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			| GR8677 #19 |  |  
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				| Problem |  |  
				|   | Quantum Mechanics
  }Bohr Theory 
Recall the Rydberg energy. QED
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				| Alternate Solutions |  
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				| casseverhart13 2019-10-01 03:22:05
 | Good Examine and fascinating. Thanks. www.trustytreeservice.com |  |  | jeka 2007-02-17 08:08:36
 | Energy spectrum of the hydrogen atom is given by the equation 
 
  , 
 where
  is the Rydberg constant. So the right answer is (E) |  |  |  
			
				| Comments |  
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				| casseverhart13 2019-10-01 03:22:05
 | Good Examine and fascinating. Thanks. www.trustytreeservice.com |  |  | ernest21 2019-08-10 03:09:33
 | Instead of doing the long division by hand, you could just say that gamma is less than 2. Solving the inequality you get that v > \\sqrt{3/4}c. E is the only answer that satisfies this. amino acid game |  |  | joshuaprice153 2019-08-08 07:30:34
 | I\'m definitely going to bookmark your site, I just love your post, thanks for such a nice sharing.. Hope to get some info on your site in future. carpet steam cleaner |  |  | jeka 2007-02-17 08:08:36
 | Energy spectrum of the hydrogen atom is given by the equation 
 
  , 
 where
  is the Rydberg constant. So the right answer is (E) 
											
											
											| FortranMan 2008-10-16 23:10:09
 | According to Griffiths, the allowed energies for a hydrogen atom are 
 
 ![E_{n} = -\left[ \frac{m}{2 \hbar^2} \left(\frac{e^2}{4 \pi \epsilon_{0}}\right)^{2}\right] \frac{1}{n^2} = \frac{E_1}{n^2}](/cgi-bin/mimetex.cgi?E_{n} = -\left[ \frac{m}{2 \hbar^2} \left(\frac{e^2}{4 \pi \epsilon_{0}}\right)^{2}\right] \frac{1}{n^2} = \frac{E_1}{n^2} ) 
 Where
  is the ground state of the hydrogen atom, 
 
  
 The Rydberg constant is defined as
 
 
 ![R_{y} = \left[ \frac{m}{4 \pi c \hbar^3}\left(\frac{e^2}{4 \pi \epsilon_{0}}\right)^{2}\right] = 1.097 \times 10^{7} m^{-1}](/cgi-bin/mimetex.cgi?R_{y} = \left[ \frac{m}{4 \pi c \hbar^3}\left(\frac{e^2}{4 \pi \epsilon_{0}}\right)^{2}\right] = 1.097 \times 10^{7} m^{-1})  
 Thus the energy levels are given as
 
 
  
 Not entirely necessary to solve the problem, but it's safer to keep your terms straight.
 
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											| VKB 2014-03-25 21:44:48
 | Its a good approach to solve problems @ home,interesting. 
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