There are $n+2$ distinct interpolation nodes but only $n+1$ free coefficients in a polynomial of polynomial degree at most $n$. Thus polynomial interpolation is possible precisely when the data satisfy one compatibility equation. To identify it, form the Lagrange interpolation polynomial through all the data:This is the unique interpolation polynomial of polynomial degree at most $n+1$. Its leading coefficient, which is also the order-$(n+1)$ divided difference, isConsequently the required condition isNecessity follows because any qualifying $p_n$ must equal $q$ by uniqueness of polynomial interpolation; sufficiency follows because $D=0$ removes the only possible term of polynomial degree $n+1$ from $q$.

When this condition holds, a slightly shorter expression uses only the first $n+1$ interpolation nodes:This polynomial agrees with $q$ at $n+1$ distinct points, so the root bound for a polynomial gives $p_n=q$ and hence $p_n(x_{n+1})=f_{n+1}$ as required. Equivalently, the compatibility condition says that this expression, evaluated at the last interpolation node, gives $f_{n+1}$.

A ground state spin-zero meson is a color singlet $q\bar{q}$ with $L=0$ and constituent spin zero, hence a pseudoscalar meson. The nine flavor combinations form $3\otimes \bar{3}=8\oplus 1$. Here $B=0$, so the vertical coordinate $B+S$ equals strangeness. The six peripheral weights in the right panel of the figure in part (a) are

At the center, $u\bar{u},d\bar{d},s\bar{s}$ all have $I_z=S=0$. A useful orthonormal flavor basis isThe neutral pion has $I=1$; ${\eta }_8$ and ${\eta }_1$ have $I=0$. The Eta octet state belongs to the octet, while ${\eta }_1$ is the flavor singlet. The physical eta and eta prime mesons are predominantly mixtures of these two isoscalar basis states,Unequal quark masses and the chiral anomaly prevent exact flavor symmetry and a degenerate nonet. The three central physical species are consequently ${\pi }^0,\eta ,{\eta }^{\prime }$, rather than three separately observable pure diagonal flavor pairs. Small isospin-breaking admixtures are neglected here.

The observed dominant two-body transition is $\boxed{{\Sigma }^0\to {\Lambda }^0\gamma }$. Both baryons have valence quarks $uds$ and the same strangeness; an electromagnetic interaction transition is therefore possible. Their mass difference is only about $77\mathrm{MeV}/c^2$, below the lightest pion mass, so no additional on-shell strongly interacting particle fits. A photon is neutral and massless, and can carry the required angular momentum through a magnetic dipole transition between the two positive-parity spin-$1/2$ states. A single neutrino would violate lepton number and would not give the required fermion parity.

At rest, four-momentum conservation gives $p_{\Lambda }=p_{\Sigma }-k$ and $k^2=0$. ThusThe result is slightly below $(M_{\Sigma }-M_{\Lambda })c^2$ because the Lambda baryon recoils; rounded masses give about $74.5\mathrm{MeV}$.

The channels $n{\pi }^0$ and $p{\pi }^{-}$ are actually kinematically open: their total rest energies are about $1075$ and $1078\mathrm{MeV}$, below $M_{\Sigma }{c}^2\simeq 1193\mathrm{MeV}$. Their suppression is therefore not a decay threshold prohibition. Each changes strangeness from $-1$ to zero and so requires a weak interaction. The much faster allowed electromagnetic transition makes such weak branching fractions extremely small. “Not observed” should be understood as overwhelming electromagnetic competition, not an exact ban in the Standard Model. The photon identification likewise uses the known low-energy spectrum and dominant transition mechanism, not neutrality alone.

Write the charge in units of $e$ as $Q=a{I}_z+b(B+S)$. The proton and neutron have $(I_z,B+S)=(1/2,1)$ and $(-1/2,1)$, so their measured charges imply $1=a/2+b$ and $0=-a/2+b$. Thus $a=1$, $b=1/2$:This is the Gell-Mann--Nishijima formula, with the full flavor hypercharge $Y=B+S$. Applying the quark baryon number $1/3$ and the isospin and strangeness assignments from part (a),Physical electric charges are these numbers multiplied by the positive elementary charge $e$.

Measured half-integer spin and fermionic behavior rule out a simple quark-antiquark pair: two spin-$1/2$ constituents have integer total spin, and adding integer orbital angular momentum does not change that. Their measured baryon number $B=1$, inferred through production and conservation bookkeeping, corresponds to $n_q-n_{\bar{q}}=3$. Their pattern of electric charges, strangeness and isospin then agrees with three light valence quarks.

Baryon magnetic moments provide a more detailed check. The additive quark model gives, for example,Approximately equal up and down constituent masses give ${\mu }_u\simeq -2{\mu }_d$, hence ${\mu }_p/{\mu }_n\simeq -3/2$, close to the measured ratio of about $-1.46$.

These observations support three valence quarks; they do not prove the absence of extra quark-antiquark pairs. For instance, $qqqq\bar{q}$ also has $B=1$ and can have half-integer spin. Ordinary quantum chromodynamics baryons also contain sea pairs and gluons. The stronger interpretation that the measurements exclude every antiquark component is therefore false.

Both central baryons have valence quarks $uds$, baryon number $1$, strangeness $-1$ and isospin projection zero. Their total isospins differ:The $ud$ flavor part is symmetric for the Sigma baryon and antisymmetric for the Lambda baryon. In the usual ground state coupling, the associated $ud$ pair has spin one for ${\Sigma }^0$ and spin zero for ${\Lambda }^0$, with the strange quark giving total spin $1/2$ in both cases. Distinct total isospin makes these orthogonal states despite their coincident position in the weight diagram.

The diagram refers to ground state, $L=0$, color singlet baryons. Their spatial wavefunction is symmetric and their color wavefunction is antisymmetric. The Pauli exclusion principle therefore requires a symmetric combined spin-and-flavor wavefunction. One cannot independently attach every three-spin coupling to each flavor assignment.

For three spin-$1/2$ particles, there are formally two ways to couple to total spin $1/2$. In the ground state quark model these are tied to the corresponding mixed-symmetry flavor functions. Their required symmetric combination yields one physical spin-$1/2$ flavor octet, not two independent species at each peripheral weight. In spin-flavor symmetry notation,under flavor $SU(3)$ and spin $SU(2)$. Thus there is one peripheral species in the baryon octet, each still possessing its two ordinary spin projections. The two abstract flavor-octet copies in $3\otimes 3\otimes 3$ do not double the physical ground state octet.

At the dotted positions, the flavor content would be $uuu$, $ddd$ or $sss$. These flavor wavefunctions are completely symmetric. The spin wavefunction must then also be completely symmetric, which means total spin $3/2$, not $1/2$. The corresponding baryon decuplet states are ${\Delta }^{++}$, ${\Delta }^{-}$ and ${\Omega }^{-}$. This exclusion is for the stated ground state spin-$1/2$ multiplet; it is not an exclusion of all excited baryons at those weights.

Each quark has baryon number $1/3$, and each antiquark has baryon number $-1/3$. A strange quark has strangeness $-1$, a strange antiquark has strangeness $+1$, and the light $u,d$ have zero strangeness. The isospin projections of $u,d,s$ are $+1/2,-1/2,0$, with the opposite values for their antiquarks. ThusFor three valence quarks, $B=1$ and $Y=B+S=1-n_s$. The baryon octet is reproduced by the following quark model assignments:

Consequently the top-left and top-right states in the source diagram are $\boxed{X=n,Y=p}$. Here the source's label $Y$ means the unknown particle at the top right; the conventional symbol $Y=B+S$ elsewhere in this solution denotes flavor hypercharge.

Model alpha decay as a preformed alpha particle making repeated escape attempts from a short-range nuclear potential well, then undergoing quantum tunnelling through the external Coulomb potential. Write the daughter atomic number as $Z_D$, the nuclear radius as $R$, and the relative reduced mass as $\mu$. For zero orbital angular momentum, the WKB approximation givesSet $v=\sqrt{2Q/\mu }$ and $r=r_2{\sin }^2\theta$. ThenWhen $R/r_2\ll 1$, the leading term isHere $P_{\alpha }$ is the alpha preformation probability and ${\nu }_{\mathrm{attempt}}$ the attempt frequency. The stated proportionality retains only the dominant barrier exponent. It assumes a thick classically forbidden barrier, nonrelativistic relative motion, a spherical unscreened external Coulomb potential, and slowly varying prefactors. Finite $R$, changes in alpha preformation probability, and a centrifugal barrier for nonzero orbital angular momentum produce significant corrections. The precise charge in the barrier is $Z_D=Z_P-2$; using the parent charge is an approximation for large $Z_P$. The measured alpha particle speed approximates $v$ only when daughter recoil is negligible.

For a parent at rest, momentum conservation gives equal and opposite daughter and alpha particle momenta. In the nonrelativistic recoil approximation,Approximating nuclear masses by their mass numbers, $M_D\simeq (A-4)m_u$ and $m_{\alpha }\simeq 4m_u$, givesThe printed phrase “atomic number $A$” must here mean mass number. Atomic number alone does not determine the recoil fraction. For example, ${}^{210}\mathrm{Po}$ and ${}^{212}\mathrm{Po}$ have the same atomic number $84$, but the approximate fractions are $206/210$ and $208/212$, respectively.

For completeness, exact relativistic two-body decay kinematics giveswhere $M_P=M_D+m_{\alpha }+Q/c^2$. Expanding for $Q\ll M_P{c}^2$ recovers the recoil-sharing expression.

Let $M_P$ and $M_D$ denote neutral atomic rest masses for the appropriate parent and daughter, and initially neglect electronic binding energy corrections and daughter excitation. The Q values areFor beta-minus decay, the daughter nucleus has one extra positive charge. Its neutral atom contains one more electron than the neutral parent, cancelling the emitted electron's rest mass in the atomic accounting. For beta-plus decay, the neutral parent contains one more electron than the neutral daughter; together with the emitted Positron, this gives the $2m_e$ subtraction. Thus beta-plus decay requires an atomic mass difference exceeding $2m_e{c}^2$, whereas electron capture need not.

For electron capture, the captured bound electron is already included in $M_P$. Capture from a shell leaving an atomic hole of excitation energy $E_{\mathrm{hole}}$, and a nuclear excited state of energy $E_x$, gives approximatelyThe hole's energy later appears as atomic radiation or Auger electrons. Subtract $E_x$ for either beta decay as well. Exact endpoints also depend on daughter ionization and small atomic binding energy changes; the boxed expressions are the conventional leading atomic-mass formulas.