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Find frequency

  1. Jul 30, 2011 #1
    1. The problem statement, all variables and given/known data
    When an electron makes a transition from (n+1) state to nth state, the frequency of emitted radiations is related to n according to (n>>1):
    (a)[itex]v=\frac{2cRZ^2}{n^3}[/itex]
    (b)[itex]v=\frac{cRZ^2}{n^4}[/itex]
    (c)[itex]v=\frac{cRZ^2}{n^2}[/itex]
    (a)[itex]v=\frac{2cRZ^2}{n^2}[/itex]

    2. Relevant equations
    [tex]\frac{1}{\lambda}=RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]


    3. The attempt at a solution
    Solving the above equation and substituting the values i get:-
    [tex]v=cRZ^2(\frac{1}{n^2}-\frac{1}{(n+1)^2})[/tex]

    [tex]v=cRZ^2(\frac{2n+1}{n^2(n+1)^2})[/tex]

    Now i am stuck, what should i do next?
     
  2. jcsd
  3. Jul 30, 2011 #2
    It's always a good idea when you're having trouble with a math problem to check whether you've used all the information given. You haven't used n>>1.
     
  4. Jul 30, 2011 #3
    I don't know how should i use n>>1? :confused:
     
  5. Jul 30, 2011 #4
    Well, try an example. Substitute in a big n, 100 for instance, and see what you get. Check whether it's closer to a, b, c, or d. Then try to figure out why.
     
  6. Jul 30, 2011 #5
    That's a lot of calculation if i substitute n=100, and i am not able to solve it.
    When i substitute n=100, i get:-
    [tex]v=cRZ^2(\frac{201}{10000(10201)})[/tex]
     
  7. Jul 30, 2011 #6
    There's nothing to solve! Just punch it into your calculator and see what you get. Or use http://www.google.com/landing/searchtips/#calculator". Then do the same for the four choices you're given.

    No, that is not a lot of calculation. That's a small amount of easy calculations. You need to get used to calculating things if you're going to be taking science classes.
     
    Last edited by a moderator: Apr 26, 2017
  8. Jul 30, 2011 #7
    Sorry, i can't use a calculator, this question is from my exam paper and in the examination room we were not allowed to use a calculator, so any other way to solve it?
     
    Last edited by a moderator: Apr 26, 2017
  9. Jul 30, 2011 #8
    Are you in the exam room now? If so, you shouldn't be asking for help. If not, then you can use a calculator.
     
  10. Jul 30, 2011 #9
    No, don't talk silly, how can i be in the examination room and post a question?
     
  11. Jul 30, 2011 #10
    So use a calculator.
     
  12. Jul 30, 2011 #11
    Ok i did it using a calculator, i get my answer to be (a) option.
     
  13. Jul 30, 2011 #12
    Good. Now see if you can figure out why it comes out so close. If it's not obvious, try a few other n's.
     
  14. Jul 30, 2011 #13
    Hi!! :smile:
    If i solve it as you said, i again get (a) option.
     
  15. Jul 30, 2011 #14
    But what's the correct method?
     
  16. Jul 30, 2011 #15

    I like Serena

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    My apologies pmsrw3, it was not my intention to interfere, so I deleted my post. :blushing:
     
  17. Jul 30, 2011 #16
    That's it. If you try a bunch of calculations, it should become obvious that when n is big, n is practically the same as n+1, and 2n is practically the same as 2n+1.
     
  18. Jul 30, 2011 #17
    Thank you so much, i got it!! :smile:
     
  19. Jul 30, 2011 #18

    I like Serena

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    Well, there is a catch...
    What happens if you apply this estimation rule to your equation?

     
  20. Jul 30, 2011 #19
    It becomes zero. :surprised
     
  21. Jul 30, 2011 #20

    I like Serena

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    Any idea why?
    And how you should handle it?
     
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