the SI definitions; the displayed decimal is rounded. For ions with signed charge number $z$, the molar electrical contribution to an electrochemical potential is $zF\psi$, where $\psi$ is electric potential. ^[ref].

The nuclear reactionconserves electric charge and total nucleon count: $3+1=2+2$ and $7+1=4+4$. Its nuclear reaction Q-value is the initial minus final rest energy. It is additional to the incoming beam kinetic energy; the accelerator triggers a rearrangement whose products can carry much more kinetic energy than the incident proton.

Use one consistent set of neutral-atom isotope masses, in unified atomic mass units, from the National Institute of Standards and Technology:These are isotope masses, not element-averaged atomic weights.^[ref] With $\mathrm{u}{c}^2=931.49410372\mathrm{MeV}$,^[ref]The four electron rest masses cancel between sides. Differences of total electronic binding energies do not cancel exactly: this atomic-mass result approximates the bare-nucleus nuclear reaction Q-value at the displayed $0.001\mathrm{MeV}$ scale, not at arbitrary precision. No mixture of rounded nuclear binding energies is needed. At chosen incident kinetic energy $T=0.500\mathrm{MeV}$ on a stationary target, the two outgoing alpha particles together carry approximately $Q+T=17.846\mathrm{MeV}$ before losses in the target and detector window.

Equal product masses imply equal kinetic energies and opposite momenta in the center of mass frame. The laboratory frame has nonzero initial momentum because the proton is moving. Conservation of momentum therefore forbids the two laboratory momenta from always being equal and opposite. The products are equal in laboratory kinetic energy only for particular emission directions, not for every event.

An exact two-body calculation makes this distinction quantitative. Write $a=m_p{c}^2$, $b=m_{\mathrm{Li}}{c}^2$, $h=m_{\alpha }{c}^2$, and $T$ for incident kinetic energy. Express every quantity below in the same energy units; $\mathcal{P}=pc$ is momentum times the speed of light. For a stationary target,Here $W$ is the center-of-mass energy, and $\beta$ is the center-of-mass speed divided by the speed of light. Each outgoing alpha particle has center-of-mass total energy $E_{\ast }=W/2$ and momentum-times-$c$ magnitude ${\mathcal{P}}_{\ast }=\sqrt{E_{\ast }^2-h^2}$. A Lorentz transformation gives the paired laboratory kinetic energies, with ${\theta }_{\ast }$ measured from the incident beam in the center of mass frame:Their sum is independent of ${\theta }_{\ast }$ and equals $Q+T$. Their difference is $2\gamma \beta {\mathcal{P}}_{\ast }\cos {\theta }_{\ast }$; the symmetry belongs to the product interchange and the center of mass frame.

For the chosen $T=0.500\mathrm{MeV}$ model, subtract $Z$ electron rest masses from each neutral-atom mass, neglect electronic binding energies, and use $m_e{c}^2=0.51099895069\mathrm{MeV}$.^[ref] At ${\theta }_{\ast }=0$, the paired laboratory kinetic energies are approximately $9.979$ and $7.867\mathrm{MeV}$; at ${\theta }_{\ast }=90^{\circ }$ both are approximately $8.923\mathrm{MeV}$. In the latter case both products have a forward laboratory momentum component, so even equal laboratory kinetic energies do not imply back-to-back laboratory trajectories. A real thick target and finite detector acceptance would broaden or select these ideal kinematic possibilities.

The quoted expansion is nondegenerate perturbation theory for a known isolated eigenstate of a time-independent Hamiltonian. A useful small-mixing condition is $|⟨k|H_1|i⟩|\ll |E_i-E_k|$ for coupled states. First order is a diagonal expectation value; second order describes virtual mixing. A degenerate or nearly degenerate coupled subspace must first be treated by diagonalization there. Symmetry can eliminate terms, and convergence still needs checking. Ground-state second-order shifts are nonpositive because all denominators are negative.^[ref]

For the rotational Stark effect, $H_1=-pE\cos \theta$, where $\theta$ is the polar angle to the field. The source calls it azimuthal, which is inconsistent with this coupling. Rotor states have parity $(-1{)}^J$, while $\cos \theta$ is odd, so every first-order matrix element within one fixed-$J$ manifold vanishes. The $m$ degeneracy therefore causes no first-order splitting.

Only $J\pm 1$ with the same $m$ contribute in second order. Using their energy differences $2BJ$ and $-2B(J+1)$,For $J\ge 1$ this simplifies to the quadratic Stark shift of a polar linear rotor,For $J=0$, only $J^{\prime }=1,m=0$ contributes, with squared angular matrix element $1/3$ and denominator $-2B$:Here $B$ is the rotational energy constant, unlike the magnetic field in Question3.

The unperturbed line gives $B/h=6.08\mathrm{GHz}$. The $J=1$ shifts are $(pE{)}^2/(10B)$ for $m=0$ and $-(pE{)}^2/(20B)$ for $m=\pm 1$. Subtracting the common lower-state shift gives transition shifts $4(pE{)}^2/(15Bh)$ and $7(pE{)}^2/(60Bh)$. Their separation determines the dipole moment from a rotational Stark doublet:The $\mathrm{D}$ unit is the debye. Using external $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{D}=3.33564\times 10^{-30}\mathrm{C}\mathrm{m}$ gives $pE/B\simeq 0.0593$, consistent with weak-field perturbation theory. The two shifted frequencies are about $12.165707$ and $12.162497\mathrm{GHz}$; polarization selects $\Delta m=0$ or $\pm 1$. The inferred value is a magnitude; these shifts do not determine the dipole's sign.^[ref][ref]

Use ground-state references and the supplied ideal-gas, rigid-rotor and harmonic-oscillator approximations. The two ions have closed electronic shells; take their electronic ground-state degeneracies and the molecular electronic ground-state degeneracy as one, and neglect higher electronic levels. Consistently treated nuclear-spin factors cancel between the species. Since the molecule is heteronuclear, its rotational symmetry number is one.

Let ${\mu }_r=m_{\mathrm{Cs}}{m}_{\mathrm{I}}/m_{\mathrm{CsI}}$ be the reduced mass, neglecting the negligible mass correction from electron transfer. The ratio of translational partition functions per volume reduces toThe molecule has additional rotational and vibrational states, whereas the monatomic ions do not. At $3000\mathrm{K}$,Hence the dimensional equilibrium quotient isThe last form makes the unassigned standard concentration explicit. For the common choice of $1\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$, the corresponding microscopic reference is $c^{\circ }=6.02214\times 10^{26}{\mathrm{m}}^{-3}$, givingEquivalently, the dimensional molar quotient is $8.27\times 10^{-7}\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$. A pressure-based gas standard is a different convention: taking $c^{\circ }=p^{\circ }/(k_{B}T)$ with $p^{\circ }=1\mathrm{bar}$ would give $2.06\times 10^{-4}$, rather than the same numerical constant.

The well-minimum convention gives the identical answer becauseConstants used here are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, $m_u\simeq 1.66053907\times 10^{-27}\mathrm{kg}$ and $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref]

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

By symmetry, the two grounded plates take equal induced surface charge and the electric field has magnitude $\sigma /(2{ϵ}_0)$ between either plate and the sheet. The grounded-plate potential of a central charged sheet is thereforeAn electron released at zero potential gains kinetic energy $e{V}_{\mathrm{sheet}}$. ThusThe planar infinite-sheet approximation neglects edge effects. Here $e>0$ denotes charge magnitude; the electron's negative charge makes it accelerate toward the positive sheet. Relativistic corrections are below0.1%. Numerical constants use rounded CODATA values, not a claim about the historical handbook: ${ϵ}_0=8.85419\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, $e=1.602176634\times 10^{-19}\mathrm{C}$ and $m_e=9.10938\times 10^{-31}\mathrm{kg}$. ^[ref]

In the time-independent Schrödinger equation, $-{\hslash }^2{\psi }^{\prime \prime }/(2m)$ is the kinetic-energy operator acting on the state, $V\psi$ its potential-energy contribution and $E\psi$ the energy-eigenvalue term. The complex wave function is a probability amplitude: the Born rule assigns density $|\psi {|}^2$ for a normalized single-particle state. A plane-wave beam uses an incident-amplitude or flux normalization instead of a unit-integral wavefunction.

For a free plane wave, ${\psi }^{\prime \prime }=-k^2\psi$, so $\boxed{E={\hslash }^2{k}^2/(2m)}$ and its time factor obeys $\hslash \omega =E$. Above the upward potential step, the bounded subthreshold solution has decay constantFor the PDF's unit incident amplitude, write $\psi =e^{ikx}+R{e}^{-ikx}$ on the left and $\psi =T{e}^{-\alpha x}$ on the right, omitting their common time phase. Continuity of both $\psi$ and ${\psi }^{\prime }$ at this finite step givesConsequentlyThe evanescent amplitude versus transmitted current distinction matters: the second quantity is a squared amplitude, not a transmitted-flux probability. The decaying tail has zero probability current, despite nonzero density and the possible value $|T{|}^2>1$.

Using $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$ and the electron mass and elementary charge specified above, the energy difference1eV gives $\alpha =5.12\mathrm{n}{\mathrm{m}}^{-1}$. Thus at the stated positionThis is relative to the unit incident density in the source figure, not an absolute normalized probability per metre. ^[ref]