GR8677 #26



Alternate Solutions 
Limerick Jim 20111014 15:34:42  way simpler guys: GRE means speed
Use: E=hf=
xrays are around 10^10 (xrays are on the order of atoms, thats why we use them for xray diffraction of crystals), You should know the speed of light or you should kill yourself, and h in eVs is quoted in the constants. plug it in to get 12 x 10^3 which aprox equals 10,000 eV   barefoot0 20061114 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.  

Comments 
Limerick Jim 20111014 15:34:42  way simpler guys: GRE means speed
Use: E=hf=
xrays are around 10^10 (xrays are on the order of atoms, thats why we use them for xray diffraction of crystals), You should know the speed of light or you should kill yourself, and h in eVs is quoted in the constants. plug it in to get 12 x 10^3 which aprox equals 10,000 eV
ryanjsfx 20141020 11:47:24 
1) I messed up this problem by doing it that way but taking Xray wavelength to be 1e11 meters (which is still Xrays) and it turned out using that gets (E) so I think that approach assumes more than is given.
2) If you just remember E is approximately where Z is the number of protons [Bohr energy approximation] then rounding 13.6 to 10 and 28 to 30 yields E = (10)(30)(30) = 1e4 much quicker than doing the (admittedly still rather simple) multiplication of h c and one over lambda.

DownWithETS 20161030 14:36:23 
I used this method also, but it doesn\\\\\\\'t really work. X rays range from wavelength=10 nm.1 nm. If you assume the highend energy limit of .1 nm, then it works, but using either 1 or 10 nm values gets us a different order of magnitude, which are both possible choices. Is there some reason you know to choose the highest energy xrays?

  engageengage 20090117 15:02:26  Im not sure why people are talking about atomic transitions. If you want to produce xrays 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 xrays along the way. Therefore, you wind up with the same equation that yosun has, where you take screening into account:
13.6eV ( Z1)^2
wittensdog 20090725 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.

  a19grey2 20081103 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.
flyboy621 20101114 19:25:17 
I think ETS expects you to make the approximation that Yosun did in the solution, i.e. that . It makes the math a lot easier, and it gets you the right order of magnitude, which is all you need for this problem.

  FortranMan 20081019 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 20081011 16:41:30  EDX (Energy Dispersive Xray Spectroscopy) is a technique to characterize materials based on the wavelength of xrays emitted from the sample when excited with an electron beam. Beam energy typically needs to be on the order of 1025 keV to produce a good signal. So, choice D.   eshaghoulian 20070915 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 . 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: . 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).
ali8 20110627 13:46:04 
Where did you get that eqn from ??
I cannot find it in my Serway Modern Physics Text.

  Furious 20070830 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.
pam d 20110923 21:19:19 
No, the electron in the Kshell is being knocked out of the atom by an incoming electron. Therefore it goes from n = 1 (definition of Kshell) to n = infinity (definition of being knocked out). The difference in energy sought in this problem is the minimum energy that an incoming electron must have to do the job.
The actual xrays produced are the result of Lshell to Kshell transitions (corresponding to n = 2 to n = 1 transitions) of the remaining electrons after the abovedescribed event. The transition could also be an Mshell to Kshell transition. It is even possible that another electron be emitted to carry off the excess energy of this transition rather than an xray photon. This is called an Auger electron and it's emission becomes increasingly probable with increasing Z.

  barefoot0 20061114 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 20070529 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.

tatitechno 20110919 03:55:48 
We know that Xray diffraction use about 40Kev in max (reality) so the correct scale is of (D)

  kevglynn 20061031 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 20061019 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 20061027 21:56:12 
Definately memorize that number.

Setareh 20111024 04:18:23 
As a physics student, you have to memorize 13.6 just like 2*2=4! I am sure if you deal more with GRE problems you will memorize it, because this number is involved in so many questions.

 

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