Past exam of the particle and nuclear physics course of the University of Cambridge / 2015 / Paper 1 / 1

A positive-parity $K=0$ nuclear rotational band in an even-even nucleus has $J=0,2,4,\dots$ and excitation energiesThe four nonzero levels give $E_J/[J(J+1)]=15.17,14.85,14.48,14.01\mathrm{keV}$. These approximately constant coefficients, together with the even-spin positive-parity sequence, support a rotational interpretation. The roughly eight-percent decline means a constant-inertia rigid rotor is only an approximation: the effective moment of inertia increases along the sequence.

Using the lowest excited level givesUsing the highest level instead gives $2.48\times 10^{-54}\mathrm{kg}{\mathrm{m}}^2$, showing the scale of the departure from a constant inertia. This is a comparison of nuclear rotational inertia relative to a rigid sphere, not evidence for an exactly spherical rotating intrinsic state.

For a conventional size estimate, explicitly take a uniform sharp radius $R=r_0{A}^{1/3}$ with $r_0\simeq 1.2\mathrm{fm}$ and mass $M\simeq A{m}_u$. These are standard scale approximations, not additional measured properties of this isotope supplied by the paper. They give $R\simeq 6.70\mathrm{fm}$ andThus the fitted rotational inertia is substantially smaller than the rigid-body scale. Nuclear pairing provides a physical reason collective nuclear inertia need not equal the classical rigid-body value. The comparison's numerical radius dependence is explicit; it is not an independently determined radius.

A physical baryon is a colour singlet. Its three-quark colour factor is antisymmetric, proportional to ${ϵ}_{abc}|abc⟩$. In the hypothetical scalar-quark theory, the quarks obey Bose-Einstein statistics, so the complete three-particle state must be symmetric under exchange. At $L=0$ the spatial factor is symmetric, and three zero spins supply no nontrivial spin factor. The flavour factor must therefore be antisymmetric to cancel the antisymmetry of the colour factor.

There is exactly one completely antisymmetric state made from three flavour basis states: the flavour singlet of three scalar quarks,Equivalently, $\dim {\bigwedge }^3{\mathbb{C}}^3=1$. A repeated flavour gives zero under complete antisymmetrization. Only one ground-state flavour singlet, containing $uds$ and having $J^P=0^{+}$, remains. Keeping fermionic exchange antisymmetry while changing the spins to zero would contradict the assumed Spin-statistics theorem. The standard antisymmetric colour factor and the ordinary-quark flavour decomposition are documented in the quark-model review.^[ref]

Use natural units and invariant squared center-of-mass energy $s$. For a massless incident neutrino and an initially stationary electron,At the production threshold the final massless neutrino can have arbitrarily small center-of-mass energy, so the lower endpoint is $s=m_{\ell }^2$. The charged-current neutrino-electron threshold is consequentlyFor the three flavours, using precisely the masses given in this part, the answers are $0$ for ${\nu }_e$, $1.0994\times 10^4\mathrm{MeV}\simeq 11.0\mathrm{GeV}$ for ${\nu }_{\mu }$, and $3.0898\times 10^6\mathrm{MeV}\simeq 3.09\mathrm{TeV}$ for ${\nu }_{\tau }$. The zero threshold for the electron flavour is an infimum for elastic scattering; it does not imply a nonzero interaction rate at zero incident energy. The large heavy-lepton thresholds arise because most laboratory energy becomes center-of-mass motion against a very light target.