$^{95}_{41}$Nb $\beta^-$-decays to the first excited state of $^{95}_{42}$Mo. This state has an excitation energy of 768 keV and the de-excitation to the ground state goes via photon emission or internal conversion.

The spin and parity of $^{95}_{41}$Nb and $^{95}_{42}$Mo can be determined from the odd proton/neutron to be $9/2^+$ and $5/2^+$ respectively (I used the energy levels from Shell model). How can I determine spin and parity of the excited state?

I also want to calculate the energy for the internal conversion electron. I suppose that it is not simply the excitation energy.


1 Answer 1


Computing the excitation spectrum for a nucleus — that is, its energy levels and their quantum numbers — is hard. Consider that the Schrödinger equation has no known exact solutions for atoms other than hydrogen: a helium atom, with three charges instead of two, is too complex to be treated except in approximation. The nuclear many-body problem is much thornier. Not only are there more participants in the system (ninety-five, for $^{95}$Nb), but in addition to the electrical interaction among the protons you have the pion-mediated attractive strong force, the rho- and omega-mediated hard-core repulsion, three-body forces, etc. (I'm impressed that you got the correct spins and parities for the ground states from the shell model; nice work!)

So when normal people want to know the spin and parity and energy of a nuclear state, we look it up. The best source is the National Nuclear Data Center hosted by Brookhaven National Lab, which maintains several different databases of nuclear data (each with its own steep learning curve). Searching the Evaluated Nuclear Structure Data File by decay for $^{95}$Nb brings up level schemes, with references and lots of ancillary data, for both niobium and molybdenum. These confirm that you've gotten the $J^P$ for the ground states correct. Two excited states are listed for $^{95}$Mo: one at $200\rm\, keV$ with $J^P = 3/2^+$, and the one you mention at $766\rm\,keV$ with $J^P=7/2^+$. You can follow the references to see the experimental arguments for assigning those spins and parities.

You can make some general, hand-waving predictions about spins by thinking about angular momentum conservation in the transitions. The matrix element for a particular transition is generally proportional to the overlap between the initial wavefunction and the final wavefunction. In nuclear decays the initial state is the nucleus, which is tiny and more-or-less spherical with uniform density, while the final state includes the daughter nucleus and the wavefunctions for the decay products. If the decay products carry orbital angular momentum $\ell$, the radial part of the wavefunction goes like $r^\ell$ near the origin. Dimensional analysis then says that the overlap between the nucleus and the decay wavefunction is proportional to $(kR)^\ell$, where $R$ is the nuclear radius and $k = p/\hbar = 2\pi/\lambda$ is the wavenumber of the decay product. (Note that nuclear decay products typically have $\lambda \gg R$, so you can treat the decay product wavefunction as roughly uniform averaged over the nucleus.)

If the probability of a decay is proportional to $(kR)^\ell$, that means that

  • decays where the product's momentum $p=\hbar k$ is large are preferred over decays where the product's momentum is small

  • decays where the orbital angular momentum $\ell$ is small are preferred over decays where $\ell$ is large

For your $\rm Nb\to Mo$ transition, the decay to the excited state is preferred over the $\frac92\to\frac52$ ground-state-to-ground-state transition, which suggests that the excited state probably has spin $\frac72, \frac92, \frac{11}2$. The most probable of these is $\frac72$, since angular momentum tends to relax during decay processes — and indeed, that's the spin of the $766\rm\,keV$ excited state.


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