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Wireless telegraphy conveys discrete symbols without a connecting message wire. Here the mechanism is radio communication: keyed excitation produces radio waves, a receiving antenna drives a selective electrical circuit, and a detector produces a readable event. This makes a beautiful bridge between resonance and information: accumulating many appropriately timed small electrical disturbances can distinguish one transmission from another. Selectivity is finite, however; resonance alone cannot guarantee immunity to strong unwanted signals or noise.

The 1909 Nobel Prize in Physics recognized Guglielmo Marconi and Karl Ferdinand Braun for contributions to wireless telegraphy.^[ref] Their apparatus should be understood as a developing collection of systems. An early coherer receiver, a later tuned receiver and a modern envelope detector are different instruments.

A WKB quantization condition selects approximate bound-state energies by requiring an oscillatory wavefunction to match both endpoint conditions. For a smooth one-dimensional well bounded by two separated simple classical turning points $a(E),b(E)$,Each soft turning point contributes an Airy turning-point connection formula phase. A hard Dirichlet wall and one soft turning point instead give $(n+3/4)\pi$, as in hard-wall and soft-turning-point WKB quantization. Two regular Dirichlet endpoints with no turning point give an integer multiple of $\pi$. The endpoint type is part of the formula.

For the quantum harmonic oscillator, the two-turning-point action is $\pi E/\omega$. The condition gives $E_n=\hslash \omega (n+1/2)$, which happens to agree with the exact spectrum. In a general potential it is a leading semiclassical estimate, and nearby or degenerate turning points can invalidate independent connection formulas.

Work measures the energy transferred by a force as its point of application moves. For a point mass following a path $C$, add the contribution of the force along each small displacement:This line integral uses the actual path, and the dot product selects the component of force along the motion. A large force perpendicular to every small displacement does zero work; holding something still also does zero mechanical work on it, even if holding it feels tiring. The SI unit is $\mathrm{J}=\mathrm{N}\mathrm{m}$. ^[ref]

The usefulness of work is that it connects a description in terms of forces and distances to one in terms of energy. The work-energy theorem gives the change in kinetic energy from the total net work, often without solving for the entire motion. The work of one selected force equals that change only when the other forces contribute zero total work.

For a fixed force $\mathbf{F}$ and endpoint displacement $\Delta \mathbf{r}$, the line integral reduces to a dot product:Here $\theta$ is the angle between the force and the displacement. For straight motion, $d$ is also the distance traveled. A $3\mathrm{N}$ force acting along a $4\mathrm{m}$ displacement does $12\mathrm{J}$ of work. Turning the same force perpendicular to the displacement gives $\cos (\pi /2)=0$ and zero work; reversing it gives $\cos \pi =-1$ and $-12\mathrm{J}$. At $\theta =\pi /3$, the work is $6\mathrm{J}$. These signs describe energy transferred to or from the moving body. ^[ref]

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]

The work function is the minimum energy needed to remove an electron from a solid to the nearby vacuum level. A junction's effective barrier need not equal either electrode's isolated work function.

The drive can supply or extract work while the quantum state stays approximately fixed. For a differentiable bounded Hamiltonian operator, the Schrödinger equation givesThe state-evolution terms cancel, leaving the external drive's power. For a finite ideal step the change in mean energy is $⟨\psi |H_f-H_i|\psi ⟩$. Thus conservation of energy for a closed system with a time-independent Hamiltonian operator does not assert conservation of the driven subsystem's energy during switching.

Specify work operationally by two energy measurements: one before the switch and one after. For the prepared initial ground state, the first outcome is definite; the second samples the new Born rule probabilities. This protocol gives a distribution of measured energy differences, rather than generally the measurement distribution of the single operator $H_f-H_i$.^[ref]and henceFor ${\Delta }_i={\Delta }_f=\Delta$ and $\theta =\pi /3$, the work is $0$ or $\Delta$, with probabilities $3/4$ and $1/4$; its expectation is $\Delta /4$ and its variance is $3{\Delta }^2/16$. The initial mean energy $-\Delta /2$ becomes $-\Delta /4$. Using different initial and final scalar offsets would also shift the measured work; the displayed Hamiltonian operators fix that convention explicitly.

For attached-state elastic energy $U=\kappa (x-x_0{)}^2/2$ and a binding window much narrower than $x_0$, the expected assisting work before $x_0$ is $\kappa p_b{x}_0^2/2$. The detachment tail produces resisting work $-\kappa p_b{v}^2/k_{off}^2$. Mean force per head is $W/D$ for site spacing $D$. The detachment-time second moment, not the square of its mean, fixes the resisting term. These are idealized prescribed-velocity kinetics, not a complete muscle energy budget.

The fixed-volume equilibrium crystal shape obtained by minimizing orientation-dependent surface free energy. Facet support-plane distances are proportional to their interfacial free energies under the construction assumptions.

Researcher in photosynthesis. Li and collaborators investigated a pigment-binding protein required for regulation of light harvesting, connecting a molecular component with the safe use of absorbed light energy.^[ref]

For a narrow beam of monoenergetic photons crossing homogeneous material of thickness $x$, the uncollided wave intensity obeys $I(x)=I_0{e}^{-\mu x}$. Here $\mu$ is the linear attenuation coefficient, counting removal from the incident beam by photon absorption and scattering of light. It depends on photon energy and material; it is not universally independent of energy.^[ref]

An original derivation makes the model explicit. If each thin layer removes fraction $\mu dx$ of the surviving beam, with the same $\mu$ throughout, thenThus equal added thicknesses multiply transmission by the same factor. The ordinary differential equation describes uncollided beam survival. To test it with a detector, keep source output, distance, selected photon energy band and detector response fixed. Use a collimator and suitable detector acceptance to exclude scattering of light into the detector. Subtract background, measure an empty-path reference, and keep the detector in its linear operating range. A changing source distance adds geometric dilution and is not an attenuation measurement at fixed geometry.

For a stipulated example, take $\mu =0.40\mathrm{m}{\mathrm{m}}^{-1}$ and $x=2.5\mathrm{mm}$. Transmission is $e^{-1}\simeq 0.368$. If the empty-path detector rate is $1010{\mathrm{s}}^{-1}$, including background $b=10{\mathrm{s}}^{-1}$, the model predicts sample rate $10+1000e^{-1}\simeq 378{\mathrm{s}}^{-1}$. Recover transmission as $(R-b)/(R_0-b)$; the raw ratio $R/R_0\simeq 0.374$ is biased. These chosen values are a calculation, not measurements of a particular material or a historical apparatus.

An X-ray binary has a compact mass-gaining star or black hole accreting from a companion, so that much of its emitted power appears as X-rays. Gravitational binding energy supplies an accretion luminosity of order $GM\dot{M}/R$ when there is a surface or an appropriate inner disk radius. The donor star may feed an accretion disk through Roche-lobe overflow or a stellar wind.

X-ray effective area is the area that converts an incident photon arrival rate per unit area into an accepted count rate. For a fixed direction and a sufficiently narrow energy interval, writeHere $A_{\mathrm{geom}}$ is projected geometric area, $T$ is transmittance through windows or optics, $\eta$ is quantum efficiency, ${ϵ}_s$ is the probability that a detected source event passes event selection, and $t$ is live observing time. $N_s$ is an expected value, not a guarantee of an integer count. Multiplication follows by counting incident photons over area and time, then applying the successive conditional acceptance probabilities. Live time already excludes intervals when the instrument cannot record events; multiplying by a live-time fraction again would count the same loss twice.

For an invented narrow-band example, choose $A_{\mathrm{geom}}=40\mathrm{c}{\mathrm{m}}^2$, $T=0.8$, $\eta =0.75$, and ${ϵ}_s=5/6$. Then $A_{\mathrm{eff}}=20\mathrm{c}{\mathrm{m}}^2$. A source with $F=0.30\mathrm{photons}\mathrm{c}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$ observed for $t=100\mathrm{s}$ gives $N_s=600$. $F$ here counts photons; converting it into observed radiative flux requires their energies. If all have energy $E_0$, the corresponding radiative flux is $E_{0}F$.

A broad spectrum instead requireswhere $F_E$ is the differential photon arrival rate per area and energy. A single band-averaged X-ray effective area depends on the assumed spectrum. This is why a count rate alone cannot give an assumption-free radiative flux.

X-ray emission spectroscopy studies the frequency spectrum of emitted X-rays. The spectral line positions identify electronic energy differences; their strengths also depend on excitation, photon absorption inside the sample, geometry and detector response. An X-ray tube may excite a sample with electrons, while X-ray fluorescence uses incident photons. The distinction matters when choosing what the experiment measures: identifying a chemical element from a line position is different from determining its abundance from a line strength.

X-ray fluorescence is Characteristic X-ray emission following excitation by incident photons. In a simple initially unexcited chemical atom, photoionization can eject an inner electron if the incident photon supplies its binding energy. An outer electron can then fill the vacancy, releasing an emitted photon with the difference of shell binding energies. Nonradiative relaxation competes with this route, so one absorbed incident photon need not yield one escaping characteristic photon.

Choose a separate stipulated electron shell model with values $B_K=10.0$, $B_L=1.5$ and $B_M=0.3$, all in $\mathrm{keV}$. Ignore subshell splitting, recoil and changes of binding energy during relaxation. The possible shell differences areThese labels denote simplified shell groups, not resolved subshell electronic transitions. With $hc=1.23984\mathrm{keV}\mathrm{nm}$, their calculated wavelengths are respectively $0.145864$, $0.127819$ and $1.03320\mathrm{nm}$. An incident $9.8\mathrm{keV}$ photon exceeds both K-line energies but cannot directly create the K vacancy in this initially unexcited model. A $12.0\mathrm{keV}$ photon can, leaving approximately $2.0\mathrm{keV}$ of photoelectron kinetic energy. This comparison fixes neither the excitation probability nor the radiative branching probability.

X-ray photoelectron spectroscopy uses X-ray excitation to investigate electron removal energies, including those of core electrons. The common name electron spectroscopy for chemical analysis emphasizes its use in distinguishing chemical surroundings. The whole recorded spectrum can also contain electrons produced by processes other than direct photoionization; a peak assignment must identify the emission process as well as the chemical element.

In a modern implementation, a quartz crystal selects aluminum characteristic X-rays according to Bragg's law and focuses them onto the sample. Vacuum reduces gas collisions and contamination; a separately pumped entry chamber limits exposure to air. Lenses collect photoelectrons; an electrostatic electron energy analyzer transmits a narrow kinetic energy band. Stepping that band and counting detector pulses produces the spectrum. An electron flood source can compensate positive electric charge on insulating samples. Ion cleaning can alter vulnerable samples.^[ref]

For a scan with count $N_j$ and dwell time $t_j$ at setting $j$, compare rates $N_j/t_j$, rather than raw counts when dwell times differ. Under an independent Poisson distribution model, the count standard deviation is approximately $\sqrt{N_j}$ for a sufficiently populated bin. Longer acquisition reduces this relative counting measurement uncertainty; it does not automatically improve separation of overlapping peaks. Photoelectron peak overlap and resolution gives an example in which the same chemical separation produces either two maxima or one.

X-ray scattering is the redirection of X-rays through interaction with matter. The angular distribution and energy of the outgoing radiation carry information about the material. In elastic scattering the photon energy is unchanged; in inelastic scattering energy is exchanged. X-ray diffraction concerns organized interference of scattered waves, rather than a separate kind of radiation.

The scattering vector $\mathbf{q}={\mathbf{k}}_{\mathrm{out}}-{\mathbf{k}}_{\mathrm{in}}$ compares outgoing and incoming wave vectors. Consider two identical scatterers separated by $\mathbf{d}$, each with real elastic scattering amplitude $f$, and neglect multiple scattering. Their amplitude sum and intensity areTwo atoms can therefore scatter strongly in one direction and cancel in another. The measured pattern depends on arrangement, not merely on the number of atoms. For repeating lattice planes in a crystal, constructive interference gives Bragg's law $2d\sin \theta =n\lambda$.^[ref]

The1913 experiment by William Henry Bragg and William Lawrence Bragg made the angular pattern measurable with a rotating crystal and an ionization chamber in place of a spectrometer telescope. Slits selected a narrow incoming beam; the detector current was measured as its angle changed. Peaks and repeated orders constrained crystal structure and X-ray wavelengths, linking an electrical measurement to atomic-scale geometry.^[ref]

X-ray scattering techniques use the change in a scattered photon's direction and, in inelastic measurements, its energy to probe matter. Inelastic X-ray scattering transfers energy to excitations such as phonons; elastic methods probe spatial structure without that energy transfer.

For a fixed homogeneous absorber in a narrow-beam geometry, define the X-ray spectral attenuation operator on L1 space over a bounded positive energy interval. A physical input $f(E)\ge 0$ is an integrable function giving incident photon rate density. Let its linear attenuation coefficient $\mu (E)\ge 0$ be a measurable function, with $x\ge 0$. The transmitted uncollided density isHere $w(E)\ge 0$ is a fixed bounded detector response described by a measurable function; background has been subtracted. This is a positive linear operator on L1 space. Since $0\le e^{-\mu (E)x}\le 1$, it cannot increase the integrated absolute incident rate. Multiplication of exponentials proves $A_{x+y}=A_x{A}_y$ for successive layers of the same material. Ignore secondary radiation entering the detector, saturation and changes to the source. Discrete spectral lines use the finite-sum version below instead of an ordinary rate density.

For a chosen filter, stipulate ${\mu }_{\alpha }=0.20\mathrm{m}{\mathrm{m}}^{-1}$, ${\mu }_{\beta }=0.80\mathrm{m}{\mathrm{m}}^{-1}$ and input line rates $(N_{\alpha },N_{\beta })=(4,1)$ in a common arbitrary rate unit. Assume equal calibrated detector weight for both lines. At $x=2.0\mathrm{mm}$,The selected $8.5\mathrm{keV}$ line's fraction rises from $4/5=0.800$ to $0.929974$, but its absolute rate falls by about $33.0\%$. Filtering improves purity at the cost of signal. It multiplies existing components and cannot create a new emitted spectral line.

With just one integrated reading, $S=N_{\alpha }+N_{\beta }$, the two unknown line rates are underdetermined. If the source is unchanged between an empty-path reading and a filtered reading, the ideal inverse problem becomesThe determinant $e^{-1.6}-e^{-0.4}\simeq -0.468424$ is nonzero, so these two ideal readings determine the two stipulated line rates uniquely. Identical coefficients would give zero determinant and destroy that separation. This small linear algebra example explains why absorber contrast supplies information, while also showing its limit: unknown scattered components or unknown detector weights add unknowns that two readings cannot determine.

An X-ray tube accelerates electrons from a negative cathode to a positive target anode. A heated cathode supplies electrons by thermionic emission; the evacuated envelope limits collisions before the target. This differs from the gas-discharge tube in the Röntgen X-ray discovery experiment.^[ref]

In bremsstrahlung, an electron changes velocity in the target's electric fields, emitting a photon while losing kinetic energy. Different encounters allow a continuous range of transferred energies. This is distinct from the discrete electronic transitions producing Characteristic X-rays.

Changing the accelerating voltage changes the available kinetic energy per electron and the X-ray tube cutoff. Changing the number of arriving electrons changes the number of opportunities for photon emission. These are distinct controls: a larger electric current does not by itself give each electron more energy. For an original idealized budget, $V=50\mathrm{kV}$ and tube electric current $i=0.20\mathrm{mA}$ give electrical power $Vi=10\mathrm{W}$ and arrival rate $i/e\simeq 1.25\times 10^{15}{\mathrm{s}}^{-1}$, where $e$ is the elementary charge. This is input power, not emitted X-ray power; obtaining the latter requires a conversion efficiency.