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Find the constant of vain (wien)

  1. Oct 26, 2004 #1
    Find a wien constant A from the equation [tex] R(\lambda,T) = \frac{2\pi h c^2}{\lambda^3} \frac{1}{e^\frac{hv}{kT} - 1} [/tex] Show that the Wien constant w = Lambda T = hc / 4.965k

    Also i know that w = 2898 micro metres Kelvin

    I'm not sure what to do here... Do i fiddle with the equation for hte spectral radiancy?? Do i expand the tern for the exp function?

    But how would you manage to get the Boltzmann constant in the denominator without expanding the exp function?

    Please do help with this!

    Part 2 of this question is

    Substitute numerical values for the constants and evalute. Compare the result with the Equation of w = 2898 [tex] \mu m K [/tex]

    Now i need to solve the first part to get this second part, i would really really appreciate your help on this matter!!

    Thank you in advance for this!
  2. jcsd
  3. Oct 27, 2004 #2


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    The Wien constant is defined as the product of the temperature and the wavelength for which the blackbody radiation is a maximum:

    [tex]w = \lambda_{max} T[/tex]

    so what you need to do is take the derivative of R with respect to [itex]\lambda[/tex], set it equal to 0 and solve for [itex]\lambda_{max}[/itex]. Remember that [itex]\nu[/itex] depends on [itex]\lambda[/itex]. Once you find that value just multiply it by the temperature and you have Wien's constant!

    It's actually easier than it looks at the outset! (It might be easier to work with [itex]\nu[/itex] and switch back to wavelength at the end.)
  4. Oct 27, 2004 #3
    Wow i would never think it was that easy! But thank you very much!

    Could you also help me with another one ?? Please?
    This one has to do with Einstein's heory for specific heat
  5. Oct 27, 2004 #4

    James R

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    w corresponds to the value of [tex]\lambda[/tex] for which R is a maximum. This maximum occurs at [tex]\lambda = w/T[/tex].

    To find the maximum you just put

    [tex]\frac{\partial R}{\partial \lambda} = 0 [/tex]

    and solve.

    The problem is that you can only solve this numerically, since you get a transcendental equation. Good luck.
  6. Oct 27, 2004 #5


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    Yes, but that's not a serious problem with modern calculators!
  7. Oct 27, 2004 #6
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