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GR8677 #26
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Alternate Solutions |
barefoot0
2006-11-14 11:17:02 | I dont know if this exactly correct but I did the following.
I know 2.5eV gives ~ 500nm (This is just an easy thing to remember and quite usefull).
X rays are about .1 nm and E is inversly proportional to Lambda so
2.5eV*500nm/.1nm ~ 10,000 eV
Someone let me know if and why this method is wrong. |  |
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Comments |
engageengage
2009-01-17 15:02:26 | Im not sure why people are talking about atomic transitions. If you want to produce x-rays in the K series, you must knock out the first electron. This will then cause more energetic electrons to transition into the newly free state, emitting x-rays along the way. Therefore, you wind up with the same equation that yosun has, where you take screening into account:
13.6eV ( Z-1)^2
wittensdog
2009-07-25 16:56:30 |
I agree, the only thing that matters for producing K series x rays is that you have a hole for other electrons to drop back into in the n = 1 shell. So the only thing your electron needs to do is pop out an n=1 electron. The only formula I ever made reference to was the hydrogenic atom approximation to the Bohr energy formula, E = - 13.6 * Z^2 / n^2. In this case n = 1 and Z = 28. There is also the issue of screening. I do not recall how much screening one of the n = 1 shell electrons should provide for the other. In chapter 7 of Griffiths' QM textbook you can see that for Helium, the screening results in around 1.69 as opposed to 2. I'm not sure how much all of the electrons in the outer shells affect that, but it hardly matters for this problem, since using 27 or 28 results in either 9,914.4 or 10,662.4, respectively, all of which are certainly much closer to 10,000 than anything else. My preferred method for the approximation is to write everything in scientific notation and pull apart the factors, so you end up with 1.36 * (2.8) ^ 2 * 1,000. With this, it's obvious that 27 vs. 28 hardly makes a difference, and the factor out front is roughly 10. If the problem involved the specific energies of the x rays emitted, then you would need the Rydberg formula, but this is not the case here.
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|  | a19grey2
2008-11-03 20:29:42 | Uh, am I the only person who calculates the value using  and  ,  and gets only ~7,400.
doing this on a test , I'd still guess the right answer, but this seems to far off for this to be the right answer since I got 7435.8eV using a calculator and on the test the approximations would make me even more off.
| | FortranMan
2008-10-19 11:03:03 | Shouldn't it be
)
Where n_{i} is the initial state (either at n_{i}=2 or  ) and n_{f} is the final state (n_{f}=1)? This would explain how the negative cancels out. | | casaubon
2008-10-11 16:41:30 | EDX (Energy Dispersive X-ray Spectroscopy) is a technique to characterize materials based on the wavelength of x-rays emitted from the sample when excited with an electron beam. Beam energy typically needs to be on the order of 10-25 keV to produce a good signal. So, choice D. | | eshaghoulian
2007-09-15 14:54:25 | Agreed. The solution above is confusing the state of the bombarding electrons with the initial state of the electron making the transition. For minimum energy necessary, the equation is *((Z-1)^2)((1/1^2)-(1/n_{i}^2))) . Minimizing this requires setting  (the initial state of the transitioning electron) equal to 2 (note that you can't set it equal to 1 since then you wouldn't have a transition). This minimizes the energy necessary for the transition, and you equate this to the KE of the incoming electron, which isn't making any transitions. So the physical process is: electron comes in from , scatters off of an atom by giving up some of its KE, and then the atom makes a transition.
Note the general formula: *((Z-b)^2)((1/n_{f}^2)-(1/n_{i}^2))) . This is a slight modification of Bohr's Law, where  and  depend on the series you are considering (e.g.  DEFINES K series and gives  ,  DEFINES L series and gives  , etc.). The constants should probably be memorized, but I don't predict them asking beyond K or L series (probably won't even ask L series). |  | Furious
2007-08-30 18:30:19 | I have to agree with kevglynn, the minimum energy comes at n_i =2 not at n_i goes to infinity. That's actually the MAXIMUM energy. | | barefoot0
2006-11-14 11:17:02 | I dont know if this exactly correct but I did the following.
I know 2.5eV gives ~ 500nm (This is just an easy thing to remember and quite usefull).
X rays are about .1 nm and E is inversly proportional to Lambda so
2.5eV*500nm/.1nm ~ 10,000 eV
Someone let me know if and why this method is wrong.
sonnb
2007-05-29 12:49:53 |
The spectrum of X rays ranges from 0.1nm to 10nm so that if you had used 1.0nm instead, you would have gotten the wrong answer.
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| | kevglynn
2006-10-31 11:14:47 | I'm pretty sure that the MINIMUM required energy refers to when the electrons come in from n_i=2. In other words, doesn't n_i refer to the initial state of the electron, not the bombarding electrons? thanks | | Healeyx76
2006-10-19 18:44:49 | Is there a way to figure out R in Ev if you haven't memorized -13.6eV ? It is not one of the given values at the start of this test.
Oh: I guess you could memorize/know R = (m_e*e^4)/(2*hbar^2) eh?
13.6 13.6 13.6 13.6 ..... have to remember that one.
herrphysik
2006-10-27 21:56:12 |
Definately memorize that number.
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