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.

 

[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.

 

[palaphys]

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.

I agree that state 1 is the intended state. Also when I thought a bit more about this question, I think I understood it now, they are just saying that the energy of the bound K shell can be thought of as -27.5KeV in a way, so we can use $E_L- E_K = E_{K_{alpha}}$ hence the "energy of the atom" in the final state mentioned in the will be magnitude of $E_L$

Please confirm if my thought process is right
Thanks

 

[Steve4Physics]

I agree that state 1 is the intended state. Also when I thought a bit more about this question, I think I understood it now, they are just saying that the energy of the bound K shell can be thought of as -27.5KeV in a way, so we can use $E_L- E_K = E_{K_{alpha}}$ hence the "energy of the atom" in the final state mentioned in the will be magnitude of $E_L$

Please confirm if my thought process is right
Thanks

That’s one way of thinking about it. But I’m not convinced that is the intended interpretation.

The ionisation energy of a molybdenum K-electron is 20.0keV, so the K-electron energy level would be -20.0 keV. Why use the value 27.5keV? I think it’s just a badly written question. I hope it’s not from an official examination.

Answering badly written questions is a problem. The best you can do is to state that you think there is a problem (with reason(s)), state any assumption(s) you need to make to answer the question, give an answer based on your assumption(s). In an exam, leave such questions to the end so you answer the 'good' questions first in the limited time.

BTW, upper case ‘K’ is kelvin (temperature unit). Lower case ‘k’ is kilo (prefix). So the correct energy unit is keV, not KeV!

 

[palaphys]

The ionisation energy of a molybdenum K-electron is 20.0keV, so the K-electron energy level would be -20.0 keV. Why use the value 27.5keV? I think it’s just a badly written question. I hope it’s not from an official examination.

well, 27.5keV is the value given in the question. Also this question appeared in a previous JEE exam

I think I am satisfied with my solution, so I will continue using this kind of approach for such questions..