--- title: Photons and Quantization module: Energy moduleNumber: 3 lessonNumber: 5 order: 305 summary: > Light delivers its energy in indivisible lumps: a photon of frequency $f$ carries exactly $hf$, and this one fact explains why a dim blue lamp ejects electrons that an intense red one cannot. We fix a photon's energy and momentum from its wavelength, follow the quanta through emission, absorption, and the photoelectric threshold $K_{\rm max}=hf-\phi$, and watch energy and momentum conservation together produce the Compton wavelength shift when a photon scatters from an electron. The recurring discipline is unit and frame care, where a stray factor of $10^9$ or a forgotten rest energy quietly ruins an answer. topics: [Energy] draft: false sources: - book: Tipler & Mosca ref: "Ch. 7 — Conservation of Energy; §7-5 Quantization of Energy" --- ## Quantized energy and photons Some systems exchange energy only in definite amounts: their allowed energies form a discrete set of states rather than a continuous interval. When such a system drops from an initial state of energy $E_i$ to a lower state $E_f$, the difference leaves as electromagnetic radiation, quantized into photons of energy $$ E_\gamma=hf, $$ where $h=6.62607015\times10^{-34}\ \mathrm{J\,s}$ is Planck's constant and $f$ is frequency. A transition emits one photon only when its energy matches an allowed difference, $hf=E_i-E_f$. The discreteness belongs to the system's permitted states, not to any rounding of the measured energy. $$ % caption: Quantized energy states. A downward transition between two allowed levels emits one photon whose energy equals the level spacing, E equals h f. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \useasboundingbox (-0.20,-0.20) rectangle (6.60,3.55); \draw[->,black] (0.60,0.35)--(0.60,3.00) node[above] {energy}; \draw[black,very thick] (1.05,0.85)--(3.05,0.85) node[right,black] {$E_1$}; \draw[black,very thick] (1.05,1.65)--(3.05,1.65) node[right,black] {$E_2$}; \draw[black,very thick] (1.05,2.50)--(3.05,2.50) node[right,black] {$E_3$}; \draw[->,black,very thick] (1.95,2.45)--(1.95,1.72); \draw[->,acc,very thick] (3.60,2.08)--(5.95,2.08); \node[above,black] at (4.75,2.10) {emitted photon}; \node[below,black] at (4.75,2.06) {$E = h f$}; \end{tikzpicture} $$ Frequency and wavelength are tied by $c=f\lambda$, so $$ E_\gamma=hf=\frac{hc}{\lambda}. $$ Shorter wavelength means higher frequency and greater energy per photon. The relation uses the vacuum wavelength. Frequency is fixed across a material boundary while speed and wavelength change, so the in-material wavelength gives the right frequency only when paired with the propagation speed in that material. One electronvolt is the energy gained by a charge of magnitude $e$ across one volt: $$ 1\ \mathrm{eV}=1.602176634\times10^{-19}\ \mathrm J. $$ In electronvolts and nanometres the relation becomes $E_\gamma[\mathrm{eV}]=1240/\lambda[\mathrm{nm}]$ and $E_\gamma[\mathrm{keV}]=1.240/\lambda[\mathrm{nm}]$. These are rounded conversions, reliable for a vacuum wavelength when the requested precision suits the rounded constant. $$ % caption: Photon wave and wavelength. One full wavelength is marked along the propagation direction; a shorter wavelength means higher frequency and greater energy, since E equals h c divided by the wavelength. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \useasboundingbox (-0.20,-0.35) rectangle (6.60,3.10); \draw[acc,very thick,smooth] plot coordinates {(0.60,1.90) (0.85,2.29) (1.10,2.45) (1.35,2.29) (1.60,1.90) (1.85,1.51) (2.10,1.35) (2.35,1.51) (2.60,1.90) (2.85,2.29) (3.10,2.45) (3.35,2.29) (3.60,1.90) (3.85,1.51) (4.10,1.35) (4.35,1.51) (4.60,1.90) (4.85,2.29) (5.10,2.45) (5.35,2.29) (5.60,1.90)}; \draw[<->,black,thick] (0.60,0.80)--(2.60,0.80); \draw[dashed,black] (0.60,1.85)--(0.60,0.80); \draw[dashed,black] (2.60,1.85)--(2.60,0.80); \node[below,black] at (1.60,0.78) {one wavelength}; \draw[->,black,thick] (4.55,2.80)--(5.60,2.80); \node[left,black] at (4.55,2.80) {propagation}; \end{tikzpicture} $$ Photons carry momentum despite zero rest mass: $$ p_\gamma=\frac{E_\gamma}{c}=\frac{h}{\lambda}. $$ Momentum conservation governs both emission and absorption. An isolated emitter recoils opposite to the photon, taking a small share of the energy as kinetic energy. An absorber receives both energy and momentum; a free object absorbing one photon must satisfy the combined energy and momentum balance, with its recoil part of that balance. In a larger apparatus the support or surrounding material carries the compensating momentum. Emission and absorption are inverse exchanges but not identical arrangements. Emission is set by an allowed energy difference. Absorption requires an incoming photon of the right energy within the relevant line width and a receiving system that conserves momentum. A spectral line marks a permitted energy difference; its observed shape also reflects motion, fields, and instrumental resolution. $$ % caption: Photon momentum in emission and absorption. The emitter recoils opposite to the photon it releases, and the absorber gains both its energy and its momentum, so total momentum is conserved. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \useasboundingbox (-0.20,-0.20) rectangle (7.40,2.90); \draw[fill=acc!12,draw=acc,thick] (1.35,1.60) circle (0.30); \node[below,black] at (1.35,1.22) {emitter}; \draw[->,black,thick] (1.05,1.60)--(0.45,1.60); \node[above,black] at (0.74,1.60) {recoil}; \draw[->,acc,very thick] (1.72,1.60)--(4.05,1.60); \node[above,black] at (2.70,1.78) {momentum p}; \draw[fill=black!10,draw=black,thick] (4.20,1.05) rectangle (5.55,2.15); \node[black] at (4.88,1.60) {absorber}; \node[below,black] at (4.88,1.05) {energy and momentum received}; \end{tikzpicture} $$ > **Worked example (532 nm photon).** A photon has vacuum wavelength $532\ \mathrm{nm}$. > > Convert nanometres to metres, then use SI values of $h$ and $c$: > $$ > E_\gamma=\frac{(6.62607015\times10^{-34}\ \mathrm{J\,s}) > (2.99792458\times10^8\ \mathrm{m\,s^{-1}})}{532\times10^{-9}\ \mathrm m} > =3.73\times10^{-19}\ \mathrm J=2.33\ \mathrm{eV}. > $$ > The rounded form $1240/532=2.33\ \mathrm{eV}$ agrees. The momentum is > $p_\gamma=E_\gamma/c=1.24\times10^{-27}\ \mathrm{kg\,m\,s^{-1}}$, and dividing $E_\gamma$ > by $h$ gives a frequency of $5.64\times10^{14}\ \mathrm{Hz}$, with $c/f$ returning > $5.32\times10^{-7}\ \mathrm m$. Agreement across wavelength, frequency, energy, and > momentum is a stronger check than repeating one formula. The equations fix the energy and momentum transfer; the experimental boundary fixes which part of that transfer is observed. Each photon calculation carries four choices: the wavelength convention, the frame when source or detector moves, the energy unit, and whether recoil is included. Mixing a nanometre wavelength with SI constants without conversion introduces errors by powers of ten, so unit labels must carry through every line. Photon momentum is small on laboratory scales yet measurable in careful experiments. At normal incidence, perfect absorption transfers momentum $p_\gamma$ to a surface, while ideal reflection transfers $2p_\gamma$ because the outgoing momentum reverses. Both follow from vector momentum conservation without any rest mass. Radiation pressure is the macroscopic result when many photons deposit momentum per unit time and area. Photon number and total energy are distinct. A single photon carries $hf$; a pulse of $N$ identical co-propagating photons carries total energy $Nhf$ and total momentum $N h/\lambda$. A broadband pulse needs each frequency component counted separately, since detector readout reports total deposited energy, not individual photons. A measured wavelength is reported with an uncertainty interval set by calibration, resolution, source motion, detector response, and background subtraction. Through $E=hc/\lambda$, a small fractional wavelength uncertainty gives an equal fractional energy uncertainty of opposite sign. State whether a reported value is a line centre, a bandwidth, or an integrated photon energy before comparing it with a calculated state difference. ## Spectral lines, momentum transfer, and thresholds A spectrum records photon energies selected by transitions between allowed states. Each emission line satisfies $hf=E_i-E_f$ for one permitted change, and the same line appears in absorption when incoming photons supply that difference. Several lines do not imply a continuously adjustable photon energy: each frequency identifies one state separation, while line intensity reports how many photons reached the detector. Intensity and photon energy are different quantities, so a bright low-energy line can deliver more total energy than a faint higher-energy line. Line position and line width answer different questions. The line centre is compared with a transition energy; the width describes a distribution of recorded energies broadened by source motion, collisions, field shifts, temperature, resolution, and observation angle. A calibration lamp sets the wavelength scale, but the reported uncertainty must still include the fit and any detector-coordinate conversion, and both centre and calculated difference must share a unit. $$ % caption: Spectral lines from distinct transition energies. Each vertical transition emits a photon of one energy difference, so the detector shows separate lines at set positions rather than a continuum. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \useasboundingbox (-0.20,-0.20) rectangle (6.60,3.45); \draw[->,black] (0.70,0.55)--(0.70,2.80) node[above] {energy}; \draw[black,very thick] (1.05,0.80)--(2.55,0.80); \draw[black,very thick] (1.05,1.60)--(2.55,1.60); \draw[black,very thick] (1.05,2.45)--(2.55,2.45); \draw[->,black,thick] (1.45,2.40)--(1.45,0.85); \draw[->,black,thick] (1.95,2.40)--(1.95,1.65); \draw[->,black,thick] (2.30,1.55)--(2.30,0.85); \draw[->,black] (3.30,0.80)--(6.20,0.80); \node[below,black] at (5.45,0.80) {detector position}; \draw[black,very thick] (3.95,0.80)--(3.95,1.55); \draw[black,very thick] (4.75,0.80)--(4.75,2.00); \draw[black,very thick] (5.45,0.80)--(5.45,1.25); \node[above,black] at (3.95,1.55) {A}; \node[above,black] at (4.75,2.00) {B}; \node[above,black] at (5.45,1.25) {C}; \end{tikzpicture} $$ Photon momentum turns optical power into force. A beam delivering energy $E$ to a perfectly absorbing surface also delivers momentum $E/c$ along the beam. For steady intensity $I$ at normal incidence on area $A$, the force on an ideal absorber is $F=IA/c$ and the pressure is $P=I/c$. A perfect reflector reverses the outgoing momentum, doubling the change to $P=2I/c$. Real surfaces fall between these limits as energy is reflected, transmitted, scattered, or absorbed. $$ % caption: Radiation pressure from photon momentum transfer. An absorbed photon gives the surface momentum p; a reflected photon reverses and gives twice as much, 2p, for the same incident energy. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \useasboundingbox (-0.20,-0.45) rectangle (6.60,3.10); \draw[fill=black!10,draw=black,thick] (3.35,0.35) rectangle (3.63,2.75); \node[above,black] at (3.49,2.75) {surface}; \draw[->,black,very thick] (0.65,2.20)--(3.31,2.20); \node[above,black] at (1.80,2.22) {absorbed photon}; \node[below,black] at (1.80,2.18) {transfers p}; \draw[->,black,very thick] (0.65,1.05)--(3.31,1.05); \node[above,black] at (1.55,1.07) {bounced photon}; \draw[->,acc,very thick] (3.31,0.60)--(0.95,0.60); \node[below,black] at (1.55,0.58) {transfers 2p}; \end{tikzpicture} $$ Threshold accounting compares one photon energy with the requirement for a stated outcome. If a material needs energy $\phi$ to release a charged particle, the maximum kinetic energy after absorption is $$ K_{\rm max}=hf-\phi. $$ For $hf<\phi$ there is no release channel, and extra intensity only adds photons without raising the energy each one carries. The threshold frequency and wavelength are $$ f_0=\frac{W}{h},\qquad \lambda_0=\frac{hc}{W}. $$ Wavelengths longer than $\lambda_0$ are below threshold; shorter ones leave a surplus that becomes kinetic energy, recoil, or other channels of the complete system. The word "maximum" is literal: a product created below the surface can lose energy before reaching a detector, so the measured distribution extends below the calculated upper limit even when the relation holds. > **Worked example (2.00 eV threshold).** A material requires $W=2.00\ \mathrm{eV}$ to > release a charged particle. Working in one energy unit throughout, the threshold > wavelength is > $$ > \lambda_0=\frac{1240\ \mathrm{eV\,nm}}{2.00\ \mathrm{eV}}=620\ \mathrm{nm}. > $$ > A $500\ \mathrm{nm}$ photon carries $1240/500=2.48\ \mathrm{eV}$, leaving at most > $0.48\ \mathrm{eV}$ for the kinetic-energy channel. Mixing eV in one term with joules > in the other would invalidate the subtraction. The relation is an energy account for one absorbed photon. It does not claim that every arriving photon is absorbed or that every absorbed photon yields the selected product; those probabilities set a count rate, not the energy of a fixed-frequency photon. Raising intensity at a subthreshold frequency adds heating and background but never the missing energy, which is why a frequency scan locates a threshold even as intensity drifts. Watch the inequality direction: photons with $f>f_0$, equivalently $\lambda<\lambda_0$, are above threshold, since a short wavelength carries the larger $hc/\lambda$. Where the receiving system can recoil, the released particle and the remaining material share momentum; a massive solid absorbs a negligible recoil and leaves nearly all the kinetic energy with the particle, while a small free system does not. Radiation pressure is the same energy-and-momentum account. For an ideal absorber, power divided by $c$ is the longitudinal force, and watt over metres per second is newton. A $1.00\ \mathrm W$ beam at normal incidence produces only $3.34\times10^{-9}\ \mathrm N$, which is why photon pressure is subtle in ordinary optics and becomes accessible only at high power, long interaction times, very light mechanical elements, or large collecting areas. ## Equilibrium radiation, photon flux, and calibrated counts Radiation in thermal equilibrium has a spectrum fixed by temperature alone. A cavity with absorbing walls and a small opening emits radiation whose distribution is independent of the wall material once equilibrium is reached. The rule $E=hf$ is what makes this work: high-frequency radiation costs more energy per photon, and the equilibrium distribution limits how many such photons are available. The resulting Planck spectrum rises from zero, reaches a broad maximum, and falls at short wavelength, a short-wavelength behaviour no classical continuous-energy model reproduces. The spectral radiance per unit wavelength is $$ B_\lambda(T)=\frac{2hc^2}{\lambda^5}\frac{1}{\exp\!\left(hc/\lambda kT\right)-1}. $$ Its units are power per area, per solid angle, per wavelength interval, and must not be read as total power or photon count; integrating over wavelength and angle gives a different observable. A hotter source radiates far more total power and shifts its peak to shorter wavelength, the trend summarized by Wien's displacement relation: $$ \lambda_{\rm peak}T=2.898\times10^{-3}\ \mathrm{m\,K}. $$ The peak location depends on the spectral variable. A maximum plotted against wavelength does not sit at the same coordinate as a maximum plotted against frequency, because the density curve changes height and position when the interval width changes even though the radiation does not. State whether a fitted spectrum is per wavelength, per frequency, or per photon energy before reading a temperature from its peak. $$ % caption: Equilibrium-radiation spectra at two temperatures. The hotter source has greater spectral radiance and its peak at shorter wavelength; radiance is spread over wavelength intervals, not carried by one photon. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \useasboundingbox (-0.30,-0.35) rectangle (6.55,3.10); \draw[->,black] (0.60,0.45)--(6.05,0.45); \draw[->,black] (0.60,0.45)--(0.60,2.85); \node[rotate=90,black] at (0.10,1.60) {spectral radiance}; \node[below,black] at (5.30,0.45) {wavelength}; \draw[acc,very thick,smooth] plot coordinates {(0.80,0.52) (1.10,1.55) (1.40,2.28) (1.62,2.46) (1.90,2.16) (2.35,1.46) (3.00,0.96) (3.95,0.64) (5.25,0.50)}; \draw[black,very thick,smooth] plot coordinates {(0.90,0.48) (1.55,0.68) (2.20,1.06) (2.80,1.34) (3.00,1.38) (3.55,1.18) (4.35,0.80) (5.55,0.52)}; \draw[dashed,black] (1.62,0.45)--(1.62,2.46); \draw[dashed,black] (3.00,0.45)--(3.00,1.38); \node[above,black] at (1.62,2.46) {hotter}; \node[above,black] at (3.35,1.40) {cooler}; \end{tikzpicture} $$ For a monochromatic beam, optical power and photon rate are directly related. With power $P$ at the detector and photon energy $hf$, the incident photon rate is $$ \dot N=\frac{P}{hf}=\frac{P\lambda}{hc}. $$ A $1.00\ \mathrm{mW}$ beam at $500\ \mathrm{nm}$ delivers about $2.52\times10^{15}$ photons per second, a large number because one visible photon holds only a few electronvolts. A broadband source needs the components summed; dividing total power by a single representative photon energy is valid only for a narrow or explicitly specified spectrum. Detection efficiency $\eta$ is the probability that an incident photon produces a selected recorded event, folding in geometric collection, optical transmission, absorption in the active region, electronic threshold, and data selection. The expected signal count in exposure time $t$ is $C_{\rm sig}=\eta\dot N t$. A count rate quoted without exposure time, detector area, and selection rule cannot be turned back into source power or photon rate. Independent arrivals fluctuate. For an expected count $C$, the Poisson standard deviation is $\sqrt C$. Background is measured separately and subtracted with its own uncertainty: after equal exposures $C_{\rm net}=C_{\rm on}-C_{\rm bg}$, the variance is $C_{\rm on}+C_{\rm bg}$, not $C_{\rm net}$, so a small net signal can carry a large relative uncertainty even when the on-source count is large. Use live time rather than clock time when dead time or vetoes remove part of an exposure, and scale a background taken over a different live time before subtracting. Calibration maps detector response to photon energy, wavelength, or power. Gain drift, nonlinearity, saturation, dark counts, and wavelength-dependent efficiency are systematic effects, listed and propagated separately from Poisson uncertainty; a calibration point far from the measurement range can fix an overall scale while leaving a local slope untested. Power location must be explicit, since power emitted, passing an aperture, reaching the detector face, and absorbed in the active region differ by spreading and losses. Apply $\eta$ once: either take power at the detector with the intrinsic detection probability, or take source power with the full collection-and-detection efficiency. Dead time limits high count rates. After one event a channel is briefly unavailable, so the recorded rate falls below the incoming rate, and overlapping pulses can register as one event. These effects change the counts-to-rate mapping and are not removed by a longer exposure, so report the calibrated rate range and any dead-time or pileup correction. A controlled attenuation series checks this: the background-subtracted rate should scale linearly with photon rate until saturation sets in. > **Worked example (detected count).** A detector receives $1.00\ \mathrm{mW}$ at > $500\ \mathrm{nm}$ for $2.00\ \mathrm s$, so the incident photon number is about > $5.03\times10^{15}$. At efficiency $0.250$ the expected selected count before > background is > $$ > C_{\rm sig}=\eta\dot N t=0.250\times5.03\times10^{15}=1.26\times10^{15}. > $$ > The Poisson relative uncertainty is only $1/\sqrt{C}$, so a percent-level power or > efficiency error would dominate the final photon-rate uncertainty. Statistical > precision and calibration accuracy are reported separately. The same split applies to a fitted Planck temperature: repeated spectra give a narrow statistical spread in the peak while a wrong wavelength scale or unmodeled detector sensitivity biases the temperature. ## Photon scattering, recoil, and the Compton shift Photon scattering tests energy and momentum conservation simultaneously. An incoming photon of wavelength $\lambda$ strikes a free electron at rest and leaves at wavelength $\lambda'$ and angle $\theta$, while the electron recoils with the momentum that balances the change in photon momentum. Photon energy alone cannot predict the shift, and photon momentum alone cannot fix the recoil energy; the four-momentum relation supplies both in one calculation. For the ideal single-scattering geometry, conservation gives the Compton wavelength shift $$ \lambda'-\lambda=\frac{h}{m_ec}\left(1-\cos\theta\right) =\lambda_C\left(1-\cos\theta\right), $$ where $m_e$ is the electron mass and $\lambda_C=2.426310238\times10^{-12}\ \mathrm m$ is the electron Compton wavelength. The shift is zero in the forward direction and reaches $2\lambda_C$ for backscattering. Within the free-electron-at-rest approximation it depends only on scattering angle, not on incident wavelength, though the fractional change is larger for longer wavelengths because the fixed absolute shift is divided by a larger initial value. $$ % caption: Compton-scattering geometry. The incoming and scattered photon momenta differ in direction and magnitude; the recoil-electron momentum closes the vector triangle set by momentum conservation. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \useasboundingbox (-0.20,-0.20) rectangle (6.60,3.10); \coordinate (VX) at (2.95,1.45); \draw[->,black,very thick] (0.55,1.45)--(VX); \node[above,black] at (1.60,1.47) {incoming photon}; \draw[->,acc,very thick] (VX)--(5.55,2.55); \node[black] at (4.15,2.72) {scattered photon}; \draw[->,black,thick] (VX)--(4.65,0.55); \node[below,black] at (4.10,0.62) {recoil electron}; \draw[black] (3.60,1.45) arc (0:23:0.65); \node[black] at (4.10,1.62) {angle}; \draw[fill=white,draw=black,thick] (VX) circle (0.07); \end{tikzpicture} $$ The photon energy after scattering follows without separately solving for the electron speed: $$ E'=\frac{E}{1+\left(E/m_ec^2\right)\left(1-\cos\theta\right)}. $$ Since $E=hc/\lambda$, a longer scattered wavelength means lower scattered photon energy. In the ideal setup the lost photon energy becomes electron kinetic energy, $K_e=E-E'$, because the electron starts at rest and no other products appear. That equality holds only for the closed two-body system: a bound initial electron, escaping radiation, or additional particles breaks it. Momentum components give an independent geometric check. Taking the incoming photon along the horizontal axis gives $$ p_e\cos\phi=p-p'\cos\theta,\qquad p_e\sin\phi=p'\sin\theta, $$ where $\phi$ is the recoil-electron angle. The electron's transverse momentum is fixed by the transverse momentum of the outgoing photon, so a reconstructed event must close both component equations within its angle and energy uncertainties, guarding against an energy-only calculation that silently violates momentum conservation. > **Worked example (x-ray backscatter).** An x-ray of wavelength $71.1\ \mathrm{pm}$ > backscatters at $\theta=180^\circ$. The wavelength shift is maximal, > $$ > \lambda'-\lambda=2\lambda_C=4.85\ \mathrm{pm},\qquad \lambda'=75.95\ \mathrm{pm}. > $$ > With $E[\mathrm{keV}]=1.240/\lambda[\mathrm{nm}]$, the incident energy is > $17.4\ \mathrm{keV}$ and the scattered energy about $16.3\ \mathrm{keV}$. The > difference, $\approx 1.1\ \mathrm{keV}$, is the electron recoil kinetic energy in the > ideal two-body account. Rounding the small shift too early distorts this difference. The shift relation rests on three assumptions: - **Free target electron**, initially at rest. - **Two-body final state**, one photon and one electron. - **Matched angles**, all referring to the same scattering event. Electrons bound in matter carry initial momentum and a binding environment, which broadens the measured wavelengths, and finite detector acceptance averages over several values of $\theta$. These effects do not break conservation; they enlarge the system the account must include. Two angular limits check the formula before any substitution. At $\theta=0$ the cosine is one, so $\lambda'=\lambda$ and the recoil momentum vanishes; at $\theta=180^\circ$ the cosine is minus one, giving the largest shift $2\lambda_C$. A result outside that interval for a nominally free electron signals an angle-convention error, a wavelength-calibration error, or physics beyond the two-body model, and such limit checks catch failures that extra digits would not. Energy and angle measurements constrain different parts of the reconstruction: a spectrometer sets $E'$ or $\lambda'$, a position-sensitive detector sets the angle, and a measured electron adds an independent closure test. The electron energy matches $E-E'$ only after correcting for loss in inactive material, and its direction matches the momentum-component equations only when both angles share a reference axis. Finite resolution blurs the prediction: for a small angle uncertainty $\sigma_\theta$ in radians, the shift formula propagates to $\sigma_{\Delta\lambda}\simeq\lambda_C\sin\theta\,\sigma_\theta$, small near the forward and backward directions where the slope vanishes. Photon-energy resolution, target-electron momentum, and detector acceptance each broaden the peak, so the prediction to compare is a distribution convolved with the response, not a single line. In a coincidence measurement the photon and electron counters must match within a time window: too wide adds accidental pairs, too narrow discards genuine ones, and off-time windows estimate the accidental background. The free-electron model is most reliable when the momentum transfer is large compared with the target electron's initial momentum; at lower transfers the initial motion smears the response. The remedy is to extend the initial-state model, not to abandon four-momentum conservation, which stays exact for the enlarged system while the wavelength formula is the approximation. ## Photoelectric thresholds, stopping voltage, and measurement checks The photoelectric effect is an energy-threshold experiment. Light on a metal releases electrons only when individual photons carry enough energy to overcome the work function $\phi$, the minimum energy to move an electron from the material into the chosen external reference region. The work function belongs to the prepared surface and its environment, not to a named metal in general, so contamination, adsorbed layers, fields, and the reference convention all shift an inferred value. The most energetic released electrons satisfy the photon energy account $$ K_{\rm max}=hf-\phi. $$ The balance is one photon to one emitted electron. The excess becomes at most kinetic energy, and electrons starting below the surface lose part of it before escaping, so the detector records a spread with an upper endpoint set by the least-loss path. The endpoint, not the average collected energy, is the quantity tied directly to photon frequency and work function. $$ % caption: Photoelectric energy accounting at a metal surface. One photon pays the work cost to free an electron, and any surplus appears as its kinetic energy, so h f equals the work function plus K. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \useasboundingbox (-0.20,-0.55) rectangle (6.60,3.10); \fill[black] (2.70,0.50) rectangle (6.15,1.45); \draw[black,thick] (2.70,1.45)--(6.15,1.45); \node[black] at (4.42,0.95) {metal}; \draw[->,black,very thick] (0.55,2.10)--(3.35,1.52); \node[above,black] at (1.75,1.98) {incoming photon}; \draw[->,acc,very thick] (3.50,1.52)--(5.15,2.70); \node[black] at (3.95,2.72) {emitted electron, K}; \node[below,black] at (3.40,0.42) {$hf$ = work function + K}; \end{tikzpicture} $$ The threshold frequency follows by setting the maximum kinetic energy to zero: $$ f_0=\frac{\phi}{h},\qquad \lambda_0=\frac{hc}{\phi}. $$ Photons with $f=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \useasboundingbox (-0.20,-0.35) rectangle (6.60,3.10); \draw[->,black] (0.70,0.50)--(6.05,0.50); \node[below,black] at (5.15,0.50) {photon frequency}; \draw[->,black] (0.70,0.50)--(0.70,2.85); \node[rotate=90,black] at (0.20,1.65) {stopping voltage}; \draw[acc,very thick] (1.55,0.50)--(5.35,2.45); \draw[fill=black,draw=black,thick] (2.35,0.912) circle (0.06); \draw[fill=black,draw=black,thick] (3.35,1.425) circle (0.06); \draw[fill=black,draw=black,thick] (4.35,1.938) circle (0.06); \draw[dashed,black] (1.55,0.50)--(1.55,0.22); \node[below,black] at (1.55,0.22) {threshold}; \node[black] at (3.20,2.35) {slope $\frac{h}{e}$}; \end{tikzpicture} $$ Intensity and frequency control different observables. At fixed frequency above threshold, raising intensity raises the photon rate and, with geometry and detector response unchanged, the collected electrons per second; it leaves $K_{\rm max}$ and $V_s$ untouched, since neither depends on photon number. At fixed optical power, raising frequency raises the energy per photon and the endpoint while lowering the photon number. A claim about "brighter light" is incomplete until intensity, photon rate, and frequency are named separately. > **Worked example (work function 2.20 eV).** A surface with $\phi=2.20\ \mathrm{eV}$ > receives $400\ \mathrm{nm}$ light. The photon energy is $1240/400=3.10\ \mathrm{eV}$, > so the ideal endpoint is > $$ > K_{\rm max}=hf-\phi=3.10-2.20=0.90\ \mathrm{eV}. > $$ > The stopping voltage is $V_s=0.90\ \mathrm V$, since one electronvolt per electron > corresponds to one volt of retarding potential. A $620\ \mathrm{nm}$ photon carries > only $2.00\ \mathrm{eV}$, below $\phi$, so it is subthreshold. Keep the wavelength in > vacuum and $\phi$ in the same energy unit; mixing joules with eV would corrupt the > endpoint. This endpoint separates the photon-energy calculation from the detector response used to observe it. Each quantity has a direct conversion and a distinct calibration requirement. | Quantity | Relation | Calibration dependence | | --- | --- | --- | | Photon energy | $E_\gamma=hc/\lambda$ | wavelength scale | | Threshold condition | $E_\gamma=\phi$ | emitting-surface condition | | Electron endpoint | $K_{\rm max}=E_\gamma-\phi$ | energy or retarding-voltage scale | | Stopping voltage | $eV_s=K_{\rm max}$ in magnitude | contact-potential offset | A threshold determination needs more than a plotted line. The frequency scale needs calibration and an uncertainty; a real beam has finite spectral width that broadens the endpoint; the applied voltage must be known at the surfaces, since contact potentials offset a distant supply reading; dark counts and stray light set a background floor; and space charge distorts collection at high electron rate. Running several intensities tests whether the endpoint stays fixed while the rate changes. The method therefore needs controls that vary rate, energy scale, and surface state separately, each with a distinct expected response under the photon model. | Controlled change | Endpoint expectation | Rate expectation | | --- | --- | --- | | Intensity at fixed frequency | unchanged within uncertainty | changes with photon flux | | Frequency at matched geometry | follows $eV_s=hf-\phi$ | may change because photon flux changes | | Surface preparation | possible shift in $\phi$ | record drift and repeatability | | Retarding-voltage calibration | fixed physical endpoint | measured voltage scale shifts if offset | The linear fit carries two measurements at once: the slope estimates $h/e$, and the frequency-axis crossing estimates the threshold and hence the work function. A voltage offset shifts the intercept strongly while barely moving the slope, whereas a frequency-scale error changes both, so the measurements must span enough photon energies above threshold to separate the two. A narrow range gives a straight-looking graph with a poorly constrained crossing, and the slope-intercept covariance belongs in the derived work-function uncertainty. Endpoint extraction needs a stated model, since the collector current approaches the background floor over a voltage interval; fitting the falling edge to a background-corrected zero and fitting an analyzed energy distribution are both valid when their response functions are given, while reading the first apparent zero on a meter adds an observer bias that cannot be propagated. Surface condition is not a minor detail. The work function refers to the actual emitting surface at the time of measurement, which can differ from a catalog value after air, heating, cleaning, or adsorption, and a drifting surface moves the threshold across a scan and adds scatter to the stopping-voltage graph. Interleaving reference-frequency points and recording vacuum, temperature, illumination history, and cleaning separates a real frequency trend from surface drift, so the result should be reported as a condition-specific measurement, not an immutable property. The measurement is best organized by the observable whose role is being tested, separating the photon-energy relation from controls that only change the number of collected electrons. | Measured response | Controlling relation | Experimental control | | --- | --- | --- | | Endpoint kinetic energy | $K_{\rm max}=hf-\phi$ | frequency or wavelength calibration | | Stopping voltage | $eV_s=K_{\rm max}$ in magnitude | contact-potential and polarity check | | Count rate | photon flux and collection efficiency | intensity scan at fixed frequency | | Work-function inference | frequency-axis intercept | surface preparation and fit covariance | Uncertainty propagation starts from the measured quantities. A fractional wavelength uncertainty gives an equal fractional energy uncertainty of opposite sign, and the work-function uncertainty combines the endpoint voltage, wavelength scale, fit covariance, and repeatability across surface preparations. Round only after the fit and unit conversion are complete: a threshold quoted with more digits than the calibration supports is a formatting choice, not physical precision.