Question

Consider an idealized (point charge) Al atom (Z = 13, A = 27). If a negative lepton or meson is captured by this atom it rapidly cascades down to the lower n states which are inside the electron shells. In the case of p-capture:

(a) Compute the energy E₁ for the p in the n = 1 orbit; estimate also a mean radius. Neglect relativistic effects and nuclear motion.

(b) Now compute a correction to E₁ to take into account the nuclear motion.

(c) Find a perturbation term to the Hamiltonian due to relativistic kinematics, ignoring spin. Estimate the resulting correction to E₁.

(d) Define a nuclear radius. How does this radius for Al compare to the mean radius for the n = 1 orbit from (a)? Discuss qualitatively what happens to the µ^- when the µ^- atomic wave function overlaps the nucleus substantially. What happens to a π^- under the circumstances? Information that may be relevant:

Answer 1

(a) We shall neglect the effects of the electrons outside the nucleus and consider only the motion of the µ in the Coulomb field of the Al nucleus. The energy levels of µ in a hydrogen-like atom of nuclear charge 2 (in the nonrelativistic approximation) are given by

(b) To take into account the motion of the nucleus, we simply have to replace the mass m of the meson with its reduced mass \(\mu = \frac{Mm}{M + m}\), M being the nuclear mass. Thus

(c) Taking into account the relativistic effects the muon kinetic energy is

The relativistic correction introduces a perturbation Hamiltonian

The energy correction AE for E₁ is then

(d) In the scattering of neutrons by a nucleus, an attractive strong nuclear force sets in when the distance becomes smaller than \(r\sim r_0A^{\frac 13}\), where \(r_0\sim 1.2 \times 10^{-13}\) and A is the atomic mass number of the nucleus. r is generally taken to be the radius of the nuclues, which for Al is

The difference between the radius of the nucleus of Al and that of the first orbit of the µ-mesic atom is not very large, so that there is a considerable overlap of the wave functions of the nucleus and the muon. This effect, due to the finite volume of the nucleus, will give rise to a positive energy correction. At the same time there's also a large interaction between the muon and the magnetic moment of the nucleus.

Under similar circumstances, for the π-mesic atom there is also the volume effect, but no interaction with the magnetic moment of the nucleus as p-ions have zero spin.