A monochromatic wave in vacuum has frequency $\nu$ and wavelength $\lambda$ related by $\nu \lambda =c$, where $c$ is the speed of light. During one oscillation period $T=1/\nu$, a constant-wave phase surface travels one wavelength, so $\lambda =cT$. Equivalently, the plane electromagnetic wave relation $\omega =ck$ with angular frequency $\omega =2\pi \nu$ and wavenumber $k=2\pi /\lambda$ gives the same result.

As a calculated example, a vacuum radio wave with $\nu =300\mathrm{MHz}=3.00\times 10^8\mathrm{Hz}$ has $\lambda =0.999308\mathrm{m}$ using $c=299792458m/s$. Doubling frequency halves wavelength, independently of wave amplitude and wave intensity. The frequency spectrum of a pulse contains multiple frequencies; the relation applies to each vacuum component, rather than assigning one exact wavelength to the whole pulse. In a homogeneous medium replace $c$ by the appropriate phase velocity; the vacuum equation does not assert that all media have the same wavelength at a given frequency.

A radio wave is electromagnetic radiation in the radio portion of the frequency spectrum. In vacuum its frequency and wavelength obey the vacuum frequency-wavelength relation.

An electromagnetic field carries momentum as well as energy. In vacuum, its SI momentum density iswhere $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$ is the Poynting vector. Thus the field momentum in a region is ${\mathbf{P}}_{\mathrm{field}}=\int \mathbf{g}dV$. ^[ref]

For a directed vacuum plane wave, $\mathbf{S}=cu\hat{\mathbf{n}}$, so a pulse with total energy $U$ carries momentum $U\hat{\mathbf{n}}/c$. Complete absorption transfers that momentum to matter. Reversing the pulse at a stationary perfect mirror transfers twice as much normal momentum, explaining radiation pressure.

Conservation joins field momentum to the mechanical momentum of matter and momentum crossing the chosen boundary. Changing a field can therefore accelerate matter even when no material particle carries the momentum across the intervening space. Static configurations can also contain nonzero field momentum; the vacuum formula does not require radiation. ^[ref]

An induction coil uses a changing electric current and Faraday's law of induction to produce a high output electric potential difference.

A spark gap separates two electrodes; sufficient electric potential difference can cause an electrical discharge through the intervening gas.

An electronic amplifier uses an electrical power supply to increase a signal's amplitude or available output power. The input controls the transfer of supply energy to the load; it does not supply all the amplified output energy itself.

A voltage amplifier, a transimpedance amplifier and a charge amplifier relate different input and output quantities. To characterize one, vary the input around its operating point and measure gain, frequency response, noise and the range before saturation. A large gain does not by itself improve signal-to-noise ratio, since input noise and amplifier noise can both be amplified.

A diode is a two-terminal component with an asymmetric current-voltage characteristic: reversing its terminal electric potential difference need not reverse its electric current with the same magnitude. Name the terminals diode anode $A$ and diode cathode $K$. Define $V_D={\varphi }_A-{\varphi }_K$ using their electric potentials, and take positive electric current $I_D$ from $A$ to $K$. The bar in the diode symbol marks $K$. Positive $V_D$ is the forward direction; negative $V_D$ is the reverse direction.^[ref]

The asymmetry makes a rectifier possible: a sinusoid can drive electric current through a resistor on one half of each cycle. An ordinary resistor obeying Ohm's law instead reverses its electric current when the applied electric potential difference reverses. The useful idea is directional selection; the ideal diode model captures it, while the Shockley diode equation describes a smooth exponential function rather than an abrupt switch.

A deoxynucleoside triphosphate is a nucleotide containing deoxyribose and three phosphate groups. DNA polymerases use these substrates to extend DNA; the four standard substrate types carry adenine, cytosine, guanine or thymine.

Serial dilution repeatedly transfers a measured portion of one well-mixed chemical solution and dilutes that portion to a new final volume. With transferred volume $v_j$ and final volume $V_j$ at step $j$, conserved solute givesThis product follows by repeated substitution. Each step conserves the solute in the portion transferred into its receiving flask, not the entire inventory in the preceding flask. Uniform mixing makes the selected fraction representative.

Continue the original solution mass and volume bookkeeping example. A $10.00\mathrm{mL}$ portion of the $0.2000\mathrm{mol}{\mathrm{L}}^{-1}$ stock contains $0.002000\mathrm{mol}$ of analytical sodium chloride, corresponding to $0.11688\mathrm{g}$ before rounding. After dilution to $100.0\mathrm{mL}$, its analytical concentration is $0.02000\mathrm{mol}{\mathrm{L}}^{-1}$. A new $5.000\mathrm{mL}$ portion of that mixed chemical solution contains $0.0001000\mathrm{mol}$; dilution to $50.00\mathrm{mL}$ gives $0.002000\mathrm{mol}{\mathrm{L}}^{-1}$. The two factors $10.00/100.0$ and $5.000/50.00$ multiply to $1/100$.

The remaining stock holds $0.05000-0.002000=0.04800\mathrm{mol}$; the remaining first dilution holds $0.002000-0.0001000=0.001900\mathrm{mol}$. Including the final flask restores the starting inventory:This is a direct check that making a dilute final sample does not destroy the solute. Most of it remains in earlier vessels.

Serial dilution makes a small final concentration accessible using measurable portions instead of a single extremely small transfer. Its limitation is that systematic errors propagate: if each transferred volume is $1\%$ below its intended value while final volumes are correct, two steps give $0.99^2=0.9801$ times the intended concentration, a $1.99\%$ deficit. Repeated readings alone do not remove that bias. The gain in manageable volume therefore depends on calibrated transfer and representative mixing.

A diatomic molecule whose two chemical atoms belong to the same chemical element.

Keep the same bond force constant $k$ and approximate the nuclear masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, and ${}^{35}\mathrm{Cl}$ by $1u$, $2u$, and $35u$, respectively. Here $u$ is the unified atomic mass unit, and $\mathrm{D}$ denotes deuterium. These are chosen mass number approximations, not measured atomic masses or natural-abundance averages. With the same chlorine isotope in both molecules,This is the isotope scaling of diatomic vibration prediction. Doubling the light partner's mass nearly doubles the reduced mass, lowering the frequency by about $28.3\%$. The factor differs from $1/\sqrt{2}$ because the chlorine partner also moves.

For the same small initial stretch $A$ and zero velocity, both models start with the same internal mechanical energy $k{A}^2/2$, but the DCl oscillation period is longer by $\sqrt{72/37}\approx 1.394972$. In the centre of mass frame, $u_{\mathrm{H}}=-35q/36$ and $u_{\mathrm{Cl}}=q/36$ for HCl; $u_{\mathrm{D}}=-35q/37$ and $u_{\mathrm{Cl}}=2q/37$ for DCl. These ratios explain which partner moves and are separate from the change in frequency.

The common-$k$ assumption belongs to the leading Born-Oppenheimer approximation: fixed nuclear positions and electric charges determine the electronic potential energy, while nuclear masses govern motion on that curve. Quantitative comparison requires actual atomic masses, departures from that approximation, and separation of anharmonicity and molecular rotation from the measured band. Even with common $k$, the quantum harmonic oscillator predicts a different zero-point energy for each isotope.^[ref]

Quantizing the stretching coordinate gives a quantum harmonic oscillator with reduced mass $\mu$:Here $v$ is the vibrational quantum number and $\hslash =h/(2\pi )$ the reduced Planck constant. Consecutive energy eigenvalues have gap $\hslash \omega$; the ground state has zero-point energy $\hslash \omega /2$ above the well minimum. The offset $V_0$ cancels from every transition energy.

A short proof uses creation and annihilation operators: with $\ell =\sqrt{\hslash /(\mu \omega )}$, write $q=\ell (a+a^{†})/\sqrt{2}$. The Quantum Hamiltonian becomes $\hat{H}-V_0=\hslash \omega (a^{†}a+1/2)$. Nonnegativity of $a^{†}a$ gives the lowest level; repeated application of $a^{†}$ raises its energy by $\hslash \omega$ each time. This also gives $⟨q{⟩}_v=0$ andThe ground state therefore has a nonzero spread of bond lengths, even though its mean stretching displacement vanishes. A stationary energy eigenstate has a time-independent position probability density; the classical oscillation curve is not a trajectory of an atomic nucleus in that state.

The parabolic model formally extends $q$ to the whole real line, whereas an actual bond length satisfies $r>0$. It is useful for low levels concentrated near $r_e$, with typical stretching displacements much smaller than $r_e$. It fails near breakup of the chemical bond and when neglected anharmonicity becomes substantial.

The bond force constant is the curvature of the effective potential energy at the equilibrium bond length. For a diatomic bond stretch, it plays the role of a spring constant. Its definition is local: two wells with the same curvature can have different bond-dissociation energies.

For a Boolean function $f$ of $n$ input bits, define its De Morgan dual by complementing every input, applying $f$, and complementing the output:

De Morgan's laws say that the De Morgan dual of logical conjunction is logical disjunction, and conversely. The constant zero Boolean function becomes constant one. This is different from merely negating the output: for AND, output logical negation gives one at $(0,0)$, whereas the De Morgan dual gives zero there.

Taking the De Morgan dual twice restores the original Boolean function:Here each input is complemented twice as well. For a Boolean formula built from AND, OR, NOT and constants, its De Morgan dual can therefore be obtained by exchanging AND with OR and zero with one throughout, leaving input names and NOT in place. A structural induction proves the rule: it holds for an input and each constant, De Morgan's laws handle the two binary logical connectives, and double logical negation handles NOT. This symmetry lets one translate a complex test systematically instead of guessing where its failures occur.