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General base catalysis uses a base to accept a proton during the reacting pathway, for example activating water while it forms a bond. It need not involve an equilibrium pool of free hydroxide.

Glutamate is an anionic form of glutamic acid, with one fewer proton. The acid and anion are different chemical species.

The three principal aqueous glycine forms differ by loss of one proton at each step:Call these glycine forms $G^{+}$, $G^0$ and $G^{-}$, respectively: a cation, a zwitterion and an anion, with net electric charges $+e$, $0$ and $-e$, where $e$ is the elementary charge. $G^0$ specifically denotes the zwitterion, not an uncharged amino/carboxyl drawing. The first dissociation removes the carboxyl proton; the second removes a proton from the positively charged nitrogen. The chemical equilibrium changes are reversible, not three permanently separate kinds of glycine. ^[ref]

For all calculations and both diagrams here, explicitly stipulate the pKa values $\mathrm{p}{K}_{a1}=2.35$ and $\mathrm{p}{K}_{a2}=9.78$. These are illustrative model inputs, not measurements from the glycine titration experiment or a claim of universal constants. Actual acid dissociation constants depend on temperature and on the convention used for thermodynamic activity; apparent concentration-based values also depend on ionic strength. For comparison, the inspected textbook tabulates $2.34$ and $9.60$ for glycine, which would give a different isoelectric point. ^[ref]

Adding sodium hydroxide to acidified glycine successively removes its two titratable protons. Two broad buffer solution regions appear around the two pKa values, separated by a steep pH rise near the isoelectric point. This is useful evidence for two separated acid-base chemistry steps even though the changing chemical species are invisible. In the approximation that added hydroxide only changes glycine protonation, $b=1-\bar{z}$ is the added base amount of substance per initial amount of substance of glycine. Then $b\approx 0.5$, $1$ and $1.5$ correspond to $\mathrm{p}{K}_{a1}$, $\mathrm{pI}$ and $\mathrm{p}{K}_{a2}$; endpoints are not exactly pure chemical species. ^[ref]

For the solid-curve model, start with $n_G=0.00200\mathrm{mol}$ of glycine in $V_0=0.0200\mathrm{L}$ and exactly $n_G$ of fully dissociated hydrochloric acid; this fixes the chloride inventory, but does not force every glycine molecule to remain $G^{+}$. Let the sodium hydroxide concentration be $C_B=0.100\mathrm{mol}{\mathrm{L}}^{-1}$ and $b=n_{\mathrm{NaOH}}/n_G$. Assume volume is additive and no other ions are present; stipulate $K_w=10^{-14}$ for self-ionization of water at a temperature of $25{}^{\circ }\mathrm{C}$. ThenBalancing all positive and negative electric charges requiresSolving this equation gives the solid curve. It remains finite at $b=0$ and beyond $b=2$, where ignoring free hydronium and hydroxide would fail. At the first pKa, the full model gives $b\approx 0.436$, rather than $0.5$: appreciable free hydronium makes that half-equivalent shortcut inaccurate. At the second pKa, $b\approx 1.502$. Known acid inventory, dilution, and thermodynamic activity assumptions are therefore part of any quantitative interpretation, not optional corrections inferred from the curve's shape.

Follow one glucose through canonical glycolysis, with no diversion of intermediates into other metabolic pathways. The investment stage uses two ATP to prepare the six-carbon substrate. Cleavage and rearrangement provide two glyceraldehyde 3-phosphate molecules, each containing three carbon chemical atoms. Each then yields one three-carbon pyruvate, one NADH, and two ATP in the payoff stage.^[ref]

The material check is $6=2\times 3$: all six carbon chemical atoms remain in the two pyruvates. No carbon dioxide has been released within this boundary. In the ATP ledger, each three-carbon route contains two ATP-forming transfers, soThe first pair of payoff transfers only restores the initial ATP inventory; the second pair creates the net gain. Four newly formed ATP is therefore compatible with only two additional ATP remaining.

For an original scaled example, stipulate complete conversion of $5\mu \mathrm{mol}$ of glucose. The initial $30\mu \mathrm{mol}$ of carbon chemical atoms becomes $10\mu \mathrm{mol}$ of pyruvate, still containing $30\mu \mathrm{mol}$ of carbon chemical atoms. The pathway spends $10\mu \mathrm{mol}$ of ATP, forms $20\mu \mathrm{mol}$ of ATP, and adds $10\mu \mathrm{mol}$ to the ATP inventory before any other ATP use. It also reduces $10\mu \mathrm{mol}$ of NAD+. These are calculated ideal amounts, not observed yields; side routes or incomplete conversion change the outcome.

A common biochemical net chemical equation isHere ${\mathrm{P}}_i$ is inorganic phosphate. The names group biochemical protonation and metal-binding states; the displayed proton and water terms use a common shorthand rather than a unique specification of every chemical species. The robust ledger records two net ADP-to-ATP conversions and two NAD+-to-NADH conversions. Summing the individual chemical equations cancels two invested ATP against four formed ATP and cancels the internal substrates, giving those coefficients.

Erwin Louis Hahn's 1950 apparatus enclosed liquid samples in the coil of a tuned LC circuit, coupled to a gated radio transmitter and radio receiver. One transmitter quadrupled a crystal oscillator's 7.5 MHz to 30 MHz. A separate proton resonance regulator stabilized the static magnetic field.^[ref]

Waiting for thermal equilibrium prepared nuclear polarization. Oscilloscope photographs compared glycerol proton echoes at different field homogeneity: echo width changed; echo time followed twice the pulse separation. Receiver saturation obscured early free induction decay; transmitter leakage required care. The delayed signal tested refocusing; diffusion could still attenuate it.^[ref]

Hahn analyzed equal pulses as well as noting improved recovery with a second pulse twice as long; the ideal sequence above isolates that latter case.^[ref]

This stable helium isotope contains two protons and one neutron. In solar proton-proton burning it is an intermediate: two such nuclei can form helium-4 and return two protons, or one can capture helium-4 to enter the second chain branch. The balance between production and destruction determines the helium-3 equilibrium abundance and its helium-3 relaxation time.

Victor Franz Hess's 1912 series used two sealed, 3-mm-walled ionization chambers with Theodor Wulf's fiber electrometers. Karl Bergwitz suggested internal plating; Günther & Tegetmeyer improved focusing. Basket-mounted ground comparisons, insulation-loss measurements, simultaneous detectors and barograph-derived heights supplied checks. The highest flight reached 5350 m; its comparisons came from Vienna and after landing, without a launch-site baseline.^[ref]

A third, thin-walled chamber admitted ambient air. Hess multiplied its readings by $750/b$, with pressure $b$ in millimeters of mercury, warning that vessel-wall contributions might violate this scaling. He recorded outside temperature but applied no absolute-temperature correction. Night flights, an eclipse and weather observations constrained alternatives to radiation from above; the eclipse argument addressed directly propagating solar radiation.^[ref]

The electrometers read gradual electric potential loss caused by continuing air ionization. The low-altitude dip and strong high-altitude rise favored an extraterrestrial contribution, without measuring proton composition.^[ref] An ionization chamber integrates electric charge: one energetic particle may create many ion pairs, several particles may contribute to one reading, and a photon can create charged particles in the gas or walls. A voltage-loss reading is therefore neither a count of primary cosmic rays nor a chemical analysis of them.

Domenico Pacini supplied independent evidence: in June 1911 he compared a copper-box-enclosed electroscope at the sea surface and 3 m underwater, more than 300 m offshore above an 8-m-deep seabed. Reduced underwater ionization, stable electrical loss between runs, and a lake comparison supported an external contribution independent of soil radioactive decay. Water provided shielding; this established a different constraint from Hess's altitude comparison.^[ref]

The ATLAS experiment and CMS experiment searched proton-proton collisions at the Large Hadron Collider at collision energies of $7$ and $8\mathrm{TeV}$. The ATLAS experiment used silicon tracking, a $2\mathrm{T}$ solenoid, liquid-argon electromagnetic calorimeters and an external muon spectrometer. Its July 2012 paper reported a combined local statistical significance of $6.0$ standard deviations near $126.5\mathrm{GeV}$; this paper result is distinct from the initial 4 July announcement.^[ref]

The CMS experiment used silicon tracking in a $3.8\mathrm{T}$ solenoid, lead-tungstate electromagnetic calorimeters and external muon chambers. Electrons combine tracks and particle showers; muons combine tracking and chamber hits; photons supply shower energies and directions. Selected $H\to \gamma \gamma$ and $H\to Z{Z}^{\ast }\to 4\ell$ candidates, with $\ell =e,\mu$, yielded invariant mass peaks. It checked lepton efficiencies with Z boson decays, estimated fake-lepton backgrounds from control samples with relaxed identification, and fitted the smooth diphoton background from data. Its combined local statistical significance was $5.0$ standard deviations near $125\mathrm{GeV}$.^[ref]

The ATLAS experiment compared signal-plus-background and background-only hypotheses with a profile likelihood, accounting for systematic uncertainties and correlations between decay channels. Diphoton and four-lepton candidates gave precise invariant masses; $W{W}^{\ast }$ decays supplied additional evidence with poorer mass reconstruction because of escaping neutrinos. A local p-value concerns one specified mass; scanning masses requires a look-elsewhere effect correction.^[ref]

Two-photon decay excludes spin one under the usual assumptions, but does not prove spin zero. A mass peak does not determine every Higgs boson coupling. Production rates, branching fractions and angular correlations require additional tests.^[ref]

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.

A hydrogen bond acceptor supplies the electron-rich atom or group $Y$ interacting with the attached hydrogen of a hydrogen bond donor in $X-\mathrm{H}\cdots Y$. An oxygen lone pair in water is the familiar example. Acceptance describes the association, not necessarily acquisition of a proton. A single molecule, such as water or ammonia, can offer both hydrogen bond donor and hydrogen bond acceptor sites. See Glasfeld: water and nitrogen-containing accepting sites.

A lone pair drawn in a Lewis structure is a useful clue, not a universal test. At an ordinary neutral amide nitrogen, the nominal lone pair participates in chemical resonance with the carbonyl group, so this nitrogen is ordinarily excluded from lists of hydrogen bond acceptors; the carbonyl group's oxygen remains an accepting site. Protonation changes the test again: ${\mathrm{NH}}_4^{+}$ has four N–H covalent bonds and no nitrogen lone pair, whereas ammonia has three and one. These distinctions follow the actual chemical form, and accepting ability also depends on the partner and whether it can approach the site. See McMurry: amide resonance and protonation and Glasfeld: limitations of nitrogen accepting sites.

Conversely, an accepting group need not supply an atom-centred lone pair. The following phenol–benzene experiment probes an O–H donor interacting with the ring's distributed $\pi$ electron density.

Saggu, Levinson and Boxer (2011) used infrared spectroscopy to compare $200\mathrm{mmol}{\mathrm{L}}^{-1}$ phenol in different solvents. The O–H stretching peak was at $3611\mathrm{c}{\mathrm{m}}^{-1}$ in ${\mathrm{CCl}}_4$ and $3555.9\mathrm{c}{\mathrm{m}}^{-1}$ in benzene. To distinguish association from a general solvent effect, they diluted the ring-containing chemical solutions with ${\mathrm{CCl}}_4$: two peaks appeared, with the free peak nearly fixed while the associated peak depended on the partner. Replacing the O–H hydrogen with deuterium reproduced the ordering of shifts for different partners. Thus the perturbed O–H group identifies the hydrogen bond donor; the ring supplies the hydrogen bond acceptor. A peak shift alone does not specify a unique molecular geometry. ^[ref]

The roles make recognition between molecules intelligible: exposed hydrogen bond donors need compatible hydrogen bond acceptors in reachable positions. A count of sites is a starting description, not a prediction of binding strength. In water, forming a contact between two partners can also replace their existing hydrogen bonds to water; the Gibbs free energy of association concerns the whole change. See Glasfeld, pp. 15–16: competition with solvent interactions.

A hydrogen bond donor is the $X-\mathrm{H}$ group supplying the attached hydrogen in a hydrogen bond $X-\mathrm{H}\cdots Y$. The donor atom $X$, or the whole molecule containing it, may also be called the donor. The terminology does not imply transfer of a free proton: in the ordinary water pair, the O–H covalent bond remains. Familiar donors include O–H groups in water and N–H groups in ammonia. See Andrew Glasfeld, Intermolecular Attraction, pp. 12–13: donor and acceptor roles.

The attached hydrogen is essential to this role. At the nitrogen site, an unprotonated amine $R_3\mathrm{N}$ can be a hydrogen bond acceptor, but has no N–H hydrogen bond donor. Protonation gives $R_3{\mathrm{NH}}^{+}$: the new N–H group can donate, while that nitrogen no longer has the accepting lone pair. This is a change of chemical form, not a permanent label attached to a chemical element. Here $R$ denotes a group attached through carbon; the statement concerns the nitrogen site and does not rule out weaker C–H associations elsewhere. See McMurry, Organic Chemistry, 24.3: amine protonation and Glasfeld: protonated amines as donors.

An amide with an N–H covalent bond can donate even though its nitrogen is ordinarily not an accepting site. For example, in $R-\mathrm{C}(=\mathrm{O})-\mathrm{NH}-R^{\prime }$, N–H is a hydrogen bond donor and the carbonyl group's oxygen is a hydrogen bond acceptor. Replacing the remaining N–H by another group attached through carbon removes this N–H donor without removing the accepting oxygen. The analogous division of roles in a peptide bond helps explain protein structure. See Glasfeld: amide roles and biological interactions. The useful question is therefore which site supplies the hydrogen, rather than whether the whole substance is simply “hydrogen bonding”.

A bubble chamber using liquid hydrogen as both target and recording medium. For a beam interaction on a proton, the target's initial momentum can usually be approximated as zero on the scale of the beam momentum. This makes four-momentum conservation especially useful: the initial state does not include the unknown motion of a proton bound inside a heavier atomic nucleus. The approximation still neglects molecular binding and thermal motion.

For the point-Coulomb ground state, radial probability density is $4r^2{e}^{-2r/a_0}/a_0^3$. Integrating to $r_p$ gives $1-e^{-2u}(1+2u+2u^2)$, with $u=r_p/a_0$, whose small-$u$ expansion is the displayed result. A finite proton charge distribution changes the interior potential, and a proton radius is not literally a sharp material boundary; the expression is an estimate of a small overlap region.

In acid-base chemistry, hydrogen ion usually refers to positively charged hydrogen, represented by ${\mathrm{H}}^{+}$. In water, this notation describes solvated species, commonly represented by hydronium, rather than an isolated bare proton. The hydrogen ion thermodynamic activity defines pH.

Hydronium is the ion formed when a water molecule accepts a proton. It provides the usual aqueous shorthand for the hydrogen ion used in pH calculations.

The inner mitochondrial membrane separates mitochondrial matrix from intermembrane space. Its selective permeability permits distinct chemical conditions, while embedded mitochondrial respiratory chain proteins and ATP synthase support oxidative phosphorylation. Cristae increase its surface area.^[ref]

A selective boundary and a productive enzyme have different jobs. Even active respiratory enzymes cannot sustain a useful electrochemical gradient if protons return freely across the boundary. Conversely, an intact boundary without an electron donor does not supply the energy for continued pumping. The isolated mitochondria respiration experiment separates some of these dependencies by additions to the same preparation.

In a mitochondrion, the intermembrane space lies between the outer mitochondrial membrane and inner mitochondrial membrane, extending into the interiors of cristae. It includes the side of the inner mitochondrial membrane from which protons return through ATP synthase to the mitochondrial matrix.

A useful sidedness check is to follow a single arrow: outward pumping goes from green to yellow in the diagram, while coupled return goes from yellow to green. These are opposing movements of protons, not a circuit in which electrons return through ATP synthase. The mitochondrial respiratory chain and chemiosmosis articles distinguish the two ledgers.

In the reaction displayed here, an electron antineutrino interacts with a proton to produce a Positron and a neutron. It turns an otherwise hard-to-observe incident particle into two products with different detection signatures. A prompt Positron signal followed by neutron capture can identify a candidate much more selectively than a single energy deposit. Inverse beta decay is a weak interaction; the incident electron antineutrino is detected through its reaction products.

The neutron and Positron together have more rest mass than the initial proton, so a finite incident energy is necessary even before recoil is considered. In the proton's rest frame, use four-momentum conservation and a negligible neutrino mass. The initial squared center-of-mass energy is $m_p^2{c}^4+2m_p{c}^2{E}_{\nu }$; the least final squared center-of-mass energy is $(m_n+m_e{)}^2{c}^4$. Equating them gives the threshold above. At threshold the final particles share one center-of-mass velocity; they are not both at rest in the laboratory.

Writing $\Delta =m_n-m_p$, the result separates intoThe positive second term is the laboratory recoil correction. With rounded rest energies $m_p{c}^2=938.272\mathrm{MeV}$, $m_n{c}^2=939.565\mathrm{MeV}$ and $m_e{c}^2=0.511\mathrm{MeV}$, the calculated threshold is about $1.806\mathrm{MeV}$. These rounded inputs follow the recommended particle masses.^[ref]

As a further ideal energy balance, suppose the Positron stops and annihilates with an electron, and all prompt annihilation gamma rays are contained. Its prompt visible energy is $T_{e^{+}}+2m_e{c}^2$. Neglecting neutron recoil gives $E_{\mathrm{vis}}\simeq E_{\nu }-(m_n-m_p)c^2+m_e{c}^2\simeq E_{\nu }-0.782\mathrm{MeV}$. Escaping gamma rays, detector response and recoil change that measured sum. The delayed capture signal is a separate energy deposit, not an extra contribution to the incident electron antineutrino's prompt energy reconstruction.