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A bond-dissociation enthalpy is the molar enthalpy change for the specified homolysis at temperature $T$, with reactant and products in specified standard states. Its usual units are $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The superscript $\circ$ specifies standard states; it does not fix $T$ to room temperature.^[ref]

Write $\delta h_i(T)=H_{m,i}^{\circ }(T)-H_{m,i}^{\circ }(0)$ for each species' molar thermal enthalpy increment. With $D_0$ an energy per parent molecule,Here $N_{\mathrm{A}}$ is the Avogadro constant. The thermal increments include translational kinetic energy, molecular rotation, thermally excited molecular vibration and electronic states when populated, and the $pV$ contribution to enthalpy. They are additional to the zero-point energies already included in $D_0$. If values of standard enthalpy of formation at the same $T$ are available, Hess's law gives the same quantity as ${\Delta }_f{H}^{\circ }(A)+{\Delta }_f{H}^{\circ }(B)-{\Delta }_f{H}^{\circ }(AB)$.

Units alone do not identify the convention. Multiplying either $D_e$ or $D_0$ by the Avogadro constant gives molar energy, but does not add thermal enthalpy. One electronvolt per dissociation corresponds to $96.485332\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. A spectroscopic wavenumber $\tilde{\nu }$ represents energy $hc\tilde{\nu }$, using the Planck constant $h$ and speed of light $c$; $1\mathrm{c}{\mathrm{m}}^{-1}$ corresponds to $0.011962657\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref]

Walther Bothe and Herbert Becker observed penetrating radiation from beryllium struck by alpha particles before James Chadwick's interpretation.^[ref]

Irène Joliot-Curie and Frédéric Joliot found that adding material containing hydrogen increased ionization, interpreting the ejected protons as Compton scattering recoils from extremely energetic gamma rays. James Chadwick tested this interpretation against nuclear recoils in other elements.^[ref]

The counter coupled a small ionization chamber to an amplifier and oscillograph: charged particles produced ions, and the electrical signal gave a deflection. With Norman Feather, James Chadwick also observed nitrogen recoil tracks in a cloud chamber.^[ref]

The apparatus used polonium on silver facing beryllium inside an evacuable vessel. Paraffin wax before the aluminium counter window supplied recoil protons; intervening aluminium screens measured their particle range. Deflections were photographed on bromide paper. The first counter contained air; later fillings included hydrogen, helium, nitrogen, oxygen and argon. Counts with wax absent included gas recoils; wax insertion added entering protons. Thick lead between source and counter hardly changed the initial counts, testing penetration against the low background.^[ref]

The diagram separates two routes to a signal: a proton entering from the wax and a nuclear recoil created inside the chamber gas. Its lower row is a logical summary, rather than a reconstructed experimental trace. In either route the observed ionization belongs to a charged secondary particle; identifying the incident radiation requires an additional collision model.

In the February letter, a photon interpretation requiring about $52\mathrm{MeV}$ for the protons predicted only about $0.4\mathrm{MeV}$ for nitrogen recoils, below the observed ionization. Here $1\mathrm{MeV}=10^6$ electronvolts. James Chadwick proposed a neutral particle near the proton mass and the nuclear reactionThese are the letter's approximate historical values, not a modern mass determination.^[ref]

Why compare two materials? A gamma ray is a photon, so its incident momentum is $E_{\gamma }/c$, where $c$ is the speed of light. A slow massive projectile instead has momentum $mv$ and kinetic energy $m{v}^2/2$. The relationship between incoming momentum and energy therefore changes. An elastic collision model must explain the recoil endpoints for both target masses with the same incident projectile population. Calling the radiation neutral supplies no such agreement by itself.

The original measurements depended on particle range calibration, material in the path, detector response and whether a recoil was fully stopped in the sensitive gas. Large deflections alone need not specify a complete recoil energy. The calculations below expose the ideal collision assumptions behind the inference; applying them to a measured spectrum requires checking those assumptions.

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.

Compton scattering redistributes photon energy and momentum in a collision with a charged subatomic particle, here an electron. For a free electron initially at rest in the chosen inertial frame, an incident photon of energy $E=hc/\lambda$ leaves with energy $E^{\prime }=hc/{\lambda }^{\prime }$ after deflection through $\theta$. The angle $\theta \in [0,\pi ]$ is between the incident and outgoing photon directions; it is not the electron recoil angle. Four-momentum conservation giveswhere $m_e$ is the electron rest mass and $T_e$ its final kinetic energy. The photon energy loss pays for the electron recoil: neither energy nor momentum disappears.

For an original worked example, stipulate $E=70\mathrm{keV}$, the rounded electron rest energy $m_e{c}^2=511\mathrm{keV}$, and the rounded Compton wavelength ${\lambda }_C=2.426\mathrm{pm}$; $\mathrm{keV}$ means $10^3$ electronvolts. Treat these rounded inputs as the example's constants. They give the following calculated values, with energies rounded to three decimal places:

Since $0\le 1-\cos \theta \le 2$, every outgoing photon has $0<E^{\prime }\le E$ and $T_e\ge 0$. At zero deflection there is no recoil; backscattering transfers the most energy. At fixed $\theta$, the absolute wavelength increase is independent of incident energy, while the fractional energy loss grows with $E/(m_e{c}^2)$. This makes Compton scattering useful for inferring energy transfers from directions in photon detectors.

The free-electron-at-rest assumption matters. An electron in a bound state has binding energy and an initial momentum distribution. Neglecting these is useful when the transferred energy is large compared with the relevant binding energy and the initial momentum is small compared with the resolved momentum transfer. Otherwise the outgoing photon energies can broaden or depart from the stationary-target formula; an initially moving electron can even give energy to the photon.

A material may also scatter X-rays through a coherent elastic scattering channel with an effectively unshifted component. When an intact chemical atom of mass $M$ takes the recoil, the corresponding Compton wavelength $h/(Mc)$ is much smaller than the electron value. The single free-electron curve does not predict that component's weight. More generally, energy versus angle is a kinematic constraint; angular wave intensity requires a differential scattering cross-section, together with target and detector information.

Tunneling through a controllable vacuum gap by Gerd Binnig and others made barrier width experimentally adjustable. Their apparatus used a pointed electrical conductor opposite a conducting plate, with fine positioning by piezoelectricity. Rubber-tube support and magnetic levitation above a cooled bowl exhibiting superconductivity suppressed vibrations; the junction itself remained near room temperature.^[ref]

The readout was electrical resistance versus displacement at a fixed electric potential difference of 60 mV of either polarity. They observed an exponential function dependence spanning four orders of magnitude. Cleaning and better vacuum increased the inferred barrier from approximately 0.6-0.7 eV to 3.2 eV. Reproducible curves and the cleaning response argued against deforming contamination layers; an appropriate slope also helped exclude an ordinary conducting contact. The paper reports residual contamination, electronics-limited range and a jump attributed to release of thermal stress; its curve origins generally do not determine absolute gap widths.^[ref]

The ideal tunnel junction model explains the slope. At fixed small electric potential difference, fixed tip geometry and approximately unchanged density of states, take an effective barrier energy $\Phi$ above the relevant electron states and approximate the current magnitude byThe coefficient $C$ includes the applied electric potential difference, available occupied and empty quantum states, and tip geometry. A single transmission probability is not an electric current: a junction combines contributions from many electron states in both directions, with the electric potential difference producing a net imbalance. Its tilted, material-dependent barrier is only approximated by the equal-exterior rectangular potential barrier; the exact single-particle formula above is not a complete junction model.

For the chosen model value $\Phi =4$ electronvolts, $\kappa \simeq 10.25{\mathrm{nm}}^{-1}$. Increasing separation by $0.1\mathrm{nm}$ multiplies $I$ by about $0.129$, a reduction by about $7.8$. The following diagram chooses $I(0.4\mathrm{nm})=1\mathrm{nA}$ solely for illustration; it contains no measured data and is not a fit to the 1982 curves.

Distance calibration is essential. In Scanning Tunneling Microscopy: From Birth to Adolescence by Gerd Binnig and others, the authors describe an initial inferred barrier near 1 eV. Recalibration found only half the quoted piezo displacement sensitivity and raised the inferred barrier above 4 eV. Thus an exponential trace alone was insufficient evidence for tunnelling through vacuum. The model makes the sensitivity explicit: the inferred $\Phi ={\hslash }^2(d\ln I/ds{)}^2/(8m_e)$ depends quadratically on the distance scale.^[ref]

For scanning tunnelling microscopy, feedback control adjusts the vertical position to hold electric current fixed during a lateral scan; the recorded height follows separation only when the surface is electronically homogeneous. Changes in the local density of states can also change apparent height. This limitation is already explicit in the authors' account of constant-current imaging.^[ref]

Combining $r=mv/(|q|B)$ with the classical kinetic energy givesHere $E$ denotes kinetic energy, not relativistic total energy. At fixed magnetic field, twice the final radius allows four times the classical kinetic energy. At fixed final kinetic energy, a larger accelerating gap voltage reduces the number of crossings needed, but does not change the required orbit radius.

For an illustrative proton with $B=1.00\mathrm{T}$ and peak gap voltage $V_0=10.0\mathrm{kV}$, take negligible injection kinetic energy, on-crest kicks, a zero-width gap, and no losses. Each crossing adds $10.0\mathrm{keV}$, so $100$ crossings give $1.00\mathrm{MeV}$, using electronvolts. With $e=1.602176634\times 10^{-19}\mathrm{C}$ and $m_p=1.67262192595\times 10^{-27}\mathrm{kg}$,The $100$th kick arrives after $99$ intervening half-turns, about $3.247\mu \mathrm{s}$ after the first kick in this idealization. The final two curvature radii differ by $r(1\mathrm{MeV})[1-\sqrt{99/100}]\simeq 0.724\mathrm{mm}$. That difference is a curvature-radius increment, not a general formula for separation of real extracted turns: successive arc centers shift, and extraction also depends on focusing and the gap geometry. These chosen settings are a calculation, not historical measurements.

Let an electron start with negligible kinetic energy and move through an increase $U>0$ in electric potential. Its electric charge is $q=-e$, where $e>0$ is the elementary charge, so the decrease in potential energy is $eU$. In a field-free region after acceleration,This is an electron formula, not a universal conversion from voltage to wavelength. It assumes negligible initial kinetic energy, negligible losses, and $eU\ll m_e{c}^2$, where $c$ is the speed of light. It describes the de Broglie wavelength outside the crystal; the potential energy inside a material can change the local kinetic momentum.

For $U=54\mathrm{V}$, take $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $e=1.602176634\times 10^{-19}\mathrm{C}$, and $m_e\simeq 9.1093837\times 10^{-31}\mathrm{kg}$. The rounded electron mass is consistent with the 2022 recommended constants.^[ref]The electronvolt is an energy unit: $54\mathrm{eV}$ must be converted to joules before combining it with $m_e$ in SI units. Dimensional analysis checks both steps:The resulting velocity is about $4.3584\times 10^6\mathrm{m}{\mathrm{s}}^{-1}$, or $0.01454c$, so the nonrelativistic assumption is consistent. For a larger accelerating voltage, the relativistic energy-momentum relation gives $pc=\sqrt{K(K+2m_e{c}^2)}$, with $K=eU$, while $\lambda =h/p$ still holds. At $54\mathrm{V}$ the relativistic wavelength is shorter by about $0.00264\%$.

Explicitly, the exact special relativity formula and its nonrelativistic quantum mechanics limit areHere $K=eU$ is the gained kinetic energy; the total energy is $E=m_e{c}^2+K$. At $80\mathrm{kV}$, the nonrelativistic quantum mechanics expression gives $4.336\mathrm{pm}$, about $3.84\%$ larger than the exact $4.176\mathrm{pm}$. This difference matters when using electron diffraction ring positions to infer lattice plane spacings.

Separate geometry from electronic response in an original one-dimensional toy tunnel junction. Write the geometric surface height as $h(x)$, tip height as $z(x)$, and positive gap as $s(x)=z(x)-h(x)$. Assume low fixed tip-sample electric potential difference: the sampled energy window is narrow enough that the tip density of states is featureless and the sample local density of states varies little across that window. Neglect thermal broadening, changes of tip shape and lateral averaging. Let $\rho (x)>0$ be the effective sample local density of states divided by a positive reference value, so it is dimensionless. Choose a spatially constant effective barrier energy $\Phi >0$, a positive coefficient $C$ with units of electric current, andHere $m_e$ is the electron mass, $\hslash$ the reduced Planck constant, and $\kappa$ the tunnelling decay constant. This is an illustrative low-bias model, not a universal calibration formula. At fixed $I=I_0>0$, taking a logarithm givesThe logarithms have dimensionless arguments. Only constant $\rho$ makes the recorded height differ from geometric height by a constant. Multiplying $\rho$ by $r>0$ raises the apparent height by $\ln r/(2\kappa )$ without moving the geometric surface.

For a concrete ambiguity, choose $\kappa =10{\mathrm{nm}}^{-1}$, $a=0.020\mathrm{nm}$, lateral period $L=1\mathrm{nm}$, $s_0=0.50\mathrm{nm}$, and $q(x)=\cos (2\pi x/L)$. CompareBoth produce exactly $z(x)=s_0+aq(x)$ at $I_0=C{e}^{-2\kappa s_0}=C{e}^{-10}$. Substitution checks this without identifying $\rho$ with geometric height: case A has constant gap $s_0$, while case B has gap $s_0+aq(x)$ and a compensating electronic factor. The chosen ${\rho }_B$ ranges from $e^{-0.4}$ to $e^{0.4}$, approximately $0.6703$ to $1.4918$; every gap remains positive. The chosen $\kappa$ corresponds to $\Phi \simeq 3.81$ electronvolts, a positive effective barrier.

At fixed tip height $z_f=s_0$, both cases instead giveTheir maximum-to-minimum electric current ratio is $e^{0.8}\simeq 2.226$. Thus even adding this fixed-height scan at the same electric potential difference cannot distinguish these deliberately chosen surfaces within the toy model. Independent structural information or additional electronic measurements are required. Changing electric potential difference can supply such information, but also changes the assumptions used to define $C$ and $\rho$.

An electronvolt is $1.602176634\times 10^{-19}\mathrm{J}$. One electronvolt per particle corresponds to approximately $96.485\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref]

For a fixed electric charge $q$, a time-independent electric potential $\Phi$ and negligible collisions or radiation losses, conservation of energy givesHere $V$ is the signed potential drop along the trajectory. A proton starting nearly at rest at $+V$ and arriving at ground gains $eV$: a chosen $500\mathrm{kV}$ gives $500\mathrm{keV}$, by the definition of the electronvolt. Several intermediate electrodes can distribute that drop along a tube, but the gains telescope to the same endpoint difference. Passing through more electrodes does not multiply the available kinetic energy beyond $qV$.

This is the distinction from a cyclotron. In a static electric field, a closed trip gives zero net work. A cyclotron changes the gap electric field between successive visits, so correctly timed particles repeatedly gain kinetic energy. Its limit involves the cyclotron resonance condition and relativistic detuning of a cyclotron; a single-pass electrostatic accelerator instead requires the full accelerating electric potential difference to be sustained at once. Neither machine's beam energy by itself specifies the energy released by a nuclear reaction.

Choose $n=10^{28}{\mathrm{m}}^{-3}$ for a uniform noninteracting electron model with two equally populated spin states. This is an original worked example, not a measured number density assigned to an actual material. Using $m_e\simeq 9.10938\times 10^{-31}\mathrm{kg}$, $\hslash \simeq 1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$, and $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ givesThe conversion uses $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$: an electronvolt is an energy unit, while a joule is its SI counterpart. The dimensions provide a quick check: $n^{1/3}$ is inverse length, $\hslash n^{1/3}$ has units of momentum, $p_F^2/m$ has units of energy, and $E_F/k_B$ has units of temperature. The ratio $p_F/(m_{e}c)\simeq 0.002574$ verifies the nonrelativistic limit for this model, where $c$ is the speed of light.

Increasing this chosen number density eightfold doubles the Fermi momentum and quadruples the Fermi energy, giving approximately $6.7701\mathrm{eV}$. The exponent $2/3$ has a geometric origin: the number of occupied states grows as the cube of a momentum radius, but kinetic energy of nonrelativistic particles grows as its square. This makes Fermi energy a useful bridge between a count of particles and a mechanical energy scale.

The bridge from detector readout to a particle resonance is four-momentum conservation. In units $c=1$, sum the measured final-state four-momenta:For two photons separated by angle $\alpha$, $|{\mathbf{p}}_i|=E_i$ gives the displayed diphoton formula. Two back-to-back photons of $60\mathrm{GeV}$ each have $m_{\gamma \gamma }=120\mathrm{GeV}$. A four-lepton candidate instead uses four four-momenta in the same sum. Collision debris outside the selected decay is not included. Since the partons carry unknown fractions of the incoming proton momenta, reconstructing the decay products is more informative than assigning the full beam energy to each candidate.

Suppose the intrinsic decay width is negligible compared with the stipulated detector spread. Let the reconstructed signal mass follow a normal distribution with mean $m_0=120\mathrm{GeV}$ and standard deviation ${\sigma }_m$. For a total expected selected signal $S=60$ and flat background density $b=8$ events per $10^9$ electronvolts, the expected count density isThis is a deliberately chosen response model, not a fit to either experiment. In the fixed window $m_0-a<m<m_0+a$ with $a=3\mathrm{GeV}$, direct integration givesThe expected signal counts are $57.2700$ for ${\sigma }_m=1.5\mathrm{GeV}$ and $40.9614$ for ${\sigma }_m=3\mathrm{GeV}$. Doubling the spread lowers the fraction in the same window without changing the total signal. Detector resolution can therefore change the visibility of a peak even with identical particle production.

As a separate counting example, stipulate an observed integer count $n=70$ and an exactly known background mean $B_w=48$. The Poisson distribution gives a likelihood $L(s)=e^{-(48+s)}(48+s{)}^{70}/70!$ for nonnegative signal mean $s$. Its maximum-likelihood estimator is $\hat{s}=22$, and the likelihood ratio statistic against $s=0$ isThis tests a specified window with a specified background, and is not either experiment's discovery significance. If the background mean is allowed to vary freely, background alone can choose $70$ and fit that count exactly. Control samples, a defensible mass-shape model, nuisance parameters and corrections for searching many masses are therefore essential to the experimental inference.

For emission take integers $n_i>n_f\ge 1$. Applying conservation of energy to the model's internal energy difference, while neglecting the much smaller whole-atom photon recoil, givesThus Johannes Robert Rydberg's Rydberg formula is a difference of two energy levels expressed as a spectroscopic wavenumber. The same constants determine every line; there is no new parameter for each electronic transition. With the chosen inputs, $R_H=1.096775834\times 10^7{\mathrm{m}}^{-1}$. All wavelengths below are calculated in vacuum, not air, and are rounded model predictions rather than historical measurements.

The conventional names commemorate Theodore Lyman and Johann Jakob Balmer. Fixing $n_f=1$ produces the ultraviolet Lyman series, beginning with the Lyman-alpha line; fixing $n_f=2$ produces the Balmer series. As $n_i\to \infty$, the wavelength approaches $n_f^2/R_H$: $91.1763\mathrm{nm}$ at the Lyman limit and $364.705\mathrm{nm}$ for the Balmer series. Crowding follows directly from $E_{n+1}-E_n\sim 2B_H/n^3$. Discreteness therefore coexists with a series limit.

Correct line positions do not give wave intensities, lifetimes or electric-dipole selection rules for hydrogen. Those require a theory of the wavefunctions and their coupling to electromagnetic radiation. The circular model also omits spin, relativistic corrections and the Lamb shift. Its successful gross energy levels do not make its assumed orbits modern atomic orbitals.

A lattice energy needs a reaction, a sign convention, and a stated amount. Using the lattice-dissociation enthalpy convention for sodium chloride,The reverse lattice-formation enthalpy is negative with the same magnitude when initial and final states match. Chemistry 2e gives $769\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for this separation, per mole of sodium chloride formula units. This is about $7.97$ electronvolts per formula unit, not per individual ion. It concerns separated gaseous ions, rather than neutral chemical atoms, molten salt, or hydrated ions in water.^[ref]

Also distinguish an internal energy change from an enthalpy change: $H=U+pV$. A static point-charge calculation omits motion and thermal contributions. Reversing a reaction reverses its energy change; replacing internal energy with enthalpy additionally changes the defined quantity. The Born-Haber cycle infers a lattice-formation enthalpy by combining other enthalpies, ionization energies, and electron affinities. These imagined stages are a thermodynamic accounting path, not a claim about the actual sequence of a chemical reaction.^[ref]

Let a projectile have mass $m$, incident speed $v$, and incident kinetic energy $E=m{v}^2/2$. The stationary target has mass $M$. In a head-on elastic collision, the signed outgoing projectile velocity $v^{\prime }$ and target velocity $u$ obey momentum conservation and conservation of energy:Factor the second equation and eliminate $Mu=m(v-v^{\prime })$. The nontrivial collision branch has $u=v+v^{\prime }$, henceThis is the maximum over scattering angles, not the energy of every recoil. To see the bound without assuming a head-on event, set the recoil momentum magnitude to $q$ and its laboratory angle to the incident direction to $\theta$. Conservation of energy with outgoing projectile momentum $\mathbf{p}-\mathbf{q}$ givesThus $T=T_{\max }{\cos }^2\theta$ for a nonzero recoil, and the endpoint occurs at $\theta =0$.

For an original numerical example, stipulate $E=4\mathrm{MeV}$ and $m=m_p$, with $m_p$ the proton mass. Approximate hydrogen, helium and nitrogen target masses by $M=m_p$, $4m_p$ and $14m_p$ respectively. Their maximum fractions and recoil energies areThese stipulated values illustrate the approximation $m_n=m_p$; they are not measurements from 1932. With $m_p{c}^2\simeq 938\mathrm{MeV}$, the incident ratio $E/(m_p{c}^2)\simeq 0.0043$ gives $v/c\simeq 0.092$, consistent with the nonrelativistic limit used here. A million electronvolts is an energy unit, whereas a mass expressed in $\mathrm{MeV}/c^2$ requires the factor $c^2$ when inserted into an energy equation.

The hydrogen endpoint permits essentially complete transfer because the projectile and target masses match. For a very heavy target, $M\gg m$, the fraction approaches $4m/M$, which is small: the target can take substantial momentum with little kinetic energy. The exact inequality follows from $(m-M{)}^2\ge 0$, which implies $(m+M{)}^2\ge 4mM$.

Mulliken electronegativity uses both the cost of removing an electron and the energy released by accepting an electron. For specified isolated-atom states, the common ground-state convention isThe free electron's energy is the zero reference. Here $I$ is the ionization energy and $A_{\mathrm{ea}}$ is the electron affinity, positive for exothermic attachment. If attachment is instead tabulated as the signed energy change $\Delta E_{\mathrm{attach}}=E(X^{-})-E(X)$, the same expression is $(I-\Delta E_{\mathrm{attach}})/2$. Stating the sign convention prevents adding the wrong quantities. The 1934 original treated suitable valence states, which are not automatically the ground states used in this example.^[ref]

Mulliken electronegativity has energy units before any additional rescaling. With $I$ and $A_{\mathrm{ea}}$ in electronvolts per chemical atom, ${\chi }_{\mathrm{M}}$ is in electronvolts; with both in $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, it is a molar energy. For hydrogen, the RSC ground-state inputs are $I=1312.05$ and $A_{\mathrm{ea}}=72.769\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, giving ${\chi }_{\mathrm{M}}=692.4095\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}\approx 7.176\mathrm{eV}$ per chemical atom. For chlorine, $I=1251.186$ and $A_{\mathrm{ea}}=348.575\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ give $799.8805\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}\approx 8.290\mathrm{eV}$. These are calculated ground-state Mulliken electronegativities, not the RSC's tabulated Pauling electronegativities.^[ref][ref]

Both schemes place chlorine above hydrogen here, but their numerical differences cannot be subtracted from one another or equated. Dividing a molar Mulliken electronegativity by an energy conversion factor changes units; it does not transform it into a Pauling electronegativity. A further empirical mapping would require its own calibration and stated domain.

Treat the incident and outgoing neutrons as free nonrelativistic particles outside the specimen. Their wave vectors, de Broglie wavelengths, momenta and kinetic energies obeyHere $k_i=|{\mathbf{k}}_i|$, $k_f=|{\mathbf{k}}_f|$, $m_n$ is the neutron mass, and $\hslash =h/(2\pi )$ uses the Planck constant. Subtracting the outgoing projectile quantities from the incoming ones gives the momentum transfer and energy delivered to the specimen:Thus $\mathbf{Q}$ has units of inverse length, rather than momentum. A positive $\omega$ means the specimen gains energy; negative $\omega$ means it supplies energy to the neutron. If the angle between incident and outgoing propagation directions is $2\theta$, the law of cosines givesKnowing the detector angle alone therefore leaves $Q$ undetermined unless the outgoing energy is known or an elastic scattering assumption is justified.

For a reproducible unit conversion choose $m_n=1.67492750056\times 10^{-27}\mathrm{kg}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$. The first is a recommended central value; the latter two conversions are exact.^[ref] With $1$ ångström equal to $10^{-10}\mathrm{m}$ and $1\mathrm{meV}=10^{-3}$ electronvolts, substitution into $E=h^2/(2m_n{\lambda }^2)$ givesSquare brackets here mean the numerical value in the indicated unit. For an original chosen example, ${\lambda }_i=2$ ångströms and ${\lambda }_f=2.5$ ångströms give $E_i\simeq 20.451\mathrm{meV}$, $E_f\simeq 13.089\mathrm{meV}$ and $\hslash \omega \simeq 7.362\mathrm{meV}$. At $2\theta =60^{\circ }$, $Q\simeq 2.879{\mathring{\mathrm{A}}}^{-1}$. These values stipulate an allowed projectile geometry; they do not assert that any particular specimen has an excitation there.

Let $\Gamma$ be coverage in $\mathrm{mol}{\mathrm{m}}^{-2}$. Then $n_s=N_A\Gamma$ and the Helmholtz surface-dipole relation iswhere a voltage in volts has the same numerical value as the electron-energy change in electronvolts. Converting the slope from square centimeters to square meters gives $-5\times 10^5\mathrm{V}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$. ThereforeEach added molecule produces an effective outward surface dipole of about $2.2$ D, lowering the electron work function. This is the incremental adsorbate-plus-surface dipole, not necessarily the isolated molecular electric dipole moment.

Use the inverse beta decay threshold, with the charged-lepton mass generalized to $m_{\ell }$. For a stationary proton and a massless incident antineutrino, the Mandelstam invariant is $s=m_p^2+2m_p{E}_{\nu }$. At threshold the final particles have no relative momentum in their centre-of-momentum frame, so $s_{\mathrm{thr}}=(m_n+m_{\ell }{)}^2$. ThereforeUsing the supplied, rounded masses givesThe second term is the recoil correction; setting the threshold equal just to the final rest-mass excess would miss it.

Reactor electron antineutrinos have energies of a few million electronvolts. They can produce a Positron through inverse beta decay above the first threshold, but cannot produce a muon. Thus reactor neutrino oscillations are measured primarily as electron-antineutrino disappearance, through the event rate and energy spectrum relative to the unoscillated prediction. An oscillated muon antineutrino cannot be identified by the proposed muon-production channel at reactor energies; disappearance alone does not distinguish all possible final flavours.

Subtract the two photoelectric effect equations $K_{\max }=hc/\lambda -\Phi$ to eliminate the work function:Here $e$ converts the energy difference from electronvolts to joules. The precision of the rounded stopping energies justifies about two significant figures, $6.7\times 10^{-34}\mathrm{J}\mathrm{s}$.