X-rays: trouble understanding

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1. Sep 28, 2016

Pavoo

1. The problem statement, all variables and given/known data
Although not a computational problem, I still have difficulties understanding emission of characteristic X-rays.

Can someone please clear up my confusions about the topic? Here's where I'm stuck, with two texts as an example:

Source for the above: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/xrayc.html

Following this explanation, let's say we shoot out an electron from the K-shell, and an electron from the L-shell fills its place. The X-ray that will then be emitted will be of equivalent energy size as the difference between binding energies of the K-shell and L-shell of the atom.

However, look at the following text:

Source: https://www.bruker.com/products/x-r...ntal-analysis/handheld-xrf/how-xrf-works.html

Here confusion begins. How can an electron have a higher binding energy if it is further away from the nucleus? Looking at this table, I can't understand how an electron further away from the nucleus can have higher binding energy: http://xdb.lbl.gov/Section1/Table_1-1a.htm

My question is this: what type of difference of energy is an x-ray made of? The difference between the shells or the difference in the energy state of an electron?

P.S. Not used the template for posting assignment, since this is not a computational problem.

2. Sep 28, 2016

Staff: Mentor

The second text is in error: it should be higher potential energy, not binding energy.

3. Sep 28, 2016

Jonathan Scott

The amount of energy that has to be supplied to remove the electron is greater when it is closer to the nucleus, so the binding energy has the opposite sign to the potential energy of the electron. I think it should refer to higher potential energy, not binding energy.

4. Sep 28, 2016

ZapperZ

Staff Emeritus
There might be a confusion due to the SIGN.

A binding energy is often quoted with the knowledge that it is negative. For example, the ground state of H atom is Eb = -13.6 eV. While the magnitude of the binding energy is large, it is considered to be "lower" than, say, the vacuum level which, by definition, has 0 eV. So mathematically, the vacuum level has a "higher binding energy" than the ground state.

Zz.

5. Sep 28, 2016

Pavoo

DrClaude and Jonathan Scott

That certainly clears up my confusion! Thanks for pointing that out!