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The dimensionless occupancy scale $n{\lambda }_T^3$, with number density $n$ and thermal de Broglie wavelength ${\lambda }_T=h/\sqrt{2\pi m{k}_{B}T}$. Increasing it approaches the regime where quantum statistics matters. This use of degeneracy must not be confused with degeneracy of an individual energy eigenspace.

One linear Bogoliubov branch $E_k=\hslash c_{s}k$ in three dimensions gives thermal energy ${\pi }^{2}V(k_{B}T{)}^4/(30{\hslash }^3{c}_s^3)$ and the displayed cubic heat capacity. There is one branch, rather than the three polarizations in a simple crystal's Debye model. The regime requires $\hslash c_s/L\ll k_{B}T\ll \mu$ for a box: the lower inequality permits a continuum integral, and the upper one selects phonons. Fixed finite boxes ultimately have discrete-level behaviour.

For electron energy transfer $\Delta$ and phonon energy ${\Omega }_q=\hslash {\omega }_q$, the symmetrized pair amplitude is $V_{\mathrm{eff}}=2|M_q{|}^2{\Omega }_q/({\Delta }^2-{\Omega }_q^2)$. It is attractive for $|\Delta |<{\Omega }_q$. This interaction is retarded, and screened Coulomb repulsion must also be considered.

The allowed phonon energies, including their wavevector, branch and phonon polarization labels when resolved. A phonon dispersion relation records the branch-resolved spectrum, whereas the phonon density of states sums over those labels.

To isolate conversion losses, suppose a converter receives 100 blue photons at ${\lambda }_b=450\mathrm{nm}$, absorbs a fraction $a=0.60$, and has fluorescence quantum yield $\Phi =0.90$: 0.90 converted photons per absorbed blue photon. Stipulate that every converted photon has ${\lambda }_c=560\mathrm{nm}$, that unabsorbed blue is transmitted, and that both outputs escape without further losses. The output is then 40 blue and 54 converted photons. The total photon yield is $0.94$, while the retained optical energy fraction isThe lower energy of each converted photon matters even if $\Phi =1$. Combining this stipulated converter with the preceding chosen light-emitting diode gives $P_{\mathrm{after}}=16.5430\mathrm{mW}$ and an overall optical/electrical power ratio $0.258485$. These chosen narrow lines and photon counts illustrate conversion accounting; they do not establish a white appearance, color quality or luminous efficacy. Those require the full spectrum and a model of human visual sensitivity.

Escape following chemical reactions that produce atoms or molecules with energies above the local gravitational escape threshold. The production altitude, collision environment and reaction energy distribution matter.

Photocurrent is the electric current attributable to illumination. Here $I_{\mathrm{ph}}\ge 0$ denotes its magnitude after background subtraction. For a photodiode without internal multiplication, let $P$ be incident radiant flux concentrated at wavelength $\lambda$, and let $\eta$ be external quantum efficiency, including losses before collection. An incident photon has energy $E_{\gamma }=hc/\lambda$, with Planck constant $h$ and speed of light $c$. Dividing energy per second by energy per photon gives the incident photon rate ${\Phi }_{\gamma }$. Multiplying by $\eta$ gives the mean collected pair rate ${\dot{N}}_{\mathrm{pair}}$; multiplying by elementary charge $q$ givesThe factor is $q$: each collected pair replenishes an electron at one contact and removes an electron at the other. A cross-section of the external wire therefore counts one transferred electron per completed pair, consistent with charge conservation. The electron and electron hole are two internal charge carriers, not two independent transfers through that wire.

Choose an illustrative photodiode model with $\lambda =650\mathrm{nm}$, $\eta =0.75$ and $P=1.00\mu \mathrm{W}$. Using the exact SI values $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$ and $q=1.602176634\times 10^{-19}\mathrm{C}$ givesA mean fraction $0.75$ does not require each fourth photon to be lost in a repeating pattern. It specifies a mean conversion rate. Nor does $0.75$ mean $0.75\mathrm{A}{\mathrm{W}}^{-1}$: quantum efficiency is dimensionless, whereas spectral responsivity compares electric current with power.

At fixed radiant flux, longer-wavelength photons have less energy, so more arrive per second. This explains the rising straight line when quantum efficiency is held constant. A real photodiode's quantum efficiency varies with wavelength; the plotted assumption cannot be extended past its band gap absorption cutoff. At fixed incident photon rate instead, $I_{\mathrm{ph}}=\eta q{\Phi }_{\gamma }$ has no additional factor of $\lambda$.

Photodetachment removes an electron from an anion using a photon. For chlorine, the reaction is ${\mathrm{Cl}}^{-}(g)+\gamma ⟶\mathrm{Cl}(g)+e^{-}$. Measuring the minimum photon energy for the specified final quantum state gives a photodetachment threshold and hence an electron affinity.

Photodissociation separates a molecule into fragments after photon absorption. For an isolated parent with a single absorbed photon, conservation of energy givesInternal excitation energies here are relative to the parent and product ground states used to define $D_0$; $E_{\mathrm{recoil}}$ is the total product translational kinetic energy in the parent's centre of mass frame. Measuring one product's kinetic energy is insufficient by itself. For two fragments of masses $m_A,m_B$, conservation of momentum gives $E_{\mathrm{recoil}}=E_A(m_A+m_B)/m_B$ when $A$ is detected. Unknown product excitation can conceal part of the available energy.

The photoelectric effect is the release of electrons from matter following photon absorption. Here the focus is emission from a metal surface into vacuum: a photon supplies energy, the work function pays the escape cost, and the remainder can become the emitted electron's kinetic energy. A photocurrent counts collected electric charge; a stopping potential probes the upper kinetic energy limit. These are different measurements.

In 1905, Albert Einstein proposed that an individual light quantum could transfer its whole energy to one electron. His §8 allowed losses before escape, distinguished the fastest emitted electrons from the rest, and predicted a straight stopping potential versus frequency relation with a material-independent slope. He separately predicted that changing light intensity changes the number emitted without changing their velocity distribution under otherwise identical conditions. He also discussed incomplete transfer and limits to the account.^[ref]

The useful modern starting model is a clean, homogeneous metal with independent single-photon events, negligible initial thermal broadening, negligible lowering of the surface barrier by an applied electric field, and negligible electric fields from the emitted electrons themselves. Specify the surface and its work function; it is not enough to name the bulk material. The Einstein photoelectric equation describes the upper edge of the emission energy distribution. It neither predicts that every absorbed photon liberates an electron nor supplies the whole photocurrent curve. Its appeal is that a complicated surface process still exposes the universal Planck constant through a simple slope.

The photoelectric saturation current is the photocurrent reached when increasing the collecting electric potential difference no longer increases the fraction of available emitted electrons collected. It is saturation with collecting voltage at fixed illumination. It does not mean that increasing incident light intensity cannot increase emission.

If $P$ is incident radiant flux on the emitting area and $\eta$ is the external quantum efficiency for producing escaping electrons, then the incident photon arrival rate is $P/(h\nu )$ and complete collection gives $I_{\mathrm{sat}}=e\eta P/(h\nu )$. For the example's $\nu =8.00\times 10^{14}\mathrm{Hz}$, choose $\eta =0.10$ and $P=1.00\mu \mathrm{W}$. The calculated photon arrival rate is $1.886488\times 10^{12}{\mathrm{s}}^{-1}$ and $I_{\mathrm{sat}}=30.224866\mathrm{nA}$. Doubling $P$ doubles this current to $60.449731\mathrm{nA}$ while leaving $V_s$ fixed. At fixed illuminated area this doubles the irradiance. Changing frequency at fixed $P$ changes the arrival rate as well as the photon energy; $\eta$ can also change, so there is no general rule that a higher frequency gives a larger photocurrent.

A missing plateau can indicate incomplete collection or an electric field altered by the emitted electric charge. Heating, surface changes and additional emission channels can invalidate a constant $\eta$. Even a small amount of unwanted higher-frequency radiation can move an apparent endpoint. A stopping potential inferred from zero net current additionally needs background subtraction and a check for emission from the collector itself.

The photoelectric threshold frequency is the lower frequency boundary allowed by the ideal single-photon energy balance. With vacuum wavelength $\lambda =c/\nu$ and speed of light $c$, its corresponding longest wavelength is ${\lambda }_0=hc/\Phi$. The chosen $2.30\mathrm{eV}$ work function gives ${\nu }_0\simeq 5.561375\times 10^{14}\mathrm{Hz}$ and ${\lambda }_0\simeq 539.062\mathrm{nm}$. These are model parameters, not a measured threshold for a named metal. Detectable photocurrent need not begin abruptly at the ideal boundary: collection, finite spectral width, thermal broadening and surface inhomogeneity matter.

An electron ejected following photon absorption. For a free residual ion with negligible recoil, the emitted electron's kinetic energy equals incident photon energy minus the ionization binding energy.

Let a conducting sample and conducting analyzer be in good electrical contact, unbiased, at equilibrium, and without appreciable charging during illumination. Their Fermi levels then coincide. Choose that common Fermi level as zero. Define $E_B>0$ as the binding energy relative to this zero: removing an electron into the electron reservoir at the Fermi level costs $E_B$. It is not the cost of removing the electron all the way to the sample's vacuum level.

Write the sample and analyzer work functions as ${\varphi }_s$ and ${\varphi }_a$. Their local vacuum levels lie at ${\varphi }_s$ and ${\varphi }_a$ above the common Fermi level. Conservation of energy for an electron emitted without additional losses gives total outgoing energy $h\nu -E_B$, where $h$ is the Planck constant and $\nu$ the photon frequency. Immediately outside the sample its kinetic energy is $K_s=h\nu -E_B-{\varphi }_s$. The contact-related electric potential changes this toThus the sample work function cancels when the measured kinetic energy is referenced to the analyzer. Substituting ${\varphi }_s$ for ${\varphi }_a$ in the final conversion would mix two different energy references. An instrument's calibrated binding energy scale can incorporate its effective analyzer work function, so the correction need not appear as a separate operator-entered number.

As an original numerical example, stipulate $h\nu =1486.6\mathrm{eV}$, ${\varphi }_a=4.5\mathrm{eV}$ and a line at $E_B=284.0\mathrm{eV}$. The predicted analyzer kinetic energy is $1198.1\mathrm{eV}$. If ${\varphi }_s=5.0\mathrm{eV}$, the surface value is instead $K_s=1197.6\mathrm{eV}$; the contact-related increase is $0.5\mathrm{eV}$. A second line at $286.0\mathrm{eV}$ gives $1196.1\mathrm{eV}$ at the analyzer. A $+2.0\mathrm{eV}$ chemical shift in photoelectron spectroscopy therefore appears as a $-2.0\mathrm{eV}$ shift in measured kinetic energy. These are chosen values, not a measured carbon spectrum.

Charging breaks the assumed reference. If a uniformly charging sample rises by electric potential $V>0$ relative to the analyzer, an escaping electron loses an additional energy $eV$, with $e$ the positive elementary charge. Applying the unbiased conversion consequently returnsTwo features from the same stable potential retain their separation. Features from regions at different potentials acquire different offsets, which can mimic chemical differences or broaden a line. A conducting holder alone does not establish electrical contact through an insulating surface layer. Compare repeat scans, electrical contact and charge-compensation settings before interpreting an offset as a change in chemical bonds.

Measurement calibration is a separate issue: a known reference line tests the scale, while charge compensation controls the specimen's electric potential. In an original offset model $E_{B,\mathrm{read}}=E_{B,\mathrm{true}}+b$, one known line determines $b$. An unknown scale factor in $E_{B,\mathrm{read}}=a{E}_{B,\mathrm{true}}+b$ requires two distinct known lines to determine both $a$ and $b$. Re-reading a reference after a sample checks drift. Treating an unidentified contamination peak as a universally fixed reference would assume the chemical identity and electrical behavior that the measurement should test.

Photoelectron spectroscopy measures the kinetic energies of photoelectrons produced by illuminating matter with photons of known energy. The photoelectric effect supplies the emission mechanism; spectroscopy turns the distribution of emitted electron energies into information about occupied energy levels and chemical bonds. A broad survey helps identify chemical elements, while small shifts within one element's lines can distinguish different chemical surroundings. These are different questions: finding carbon does not establish which chemical bonds its chemical atoms form.

The useful inversion is an energy balance. A known incoming photon energy pays for removal of an electron and for its outgoing kinetic energy. To apply that balance to a solid, specify the zero of energy and the work function of the instrument. Photoelectron binding energy reference derives the conversion for a conducting sample in electrical contact with its analyzer. A free molecule instead has an ionization energy referenced to an electron at rest in vacuum; it need not share a Fermi level with the analyzer.

A redshift estimated from multi-band fluxes or an inferred spectral energy distribution rather than directly resolved spectral-line wavelengths. calibration, templates, dust and population differences can produce biases and catastrophic outliers. In lensing or clustering analyses, the inferred source redshift distribution affects the geometric kernel as well as assignment to redshift bins.

Photometry weights visible electromagnetic radiation by standardized visual response functions; radiometry measures its energy and power without visual weighting.^[ref]

The question is how much light a source supplies under a stated visual weighting. A beam containing $1\mathrm{mW}$ at the defining frequency $540\mathrm{THz}$ has luminous flux $0.683\mathrm{lm}$. Moving the same radiant flux to a region of smaller spectral luminous efficacy reduces its luminous flux. Changing the optical spectrum changes this conversion even when its total power stays fixed. This makes photometry useful for comparing lighting, while radiometry remains necessary for an energy budget. The lumen counts luminous flux; the candela additionally specifies its distribution over solid angle.

Two measurement routes implement the weighting: measure an optical spectrum and evaluate a weighted integral, or use a photometer whose spectral responsivity approximates the prescribed weighting. Both require a specified collection geometry. Neither is a complete prediction of an individual's perception of a scene.^[ref]

Photon absorption removes a photon and transfers its energy and momentum to matter. It contributes a negative photon-number collision source. It differs from Compton scattering, which redistributes a photon's energy and direction while preserving the count. The integrated source can vanish if photon emission and absorption balance.

An initially stationary isolated chemical atom of rest mass $M$ absorbing a single photon must recoil. If the internal excitation changes its rest mass by $\Delta M=\Delta E/c^2$, conservation of energy and conservation of momentum give $M{c}^2+h\nu =\sqrt{(M{c}^2+\Delta E{)}^2+(h\nu {)}^2}$ and hence the exact formula in the title. The additional positive energy supplies recoil. An external body sharing the momentum, or an initially moving chemical atom, changes this laboratory resonance condition.

For photon energy $\epsilon =h\nu$, equating radiated energy in corresponding spectral intervals gives $L_{\nu }d\nu =\epsilon L_{\epsilon }^{N}d\epsilon$. Thus $L_{\epsilon }^N=L_{\nu }/(h\epsilon )$. A quoted rate per joule must be multiplied by the number of joules per electronvolt or per $\mathrm{keV}$ to express it in that spectral unit. For homogeneous optically thin emission, $L_{\nu }=4\pi V{j}_{\nu }$ if $j_{\nu }$ is defined per solid angle.