Spectral analysis and displacement law

[Ruven]

He everyone

first of all, I am not quite sure, if I put this post in the correct topic, so sorry if that's the case.

The things is, altough I don't go to school anymore I recently stumbled upon some theories and wanted to check them myself using the formulas that I could find on the net. I also had to figure out the theory behind all that myself, so I would really be glad if someone could "check" on my research or correct it if I simply misunderstood a big part of it.

As far as I figured some guy named Max Planck formulated a theory in which he states that certain materials would absorb some specific amount of energy/light according to the wavelenght(s). Also that there are specific wavelenghts for each material which only can be absorbed. So if there's a meterial that could absorb wavelenghts of 2.8 x 10^-5, 4.5 x 10^-5 and 15 x 10^-5 meters of lenght, that would mean, that every other lightradiation would simply pass by not affecting the material at all (?).

Further I came across the Displacement Law by Wien. which gives a formula to calculate the temperature of that specific wavelenght. (Lmax= b/T). So would that also mean that if there's a black body which is radiating a temperature of 10 degrees celsius, that this temperature would have a weavelenght of 1.101 x 10^-5 Meters (since b/-263.15 K= 1.101x10^-5 m)
thus making it impossible for the above material to absorb any of the energy?
or did I make some major mistake in my calculations or did I simply make some major faults using the wrong formulae etc.?

I'd be really thankful for your help, even if I'm not a physician or anything like that.

Greetings
Robin

 

[mhsd91]

I divided my answer into two. This is the first part considering:

So if there's a meterial that could absorb wavelenghts of 2.8 x 10^-5, 4.5 x 10^-5 and 15 x 10^-5 meters of lenght, that would mean, that every other lightradiation would simply pass by not affecting the material at all (?).

Well, it depends on multiple factors (and I welcome all to correct or improve my answer). I suggest you take a look at Einstein's theory of Photoelectricity. Light (photons) are absorbed by an atom if the photon's energy corresponds to the energy level/gap of electrons within the atom. This is the principle of solar cells: Say we have such a piece of Silicon solar cells with energy gap $E_g$. Note that the energy levels separated by gaps are a direct result of the quantized energy states of the electrons within the atom. Then, let light/photons with energy $E_p = h \nu$ hit the material. Here, $\nu$ is the incoming light's (wave) frequency, and $h$ is Planck's constant. Now, this is what happens:

if $E_p < E_g$, nothing happens and the photons just pass through the material or is reflected away.
if $E_p \geq E_g$, an electron absorbs the photon and is lifted over the energy gap: it is excited! If $E_p$ was much greater than $E_g$, the remaining photon energy becomes kinetic energy for the electron (or even lifting the electron mulitple energy levels (gaps) are also possible). Such that in theory any wavelength corresponding to $E_p > E_g$ could in principle be absorbed. However, in reality its more complicated. For instance, the material will have a finite amount of electrons to be excited from a specific energy state corresponding to a specific wavelength. Be aware that energy levels for electrons in solids can be really complicated.

So to answer first your question, other (shorter) wavelengths (with higher energy) may(!) also be absorbed. If you know what what specific material, and what kind light it's exposed by, it shoudn't be too hard to find experimental data/tables on this.

 

[mhsd91]

I divided my answer in two for multiple reasons, I hope is is okay, and here follows the second part, regarding Wiens Displacement law.

I'm not entirely sure if I understand your question, as a temperature does not have a wavelength. What Wien's Disp. law states, is that a black object (as defined by Planck) will radiate radiation (emit photons/light ) of multiple wavelegths! However, the emitted light will be distributed about some specific wavelength, given by the formula you state! That is, the (absolute) majority of emitted light from a black body of temperature $T$ will have a wavelengths close to

$\lambda = \frac{b}{T}$

However, if you study the graphs on wikipedia, you see that the peak is sharper for higher temperatures, meaning that for really high $T$, almost all emitted photons have the wavelegth from this forumla. For really low $T$, $\lambda$ is more uniformly distributed, and lots of different wavelengths will be emitted, even though $\lambda$ will still stand for the most frequent wavelength..

I hope this was to some help for you..

PS: I think you meant a Physicist(?), who work with math and physics .. a physician is a doctor who treats and heals injuries/diseases.