Wiens law to calculate max frequency?

[PsychonautQQ]

Homework Statement

so λT = C = 2.898*10^-3 ,, where lambda Is the most common wavelength in a spectral distribution. How do I find the maximum FREQUENCY?

The Attempt at a Solution

Maybe this is a fools hope... but could I just put

T/ƒ = 2.898*10^3 ? or something along these lines? Idk.. this is my best guess so far.. anyone want to help me understand?

EDIT: Okay.. ugh.. I wasn't understanding this properly.. The maximum wavelength in the distribution isn't the MAGNITUDE of the wavelength rather the most common wavelength, so the most common wavelength is easy to find with wien's law, and the most common frequency is just c/lambda... However the question goes on to say that the max frequency in the distribution and the max wavelength correspond to different photon energies and asks me to explain why.. this seems completely counter intuitive.. any advice?

 

[NascentOxygen]

This sounds like a trick question.

Let me hint strongly, rather than give you the answer outright. Blue corresponds to visible light with maximum frequency. But rays of visible light having maximum wavelength definitely are not blue, not even close to blue.

 

[NascentOxygen]

Homework Statement

so λT = C = 2.898*10^-3 ,, where lambda Is the most common wavelength in a spectral distribution. How do I find the maximum FREQUENCY?

Are you sure it isn't simply asking you to determine the frequency of that spectral maxima, where you are told its wavelength?

EDIT: On reading again, I see you do agree.

 

[PsychonautQQ]

Yeah, but why would the photons have different wavelengths? Is she just asking that to trick us? Sure seems like it..

 

[BvU]

Check out Wien's law and Planck.

 

[PsychonautQQ]

I mean why would the photons have different energies*, whatever the most common wavelength is will have a most common frequency and they will have the same energy of course..?

 

[BvU]

Check out Wien's law and Planck.

What did you find ?

 

[PsychonautQQ]

Hm. I read the thing but I'm not still comprehending this fully. I don't understand why I can't use the frequency*wavelength = c relationship... Something about the domains of each being different, yet it gives no explanation as to WHY they are different (that I was processed by my brain anyway)

 

[BvU]

λ_max was found by differentiating wrt λ. f_max needs to be found the same way. $\frac {dR}{d\lambda}$ is not identical to $\frac {dR}{df}$ because $\frac {d\lambda}{df}\ne 1$

 

[NascentOxygen]

I mean why would the photons have different energies*, whatever the most common wavelength is will have a most common frequency and they will have the same energy of course..?

Of course. The same λ ⇔ same photon energy ⇔ same colour (if visible)

But the case of photons having the same energy is not what the question is concerned with.

However the question goes on to say that the max frequency in the distribution and the max wavelength correspond to different photon energies and asks me to explain why..

Are you still mulling over this? Did my hint in my first post not answer it?

BTW, I think BvU is on another wavelength.

 

[BvU]

Come on, both of you. This is an FAQ in introductory QM. It's even worked out in he link I gave. perhaps a more literal rendering of the OP will help here, because I would be real surprised if they really asked for "the maximum frequency".

 

[BvU]

Re post #8: You certainly can use $c = \lambda f$ at $\lambda_{max}$, but what you get is the frequency at the peak in the $\lambda$ spectrum. The whole thing about this exercise is the "discovery" that the peak in the frequency spectrum is at a different point. And the mathematical background is revealed to you in post #9. Perhaps a little premature, so if you want to take a step back (pour mieux sauter), you check how a spectrum is defined. Something with $dR = ... d\lambda$ in a $R(\lambda)$ spectrum. And $dR = ... df$ in a $R(f)$ spectrum. Not the same R, not the same peak point (using $c = \lambda f$, which of course is correct). So:

..whatever the most common wavelength is will have a most common frequency...

A definite, most sincere: No way!

 

[PsychonautQQ]

okay. Many thanks to you :D. I feel you have done all you can to help me understand what is going on here, but I still feel like I'm missing something conceptually that is going to make all of this make sense... so R(f) peaks at a different place than R(lambda)... as in the most common frequency for power being radiated does not correspond the most common wavelength for power being radiated.. still seems odd to me.. and yes I derived u(f) starting with u(lambda) but I'm still not completely sure why the derivation took more steps than using lambda = c/f... I'm sure in time this will make sense... I'm only in the second week of my 300 level quantum undergrad class :P

 

[BvU]

The crux is in the spectrum definition. A spectrum is a probability density distribution. Between $R(\lambda)$ and $R(\lambda + d\lambda)$ the probability is $R(\lambda) d\lambda$. Mutatis mutandis f. Since $df \ne d\lambda$, the spectra look different and peak at a different point.

Mind you, if you sit at the peak in $R(\lambda)$ and change the frequency, you do move away from the highest point in the $R(\lambda)$ spectrum, of course.

There is another case where this kind of thing has even more influence and is perhaps easier to grasp: particle size distributions are mostly plotted as function of a length (e.g. particle diameter). But they can be length density distributions or volume density distributions (or mass, if you multiply by $\rho$).

An often used length distribution is $N(x) = N_0 e^{-x}$. "Maximum" at x = 0. Converting that to a volume distribution (while still plotting as a function of length) brings in a factor x²: $N\left(V(x)\right)dx = N(x)dx \frac{dV}{dx}$ which makes the distribution look completely different !

Not quite the same, but $dV\ne dx$ causes a difference, just like $d\lambda\ne df$.

 

[BvU]

No reason to worry as in your post #13. Remember: the spectra show energy density per unit wavelenth (or frequency) ! So energy in a certain range $\Delta \lambda$ is $R(\lambda)\Delta \lambda A$ .

$\Delta f$ corresponding to $\Delta \lambda$ is $\frac {c}{\lambda^2} \Delta \lambda$ and with that $\Delta f$ you get the same radiated power.

Spectra are always "per something", just like histograms, probability distributions and what have you. Basically they are plots of derivatives. Only become physical observables after integrating over the "something".