Assume independent arrivals at constant rate $r=2000{\mathrm{s}}^{-1}$, so the count follows a Poisson distribution. In $1\mathrm{ms}$ its mean is $\mu =2$. ThereforeOver observation time $T$, the rate estimate is $\hat{r}=N/T$ and its standard deviation is $\sqrt{r/T}$. Thus the fractional one-standard-error uncertainty is $1/\sqrt{rT}$. To make this strictly less than ten per cent requiresThe mean current magnitude is $er\simeq 3.20\times 10^{-16}\mathrm{A}$. The accuracy statement needs a confidence convention: the boxed time uses one standard error, not a guaranteed bound on every realization. A Gaussian 95-percent relative interval would instead require approximately $(1.96/0.1{)}^2$ events, or $T\gtrsim 0.192\mathrm{s}$. Dead time and correlated arrivals would need another count model; the mean rate alone does not establish Poisson statistics.

An element at $a(\cos \varphi ,\sin \varphi ,0)$ carries charge $dq={\lambda }_{0}a\sin \varphi d\varphi$. Its field at the origin points opposite its radial position. By Coulomb's law,ConsequentlyFor $q{\lambda }_0>0$ the particle starts along $-y$; for the opposite sign it starts along $+y$. Reflection symmetries make the $y$ axis invariant for motion released there from rest. The ring has zero net charge but a nonzero dipole moment, so zero total charge does not imply zero field at its centre.

Let $x$ be the cylinder centre's small displacement away from the wall. Rolling without slipping gives rotation angle of magnitude $x/r$. The material point initially at the top moves horizontally by another $x$ relative to the centre to first order, so the spring extension is $2x$, not $x$. Its small vertical displacement affects the extension only at higher order.

The kinetic and potential energies areThe linearized equation is $(3m/2)\ddot{x}+4kx=0$, givingStatic friction enforces the rolling constraint and does no work at the instantaneous contact point. Omitting rotational kinetic energy, or attaching the spring to the centre instead of the top material point, would change the result.

For total relativistic energy $E=\gamma m{c}^2$, the speed is $v=c\sqrt{1-(m{c}^2/E{)}^2}$. Both beam energies are enormously greater than the rest energy, so expanding the square root avoids obscuring the small difference:With all energies in GeV, $m{c}^2=0.9383$, $E_1=4000$ and $E_2=6500$, givingThe deficits below the speed of light are about $8.25\mathrm{m}{\mathrm{s}}^{-1}$ and $3.12\mathrm{m}{\mathrm{s}}^{-1}$. The large energy increase therefore produces only a small speed increase; the energy mainly changes the Lorentz factor. These numbers use the quoted beam energies as total energies. Interpreting them as kinetic energies requires adding the rest energy, with only a tiny correction at these scales.

A spin-half gas has two spin states. Filling the Fermi sphere at zero temperature gives $n=k_F^3/(3{\pi }^2)$, soWith $m_3\simeq 5.008\times 10^{-27}\mathrm{kg}$, the number density is $1.62\times 10^{28}{\mathrm{m}}^{-3}$ and $k_F\simeq 7.82\times 10^9{\mathrm{m}}^{-1}$. ThusThe associated Fermi temperature isFermionic degeneracy matters on cooling to a few kelvin or below in this model, and is pronounced for $T\ll T_F$. Liquid helium-3 is interacting; its quasiparticle effective mass and thermodynamics need not equal the bare-mass free-gas predictions. The requested calculation is the ideal Fermi-gas scale.

For a homogeneous ideal gas in a two-dimensional box of area $A$, with dispersion $ϵ=p^2/(2m)$ and one internal state, the density of states is $g(ϵ)=Am/(2\pi {\hslash }^2)$. The excited-particle density at temperature $T>0$ isAs $\mu \to 0^{-}$, this diverges. Hence at every fixed positive temperature the excited states can accommodate any finite areal density with $\mu <0$; they never saturate at a finite critical density. There is no finite-temperature Bose-Einstein condensation in the homogeneous two-dimensional ideal gas thermodynamic limit.

The low-energy logarithmic divergence is the reason, not merely the small number of spatial dimensions. A finite box has a low-energy cutoff and a crossover; a trap changes the density of states and can change the conclusion. Condensation at zero temperature and an interacting gas's Berezinskii–Kosterlitz–Thouless transition are separate issues. The proof assumes the uniform quadratic-dispersion model, rather than every conceivable two-dimensional trapping potential.

In the sudden approximation, the nuclear charge changes before the electronic wavefunction has time to evolve. Immediately afterward the state is still the original hydrogenic 1s wavefunction, now expanded in the eigenstates of the new Coulomb Hamiltonian. The final-ground-state probability is the squared overlap, rather than one minus an energy change.

For initial and final charges $Z_1,Z_2$, the normalized 1s overlap isUsing $Z_1=1,Z_2=2$ givesThe remaining probability includes higher bound $ns$ states and the $s$-wave ionization continuum. In this central, instantaneous model the initial orbital has $l=0,m=0$, and angular orthogonality excludes states with $l\ne 0$. The electronic spin is unchanged. Nuclear recoil and interactions with the emitted decay particles are outside this leading sudden-charge-change approximation.

The hard wall imposes $\psi (0)=0$, while the increasing linear potential requires decay as $x\to \infty$. SetThe stationary Schrödinger equation becomes $d^2\psi /d{\xi }^2-\xi \psi =0$. The growing Airy function solution is excluded, leavingThis is the quantum bouncer. For the first four states, $z_n\simeq 2.3381,4.0879,5.5206,6.7867$. Their interior node counts are respectively zero, one, two and three. Each wavefunction starts at zero at the wall, oscillates in the classically allowed interval, and crosses smoothly into an exponentially decaying tail beyond the turning point $x_n=E_n/\beta =z_n\ell$. The turning point is not an extra hard wall: the wavefunction generally remains nonzero there and beyond it.

Normalize the central slit amplitude to one. The two outer amplitudes are $1/2$ each, not $1/\sqrt{2}$, because the specified attenuation concerns amplitude. For centre spacing $d=5\lambda$, adjacent slits have phase difference $\delta =(2\pi d/\lambda )\sin \theta =10\pi \sin \theta$. The far-field amplitude and intensity are thereforewhere $I_0$ is the central slit's intensity alone in the same narrow-slit approximation. Principal maxima have $\sin \theta =m/5$ and height $4I_0$; between them, zeros occur at $\sin \theta =(2m+1)/10$. Only values in $[-1,1]$ are physical. The endpoints are grazing-direction limits of this ideal angular model.

The amplitude weights form a binomial pattern. They remove the secondary maxima that an equal-amplitude three-slit array would have between its principal peaks. No phase shift is needed: changing the coherent weights already changes the interference structure.

For an infinite cubic potential well of side $L$, the one-particle energy eigenvalues are $E_{n_x{n}_y{n}_z}=E_0(n_x^2+n_y^2+n_z^2)$, where $E_0={\pi }^2{\hslash }^2/(2m{L}^2)$ and each quantum number is a positive integer. The lowest spatial orbital is $(1,1,1)$, with energy $3E_0$ and orbital degeneracy one. The next energy is $6E_0$, with three orbitals given by permutations of $(2,1,1)$.

Spin-zero identical particles are bosons. Both can occupy the lowest orbital, so the two-boson ground state is nondegenerate. The first excited total energy is $9E_0$: one particle occupies the lowest orbital and one an excited orbital. Symmetrizing the two-particle wavefunction gives one state for each of the three excited orbitals, so the first-excited degeneracy is three.

Spin-half identical particles are fermions, so the total state must be antisymmetric. Two particles in the same lowest orbital have a symmetric spatial wavefunction and must form the unique spin singlet state. Thus the fermion ground state is also nondegenerate. At the first excited energy, each choice of excited spatial orbital permits a symmetric spatial state with a singlet state, or an antisymmetric spatial state with a three-dimensional triplet state. ThereforeThe spin multiplicity is essential: counting only the three orbital choices would omit allowed triplet states. The result assumes ordinary spin-statistics, an unsplit spin degeneracy and the hard-wall cube model.

Treat the transparent plate as a phase plate at normal incidence. Its extra optical path is $(n-1)50\lambda$. An unchanged interference pattern requires an integer number of full phase cycles:For full-cycle phase-plate invisibility in an ordinary nontrivial material with $n>1$, the two smallest are $n=1.02$ and $1.04$. If the optically null case $n=1$ is admitted, the first two nondecreasing-$n\ge 1$ values are $1$ and $1.02$. A different allowed range of refractive indices would need to be specified before calling these the absolute smallest possibilities.

At the centre, equal illumination gives slit amplitudes proportional to widths, so $A_2=2A_1$. The local interference extrema are $(A_1+A_2{)}^2$ and $(A_2-A_1{)}^2$, givingThese are the central-envelope extrema as relative phase varies, or equivalently the near-centre fringe limits where the envelope is nearly constant.

A slit of width $w$ has its first zeros at $\sin \theta =\pm \lambda /w$. Thus the full central-envelope widths areThe narrower envelope determines the overlap, so it spansThis is the usual approximate “50 fringes” count. If counting nominal bright fringe centres strictly inside both first zeros, they have indices $-24,\dots ,24$, giving 49 centres. The endpoints $\pm 25$ lie at the wider slit's first zeros, where two-slit visibility vanishes; they should not be counted as fully interfering bright centres.

The relative slit-centre phase is $2\pi d\sin \theta /\lambda$. Near the optical axis, $x\simeq L\theta$, so adjacent interference maxima have spacingSlow diffraction-envelope variation can slightly shift the exact local maxima.

For proper-time acceleration by reflected laser light, the spacecraft receives momentum $(hf+h{f}_r)/c=2hf/[c(1+\beta )]$ per incident photon. Combining with the impact rate givesBecause $dt/d\tau =\gamma$ and $d(\gamma \beta )/d\beta ={\gamma }^3$,Integrating from rest gives $1/(1-\beta )-1=\tau /{\tau }_0$, henceThe mirror's rest mass is assumed constant and the laser illuminates it continuously; diffraction losses, incomplete interception and microscopic recoil are neglected.

A photon reaching spacecraft position $x(t)$ at Earth time $t$ was emitted at $t_e=t-x(t)/c$. Differentiating gives $d{t}_e/dt=1-\beta$. Since $Rd{t}_e$ photons arrive in $dt$,The emission rate and impact rate refer to different events, even though both are expressed in Earth coordinate time.

In the instantaneous mirror frame the incident frequency is $f_m=\gamma (1-\beta )f$. Ideal reflection reverses propagation without changing that frame's frequency. Transforming the backward photon to Earth gives another factor $\gamma (1-\beta )$:This uses a classical mirror with negligible recoil per individual photon.

The first dipole remains perpendicular to the separation and is now also perpendicular to the second dipole. Both terms in the point-dipole interaction vanish, giving $U=0$. This is the electrostatic point-dipole contribution, not a claim that real finite water molecules have no other intermolecular interaction.

The point-dipole interaction isHere ${\mathbf{p}}_1\cdot \hat{\mathbf{r}}=0$. For parallel axes, ${\mathbf{p}}_1\cdot {\mathbf{p}}_2=p^2$, henceThis side-by-side parallel arrangement is repulsive in the point-dipole model.

Substitution of a plane wave in the Schrödinger equation gives $\hslash \omega ={\hslash }^2{k}^2/(2m)+V_0$. For a stationary step-scattering state, remove the common phase $e^{-iEt/\hslash }$ and write ${\psi }_L=e^{ikx}+r{e}^{-ikx}$, $k=\sqrt{2mE}/\hslash$. If $E>V_0$, the right-side outgoing wave is ${\psi }_R=t{e}^{iqx}$, $q=\sqrt{2m(E-V_0)}/\hslash$. A finite step with no delta-function potential in the Schrödinger operator requires continuous $\psi$ and ${\psi }^{\prime }$: a discontinuous wavefunction would generate a delta derivative, and integrating the equation over a vanishing interval gives derivative continuity. Matching yieldsFor $0<E<V_0$, take $q=i\kappa$ and the bounded right-side state $t{e}^{-\kappa x}$. It carries no transmitted flux and $\mathcal{R}=1$. The threshold limit also has reflection probability one. Thus at $V_0=1\mathrm{eV}$,For the two-step profile, set $k_0=\sqrt{2mE}/\hslash$, $k_1=\sqrt{2m(E-V_1)}/\hslash$ and $k_2=\sqrt{2m(E-V_0)}/\hslash$. With all regions propagating, the net reflection amplitude isThe given quarter-wave condition makes $e^{2i{k}_{1}a}=-1$. Zero reflection then requires $r_{01}=r_{12}$, equivalent to $k_1^2=k_0{k}_2$. ConsequentlyFor an increasing intermediate step this needs $E>V_0$ and $V_0/2<V_1<V_0$, together with the stipulated thickness condition. In this quarter-wave matching layer for a quantum potential step, two reflected amplitudes cancel while the transmitted flux equals the incident flux.