How do I differentiate $K_alpha$ from $L_alpha$ transitions of X-ray emissions?

Question

If I shine an electron beam in a sample and it returns, for an example, an X-ray wavelength compatible with both $K_alpha$ of copper and $L_alpha$ of hafnium. How can I tell the difference between these elements?
Edit: Using Moseley Law here I get copper for $K_alpha$ and hafnium for $L_alpha$ for an wavelength of 0.155 nm.

Farcher · Accepted Answer

Strange, since $rm K_{alpha}$ for Copper is $8.046,rm MeV$ and $rm L_{alpha}$ for Europium is $5.849,rm MeV$.
The energies of the X-rays are well defined, as are the wavelengths of light in the visible part of em spectrum, so differentiation is not normally a problem.
With the error corrected by the author of the post from Europium to  Hafnium, $rm L_{alpha}$ is $7.899,rm MeV$ which is significantly different (ie relatively easily measurable) from the Copper.
It probably shows that the author of the OP was not working to sufficient significant figures?

user137289 · Answer

One looks at the spectrum. There should also be a weaker $beta$ line at slightly higher energy (shorter wavelength). An L$_beta$ line is almost as strong as the L$_alpha$ but a K$_beta$ would be four of five times weaker.

Answered by user137289 on May 19, 2021

How do I differentiate $K_alpha$ from $L_alpha$ transitions of X-ray emissions?

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