Consider characteristic X-ray emission from copper. The $K_{\alpha}$ line is a doublet because of the spin-orbit interaction.

But why is the $K_{\alpha,3/2}$ line always more intense than the $K_{\alpha,1/2}$?


1 Answer 1


The basic reason is the multiplicities of the L-shell levels $2p_{3/2}$ and $2p_{1/2}$ that the K-shell hole makes a transition to. There are four states with $j=3/2$ ($m_j= -3/2, -1/2, 1/2, 3/2$) and only two with $j=1/2$ ($m_j=\pm 1/2$). So the intensity ratio is expected to be $2 : 1$.

And since you asked specifically about copper, I include a level diagram with the transitions:

enter image description here


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