Subatomic question - K alpha lines

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raintrek
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Homework Statement



Light from the K-alpha line of an unknown material X is compared with the K-alpha line of Carbon. The wavelength ratio is [tex]\frac{\lambda_x}{\lambda_c} = 0.148[/tex]. What is the matieral X?

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The Attempt at a Solution



I've got no idea where to go with this one. The only time I've used K-alpha lines is in regards to changes of energy levels. I know E=hf=hc/lamba, but i don't see how that helps :confused:
 
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Answers and Replies

  • #2
Dick
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Your first question to answer is which energy levels is a K-alpha line related to and what is the relation of those levels to atomic number?
 
  • #3
quantumdude
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It looks like you're supposed to consult a table. You would have to find the entry in the table for which [itex]\lambda_x=0.248\lambda_C[/itex].

Do you have such a table in your book?
 
  • #4
raintrek
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^ Definitely no table, Tom. It's an isolated question, so I'm thinking I need to approach it a different way.
 
  • #5
Dick
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You didn't answer my questions yet? :frown:
 
  • #6
raintrek
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Sorry Dick, missed that post! I'm pretty sure the K-alpha line arises when an electron transfers from the L shell (2nd) to the K shell (first), not sure about the relation :confused:
 
  • #7
Dick
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Ok! You have the levels. Now if you ignore the other electrons around, what is the relation between the energy of these levels and the nuclear charge?
 
  • #8
raintrek
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OK, I'm looking through some information on the Bohr Atom and I've found this relation which seems to link the two:

[tex] E_n = -Z^{2} \frac{E_0}{n^{2}}[/tex]

Not sure if that's the one you're thinking about ...
 
  • #9
Dick
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That's the one! So the energy difference is proportional to Z^2. But there are other electrons around. There is one in particular to worry about. Which one? Isn't this fun!?
 
  • #10
raintrek
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OK, so the transition between the n=2 and n=1 shell will result in the emission of a photon, i'd imagine, which would have a wavelength associated with it [tex]\lambda = \frac{c}{f} = \frac{hc}{E_i - E_f}[/tex]
 
  • #11
Dick
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Yes. And the energy difference is ROUGHLY proportional to Z^2. But I'm worried about the effect of another electron besides the one making the jump.
 
  • #12
raintrek
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^ Not quite sure what you mean there. I'm looking through my notes on the Bohr Model and some worked examples of Lyman series transitions (n=2 -> n=1) and they only seem to be concerned with the energy of the photon and the difference in energies of the final and initial atomic state :confused:
 
  • #13
Dick
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Fair enough. But they are talking about fully ionized atoms. In our case we should assume that the K shell probably won't be empty before the transition occurs. How many electrons might be there? What effect would this have on the problem?
 
  • #14
raintrek
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Just another thought, if I use Mosely's law, and turn it into a ratio of the wavelengths , that'll let me solve for Zx which comes out as 13.9999 = 14 = Si. Not sure if that's ok...
 
  • #15
Dick
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Dang. I was one step away from having you (at least partially) derive Moseley's law. The other K electron will screen the nuclear charge - so you only get an effective charge of (Z-1) (as you see in his formula). The only thing you can't really easily compute is whether the screening is 100% effective. But it turns out to be pretty close. So you now understand the formula as well as you use it right? Si sounds right.
 
  • #16
raintrek
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Yeah, i think i understand Mosely's formula - from what I've just read, the 1 accounts for the shielding due to other electrons (and is always roughly 1 for innermost levels)
 
  • #17
Dick
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Yep! That's pretty much it.
 
  • #18
raintrek
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OK, thanks for the help Dick! :)
 

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