2
$\begingroup$

So one of the basic selection rules for nuclear transitions is for initial and final spins $I_i$ and $I_f$ that the possible values of L of the emitted photon are $|I_i - I_f| \le L \le |I_i + I_f|$.

However in a recent lab I undertook, we investigated Barium-133 which has a 79.6 keV transition from a state at 160.6 keV and angular momentum of 5/2+ and 81.0 keV and 5/2+. Surely this transition is forbidden and cannot occur?

$\endgroup$
1
  • $\begingroup$ Why do you say it is forbidden? $\endgroup$
    – Charlie
    Nov 6, 2016 at 12:28

2 Answers 2

1
$\begingroup$

Transitions where $J$ changes from zero to zero are forbidden, because the photon must carry away at least one unit of angular momentum.

However if $J\neq 0$, transitions which preserve $J$ are allowed. The angular momentum carried away by the photon goes into changing the projection of $J$ onto your favorite axis --- that is, $J_z$ changes --- but the total angular momentum is conserved in all reference frames. Working this out is a nice QM problem.

$\endgroup$
1
$\begingroup$

First of all each transition can be mediated by one or more terms of the multipole expansion. For each term (electric dipole E1, magnetic dipole M1, electric quadrupole E2, etc.) you have some selection rules that relates the initial to the final state, indicating the allowed transitions. For your specific case take a look at the image below [Reference]. Ba133 Focusing on the transition of interest ($5/2^+\to5/2^+$), you see that it is labeled with M1+E2. Indeed, checking the selection rules for these two (see previous link), you have that both satisfy $$ \Delta J=0\;\;\;\text{and}\;\;\;\pi_f=\pi_i, $$ which is exactly what you need in order to make the transition possible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.