Confusion on sign convention in X-ray problems

[palaphys]

Homework Statement

The Kα X-ray of molybdenum has wavelength 0.071 nm. If the energy of molybdenum atoms with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be ______ keV. (Round off to the nearest integer)

Relevant Equations

Energy conservation

The question seems very simple, and its solution is also very simple. But what bothers me is the wording- "energy of Molybdenum atoms" what does that mean?? energy relative to what? also how can the energy of a bound system be positive? and I have never seen "energy of atoms" being used as a terminology before.

Please help me understand what the question is asking for.

Thanks.

 

[realJohn]

Do you know the equation for energy conservation?

Assume we can say things such as "the energy level of an electron" or "binding energy of a specific electron in an atom".

 

[Steve4Physics]

The question seems very simple, and its solution is also very simple. But what bothers me is the wording- "energy of Molybdenum atoms" what does that mean?? energy relative to what? also how can the energy of a bound system be positive? and I have never seen "energy of atoms" being used as a terminology before.

Please help me understand what the question is asking for.

This is (IMO) a badly written question for another reason. The final state of the atom is ambiguous. Is it:
a) missing only an L-electron? (call this state-1)? or
b) missing a K-electron and an L-electron (call this state-2)?

For an answer to be possible with the given data, I think we need to assume state-1 is intended.

27.5 keV seems to be an arbitrary energy value with an arbitrary zero reference – think of it simply as the kinetic + potential (external and internal) energy of the singly-ionised molybdenum atom in (let’s call it) state-0.

The atom starts in state-0 with a K-electron missing. Note that if an L-electron drops into the K-vacancy, the atom loses some energy ($E_{photon}$) and is now in state-1.

So, to find the atom’s final (state-1) energy, we just need to subtract the energy lost (the photon energy) from 27.5keV.

Edit - typo's.