Substitution in the moment of inertia givesFor independent errors interpreted as standard uncertainties, logarithmic differentiation and propagation of uncertainty giveConsequentlyThis is a $0.515\%$ statistical precision requirement. If “accuracy” instead means hard bounds whose worst cases must be guaranteed, first-order absolute errors add: $0.01+0.01+4{ϵ}_R\le 0.025$, giving $\boxed{{ϵ}_R\le 0.125\%,|\delta R|\le 1.25\mathrm{mm}}$ to first order. For an exact symmetric hard bound about the nominal values, the upper extreme is slightly more restrictive: $(1.01{)}^2(1+{ϵ}_R{)}^4\le 1.025$ gives ${ϵ}_R\le 0.1199\%$, or $1.20\mathrm{mm}$. The question does not distinguish a statistical uncertainty from a worst-case tolerance, so the convention matters.

In the orthonormal $S_z$ eigenbasis $(\alpha ,\beta )$, the spin one-half operator is $S_x=(\hslash /2)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. An eigenvector $a\alpha +b\beta$ with eigenvalue $s_x$ obeys $b=(2s_x/\hslash )a$ and $a=(2s_x/\hslash )b$. A nonzero vector requires $(2s_x/\hslash {)}^2=1$. Choosing normalized quantum phases givesThese spin eigenstates are orthogonal and have unit norm because $\alpha$ and $\beta$ do.

The infinite square well has energy levels $E_n=n^2{h}^2/(8m_e{L}^2)$, with $n=1,2,\dots$. The longest absorption wavelength corresponds to the smallest available energy difference from the ground state, namely $n=1\to 2$. Its position matrix element is nonzero because the two states have opposite parity about the center of the well, so it is an allowed electric dipole transition.

Equating the photon energy to this gap givesHere the photon gap is approximately $1.55\mathrm{eV}$, not the ground-state energy itself.

For a neutral atom, count both the nucleons and the electrons. Protons, neutrons and electrons are all fermions, and exchanging two identical intact atoms produces the sign $(-1{)}^{N_f}$ from exchanging their fermionic constituents. An odd number gives a composite fermion; an even number gives a composite boson, consistently with the Spin-statistics theorem.

A lithium atom has three protons and three electrons. Lithium-6 also has three neutrons, giving $3+3+3=9$ fermionic constituents; lithium-7 has four neutrons, giving $3+4+3=10$. ThereforeThe composite statistics of neutral atoms cannot be inferred from the mass number alone: counting only nucleons would classify the nuclei and give the opposite answers here. The atom's total spin includes its electrons as well as its nucleus.

Circular apertures and circular obstructions illustrate both the usefulness and the limits of geometric shadow arguments. In scalar diffraction theory, each point of a transmitted wavefront contributes a complex secondary field. Their coherent sum, described by the Huygens-Fresnel principle, determines the intensity. Circular symmetry simplifies the integral, but the answer depends on whether propagation is in the Fraunhofer diffraction or Fresnel diffraction regime.

For aperture radius $a$ and axial propagation distance $z$, the Fresnel number $N_F=a^2/(\lambda z)$ measures the importance of quadratic phase across the aperture. The far-field condition is $N_F\ll 1$; a lens can produce the same angular spectrum in its back focal plane. Fresnel propagation retains that quadratic phase. For an axial point source a distance $s$ before the object and an observer a distance $r$ after it, the phase uses the effective distance $z_{\mathrm{eff}}=sr/(s+r)$, because both incident and outgoing spherical paths contribute curvature. Scalar, paraxial propagation and an opening large compared with wavelength are assumed throughout.

A circular hole in the far field gives an Airy pattern. For uniform plane illumination the field is the two-dimensional Fourier transform of the aperture. In polar coordinates, let $Q=k\sin \theta$; the angular integral gives the Bessel function $J_0$:ThusThe limiting central value is one because $J_1(u)\simeq u/2$ near zero. A bright central Airy disk is surrounded by progressively weaker concentric rings. The first dark ring has $ka\sin \theta =3.8317$, or $\sin \theta \simeq 1.22\lambda /D$ with diameter $D=2a$. This scale gives the Rayleigh criterion for a circular optical instrument. Larger apertures narrow the angular pattern; a lens of focal length $f$ maps the first dark ring to radius $1.22\lambda f/D$ in its focal plane.

In the Fresnel regime, the axial intensity oscillates as aperture size or distance changes. Fresnel half-period zones have boundaries at path excesses $m\lambda /2$ and approximately opposite phases in successive zones. For a plane wave, their radii obey $r_m^2=m\lambda z$. A circular opening can therefore expose alternating constructive and cancelling zone contributions. More precisely, Fresnel diffraction by a circular aperture gives, normalized by the unobstructed field,An odd integer number of zones gives a bright axial point with ideal intensity $4I_0$; an even number gives an axial zero. Intensity can locally exceed the unobstructed value because interference redistributes energy; total transmitted power still obeys conservation of energy. The relative normalisation here differs from the Airy expression, which normalises each far-field pattern by its own central intensity.

Off axis, the circular Fresnel field isIts concentric structure evolves with $N_F$ and includes fringes near the geometrical edge. The axial zone argument does not determine the complete off-axis pattern.

Complementary holes and opaque disks are related at the field level. Babinet's principle states $U_a+U_d=U_0$ for complementary aperture and disk transmissions under the same illumination. In the Fraunhofer region away from the direct transmitted beam, the background field is absent and the diffracted fields are opposite in sign. Their scattered intensity profiles therefore have the same Airy angular dependence. Exactly in the forward direction the unobstructed field must be retained; subtracting intensities is incorrect. Finite incident beams also give a finite-width direct beam rather than an ideal angular delta function.

For Fresnel diffraction by an opaque disk, the axial complement is particularly striking:The ideal circular disk has an Arago spot at the centre of its geometric shadow. Circular edge contributions reach the axial point with the same phase. This does not mean the whole shadow is illuminated: off-axis cancellation produces the surrounding shadow and rings, and the axial spot becomes narrower for larger obstruction or shorter propagation distance. An aperture's axial intensity oscillates, while its complementary disk's axial intensity stays constant in this ideal paraxial model.

A practical observation uses a compact coherent source, an accurately circular hole or disk, and an aligned screen or camera. Varying the propagation distance or aperture radius reveals the transition between regimes. A point source, or spatial filtering, prevents different source positions from washing out the fine structure. Finite source size, incomplete coherence, edge roughness, misalignment, vector effects for wavelength-sized objects and lens aberrations reduce the ideal contrast. These experimental limitations explain why a geometrically unexpected axial bright point requires a suitable setup even though its scalar derivation is simple.

For fixed electron number $N$, the specified electron degeneracy pressure scales as $P\propto V^{-5/3}$. Differentiating at fixed low temperature gives the degeneracy-pressure bulk modulus,A monovalent sample has one conduction electron per atom. Its carrier density isusing the Avogadro constant $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$. Substitution with reduced Planck constant $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$ and electron mass $m_e=9.10938\times 10^{-31}\mathrm{kg}$ givesEquivalently, $P=2n{E}_F/5$ and $K_T=2n{E}_F/3$ for this free electron model. This is the assumed electronic contribution, not a calculation of the complete equation of state of real potassium: ionic cohesion and interaction corrections have been excluded as instructed. The supplied pressure is the zero-temperature form, appropriate to leading order for $T\ll T_F$.

In a conventional face-centered cubic lattice cell of side $a_F$, touching nearest-neighbour hard spheres lie along a face diagonal. Four radii span that diagonal, so $\sqrt{2}{a}_F=4r$ and $a_F=2\sqrt{2}r$. In a conventional body-centered cubic lattice cell, the contact lies on the body diagonal: $\sqrt{3}{a}_B=4r$, giving $a_B=4r/\sqrt{3}$. ThereforeThese are the nonprimitive cubic unit cells requested. They contain four and two atoms respectively, so their volume ratio should not be confused with the volume-per-atom ratio, which is half as large, $3\sqrt{6}/8\simeq 0.9186$. The common sphere radius cancels.

The original diagram puts the upper edge of the obstruction exactly on the straight line from source to detector. The detector is therefore at the geometric shadow boundary, not deep inside the shadow. In the optical convention, the Cornu spiral iswith limiting points $(-1/2,-1/2)$ and $(1/2,1/2)$. The unobstructed Fresnel diffraction amplitude is the chord joining both endpoints, proportional to $1+i$. The open half-plane supplies only the chord from $(0,0)$ to $(1/2,1/2)$, proportional to $(1+i)/2$. The coordinate perpendicular to the page contributes the same factor in both cases.This is the half-plane shadow-boundary intensity. The missing half of the aperture removes half the field here, so taking half the intensity would be incorrect.

For an object near the front focal plane, the objective's numerical aperture in air isThe Rayleigh criterion for a circular objective gives object-space diffraction-limited resolutionThe factor $1.22$ comes from the first zero of the Airy disk. Using the exact $\sin (\arctan (D/(2f)))$ changes the result by less than $1\%$. A coarser $\lambda f/D$ estimate gives $2.4\mu \mathrm{m}$; the chosen criterion must be stated. This estimates the aperture limit, before aberrations and detector sampling.

The characteristic impedance of a stretched string is $Z=T/c=\rho c$, since its wave speed obeys $c^2=T/\rho$. Put the endpoint at $x=0$ and use displacement complex amplitudes $A$ incident and $B$ reflected:At the massless end, the string's transverse force $-T{\Psi }_x$ balances the damper's force $-\alpha {\Psi }_t$. Thus $T{\Psi }_x=-\alpha {\Psi }_t$, or $Z(A-B)=\alpha (A+B)$. This viscous termination of a stretched string hasHalf the incident mean power is reflected when $r=\pm 1/\sqrt{2}$, giving two positive solutions:They are approximately $0.1716\rho c$ and $5.828\rho c$, in $\mathrm{kg}{\mathrm{s}}^{-1}$. The smaller coefficient produces no reflected displacement-phase reversal; the larger produces a $\pi$ reversal. Both dissipate the remaining half of the incident power. Choosing the impedance-matched value $\alpha =\rho c$ would instead absorb all of it.