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Homework Help: Planck's Radiation Law?

  1. Sep 22, 2007 #1
    1. The problem statement, all variables and given/known data

    The dependence on wavelength [tex] \lambda [/tex] of the intensity [tex]I(\lambda)d\lambda[/tex] of the radiation emitted by a body which is in thermal equilibrium with its surroundings at temerature T is given by:

    [tex]I(\lambda)d\lambda = \frac{2 \pi h c^{2}/\lambda^5}{e^{hc/kT\lambda}-1}d\lambda[/tex]

    in the interval of wavelength between [tex] \lambda [/tex] and [tex] \lambda+d\lambda [/tex]. In this expression, h is Planck's constant, k is Boltzmann's constant, and c is the velocity of light.

    Sketch and clearly label on one figure the dependences of [tex]I(\lambda)d\lambda[/tex] on [tex] \lambda [/tex] for three different temperatures [tex] T_{1} < T_{2} < T_{3} [/tex].

    Simplify the above expression in the limit of (i) short wavelength ([tex]\lambda\rightarrow0[/tex]) and (ii) long wavelength ([tex]\lambda\rightarrow\infty[/tex]).

    (The binomial expansion [tex]e^{x} = 1+x+x^{2}/2+...[/tex] may be useful.)

    2. Relevant equations

    All given in the problem i think.

    3. The attempt at a solution

    I found Planck's Radiation Law was almost exactly the same as this i searched for it on wikipedia for more information:

    http://en.wikipedia.org/wiki/Planck's_law

    On that page is a graph which i thought was showing what the first part of the question is asking but i don't understand what the question means when it says "clearly label on one figure the dependences of [tex]I(\lambda)d\lambda[/tex] on [tex] \lambda [/tex]"?

    For the second part i tried to make the formula look simpler first:

    [tex]\frac{A}{\lambda^{5}(e^{B/\lambda} - 1)}[/tex]

    I think as [tex]\lambda\rightarrow0[/tex], [tex]e^{B/\lambda} - 1[/tex] can be simplified to [tex]e^{B/\lambda}[/tex] because the latter expression will be very large giving:

    [tex]\frac{A}{\lambda^{5}e^{B/\lambda}}[/tex]

    I'm having some trouble posting the rest of my thread but i thought for the last part as lambda goes to infinity the expression would simplify to A/lambda^5 but i'm not sure how to work these out for definite i think this is probably wrong.
     
    Last edited: Sep 22, 2007
  2. jcsd
  3. Sep 22, 2007 #2

    Gokul43201

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    For large [itex]\lambda[/itex], you can expand [itex]e^{B/ \lambda } [/itex] to first order, using the Taylor series provided to you. You will get a different answer than [itex]A/ \lambda ^5 [/itex]. Your answer for the short wavelength limit is correct.

    As for "labeling", I believe that is asking you to make it clear which curve represents which temperature (T1,T2,T3).
     
  4. Sep 22, 2007 #3
    Thanks for the help this is what i've got so far:

    [tex]\lambda^{5}(e^{B/\lambda} - 1) = \lambda^{5}(1 + \frac{B}{\lambda} + \frac{B^2}{2 \lambda^2} + \frac{B^3}{6 \lambda^3}... -1)[/tex]

    [tex]\frac{A}{\lambda^{5}(\frac{B}{\lambda} + \frac{B^2}{2 \lambda^2} + \frac{B^3}{6 \lambda^3}... )}[/tex]

    [tex]\frac{A}{B \lambda^4 + 1/2 B^2 \lambda^3 + 1/6 B^3 \lambda^2 + 1/24 B^4 \lambda + 1/120 B^5}[/tex]

    I think all the other terms in the expansion would tend to 0 so they can be ignored, if this is right it's as far as i can get though i can't see how to simplify any further although this makes it look longer and maybe even more complicated?
     
  5. Sep 22, 2007 #4

    Gokul43201

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    Actually, it's better than that!

    [itex]B/ \lambda <<1 \implies B^2 \lambda^3 << B \lambda ^4 [/itex], etc. So feel free to throw away all terms after the first one.
     
  6. Sep 23, 2007 #5
    Thanks for the help again :)

    I was wondering i can see the difference between this:

    [tex]I(\lambda)d\lambda = \frac{2 \pi h c^{2}/\lambda^5}{e^{hc/kT\lambda}-1}d\lambda[/tex]

    and Planck's radiation formula:

    [tex]I(\lambda)= \frac{2 h c^{2}/\lambda^5}{e^{hc/kT\lambda}-1}[/tex]

    is [tex]d\lambda[/tex] and [tex]\pi[/tex]. How do you transform one into the other? What does it mean to write it like the question did with [tex]d\lambda[/tex] either side or more specifically what does the [tex]d\lambda[/tex] mean in that context?

    If you integrated both sides to get rid of the [tex]d\lambda[/tex] s would the [tex]\pi[/tex] dissapear? I really should try and integrate it myself rather than just asking but it looks complicated :S i'll give it a shot though.
     
  7. Sep 23, 2007 #6

    dextercioby

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    There' a [itex] \pi [/itex] difference. But only because you forgot to put it in the second formula. The difference between the 2 formulas is important only when you try to change the variable from wavelength to frequency [itex] \nu [/itex] or angular frequency [itex] \omega [/itex]. In that case, the formula involving differentials should be used.
     
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