nirav_605 20120319 17:31:20  Isn't is suppose to be..
E_n = E_1 * (Z)^2/n^2
Not
E_n = Z^2/(n^2 * E_1)  
Poop Loops 20081005 12:40:06  Basically, you are using Moseley's Law:
http://en.wikipedia.org/wiki/Bohr_model#Moseley.27s_law_and_calculation_of_Kalpha_Xray_emission_lines
Where a K Xray is emitted when you go from 2nd shell down to 1st shell.
So you take with some constant out front that will cancel to get the energy for an atom, and Z = 6 for one and Z = 12 for the other, so you get
And of course cancels, so you get which is almost 1/5, not quite the right answer, but close enough to 1/4.  
Jeremy 20071030 12:18:12  Typo: should be (...though I don't see why this is needed in solving the problem).
A nice, concise background is given on Wikipedia about Moseley's " target="_blank">http://en.wikipedia.org/wiki/Moseley%27s_law">Moseley's Law. In particular, the derivation section ties this back into the more familiar realm of Bohr theory.
Jeremy 20071030 12:28:40 
Link

 
Mexicana 20071004 13:38:09  The way to justify the method of Yosun is to realise that K shell Xrays are produced by the transition of atomic electrons which were on the highest energy levels of that atom. These jump to the lowest level () and since is small for large , one can readily neglect its contribution to the total transition energy and only account for the final level they jumped to, i.e. for  
Mexicana 20071004 13:36:36  The way to justify the method of Yosun is to realise that K shell Xrays are produced by the transition of atomic electrons which were on the highest energy levels of that atom. These jump to the lowest level () and since is small for large , one can readily neglect its contribution to the total transition energy and only account for the final level they jumped to, i.e. for  
Mexicana 20071004 13:35:02  The way to justify the method of Yosun is to realise that K shell Xrays are produced by the transition of a atomic electron which were on the highest levels of that atom. These jump to the lowest level () and since is small for large , one can readily neglect its contribution to the total transition energy.  
kolahalb 20070925 11:08:54  Bohr's law fits here well.But C or, Mg are not Hydrogen like systems.Preferably we can use Moseley's law.
$\sqrt{ν}$=a(Zb) where for K lines,b~1.
And,E=hν of course
This calculation shows that E(C)/E(Mg)=25/121
~(1/5)
Which rather matches (A)
sidharthsp 20081001 21:36:27 
that seems very just....

 