How to prove this ?

Discussion in 'Homework Help' started by vick5821, Mar 6, 2012.

  1. vick5821

    Thread Starter Member

    Jan 27, 2012
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    0
    How to prove that the circle part is equal to the square part ?

    [​IMG]

    [​IMG]


    Thank you
     
    Last edited: Mar 6, 2012
  2. Papabravo

    Expert

    Feb 24, 2006
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    Well the quantity in the circle part has units of energy. The quantity in the square part has units of frequency. Do you know of a relationship that relates Energy and frequency? Could Planck's constant be involved? Is R the Rydberg constant? Have you ever heard of dimensional analysis?
     
  3. vick5821

    Thread Starter Member

    Jan 27, 2012
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    yes..I know about dimensional analysis..Yea..I realised that both of them have different unit :) So how to prove them ?
     
  4. Papabravo

    Expert

    Feb 24, 2006
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    The key to doing the proof is to find the precise relationship between the frequency or wavelength of the spectral line and the chnage in energy level of the electron that emits a photon at that frequency (wavelength)

    Hint(s): The frequency times the wavelength is equal to the speed of light. Now go find the "constant of proportionality" between Energy and frequency. This constant is named after a famous Physicist.
     
  5. vick5821

    Thread Starter Member

    Jan 27, 2012
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    Planck constant ,h ?
     
  6. Papabravo

    Expert

    Feb 24, 2006
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    That would be the one.

    I still would like to know what R stands for. I think it might be the Rydberg constant from the Ideal gas law, but I could be wrong.
    [Edit: R is definitely the Rydberg constant]

    Does K stand for Coulomb's constant or something else?
    [Edit: It looks like the reciprocal of Coulomb's constant]

    If those assumptions are true then the solution falls out with two steps.
     
    Last edited: Mar 7, 2012
  7. vick5821

    Thread Starter Member

    Jan 27, 2012
    54
    0
    [​IMG]

    R is rydberg contant yea..

    so how ?
     
  8. steveb

    Senior Member

    Jul 3, 2008
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    The first step in a proof is to accurately state exactly what you are trying to prove. Your first statement is inaccurate because the circled part is not equal to the boxed part.

    So, do you know what you are trying to prove? Is it supposed to be a mathematical proof or a scientific one? Are you trying to show equality or proportionality? Are you trying to show it numerically or symbolically?

    Once you clearly state to yourself what you are trying to do, it will be much easier to do it.
     
  9. vick5821

    Thread Starter Member

    Jan 27, 2012
    54
    0
    I think prove the circled part to be in hcR ?

    hv = E

    So both of them is equal already..
     
  10. Papabravo

    Expert

    Feb 24, 2006
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    They are not numerically or dimensionally equal but they state the same principle. Once in terms of energy difference, and once in terms of frequency. Energy and frequency are related proportionally by Planck's constant.

    The actual 2-step proof is detailed on Wikipedia.

    PS doesn't it feel better to have discovered that yourself with some help and a hint rather than seeing the result?
     
    Last edited: Mar 8, 2012
  11. vick5821

    Thread Starter Member

    Jan 27, 2012
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    0
    Ok..will try..thank you
     
  12. vick5821

    Thread Starter Member

    Jan 27, 2012
    54
    0
    so any hint?
     
  13. Papabravo

    Expert

    Feb 24, 2006
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    Yeah. We know that K is 4*pi*epsilon sub 0. We know that E is h times nu and we know that the Rydberg constant can be expressed in terms of h, c, epsilon sub 0, mass and electron charge.

    Pick either the energy or the frequency expression and transform one into the other by applying the simple rules of algebra: substitution and multiplying both sides of an equation by the same thing.

    For example:

    1. Start with the expression for nu (frequency)
    2. Substitute an expression for R
    3. Multiply both side by h
    4. The LHS (Left Hand Side) should be Energy Difference and the RHS (Right Hand Side) should be the other expression.
    Also remember that h-bar, the reduced Planck's constant is h over 2*pi
     
  14. vick5821

    Thread Starter Member

    Jan 27, 2012
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    0
    So I proved by equating the two equation ? and cancel out those same terms ?
     
  15. eng.mustafasalah

    Member

    Nov 10, 2011
    41
    2
    hi
    the total E of the electron in the n'th orbit is

    En = KEn + PEn = - m q^4 / 2 k^2 n^2 h^2 ....... 1
    when
    KE : kinetic energy
    KE : potential energy
    h : is not planck constant but is = h/2*3.14

    The energy difference between orbits n1 & n2
    En1 - En2 = ( m q^4 / 2 k^2 h^2 )*(1/n1^2 - 1/n2^2)
    the value of this eq ( m q^4 / 2 k^2 h^2 ) = 13.6 eV

    so
    En1 - En2 = 13.6 *(1/n1^2 - 1/n2^2)

    also the energy = hf
    when
    f = frequency
    h = Planck constant
    so the F of a light given off by a transition between n1 & n2 is

    hf21 = En1 - En2 = ( m q^4 / 2 k^2 h^2 )*(1/n1^2 - 1/n2^2) eV

    f21 = ( m q^4 / 2 k^2 h^2 h ) *(1/n1^2 - 1/n2^2) eV

    ( m q^4 / 2 k^2 h^2 h ) = C.R

    so f21 = C.R *(1/n1^2 - 1/n2^2) n2>n1
     
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