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Drakkith

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So lets say for example the hydrogen absorption corresponding to a jump from n=1 to n=2. Is it a band of frequencies that gets absorbed there? I thought it was only a single frequency corresponding to the exact energy difference between n=1 and n=2 that gets absorbed. A band being a range of frequencies would be a range of energies.

EDIT: Ah wait, I just read there that if you remove a single wavelength from white light that the eye will perceive the light as the missing wavelengths complementary colour. If thats true that would explain how single wavelength absorptions cause such profound colour changes. I was thinking of it as the perceived colour being the product of all the remaining colours kinda like mixing paint together. I also read that if you combine two single complementary wavelengths of light your eye will perceive it as white light because there are no missing complementary wavelengths. I'm a bit skeptical about that but if thats true that is extremely interesting and I bet theres a mathematical relationship between one wavelength and its complementary wavelength.

EDIT: Ah wait, I just read there that if you remove a single wavelength from white light that the eye will perceive the light as the missing wavelengths complementary colour. If thats true that would explain how single wavelength absorptions cause such profound colour changes. I was thinking of it as the perceived colour being the product of all the remaining colours kinda like mixing paint together. I also read that if you combine two single complementary wavelengths of light your eye will perceive it as white light because there are no missing complementary wavelengths. I'm a bit skeptical about that but if thats true that is extremely interesting and I bet theres a mathematical relationship between one wavelength and its complementary wavelength.

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Drakkith

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Yeah but the energy difference between n=1 to n=100 would be very different to the difference between 1 and 2 so the absorption would appear at a different area of the spectrum. I can't remember the Rydberg formula off hand but I remember seeing that the energy differences get smaller and smaller the further you move away from the nucleus so there would be much smaller energy difference between 99 and 100 then wouldn't there. Theres something really boggling my mind now. On a continuous spectrum like that what is a unit? Whats a single wavelength? The range between 400 and 700 nm is also the range 400 x 10

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Drakkith

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I can't answer that as I don't know. I don't know if there are actually infinite wavelengths between 400-700 nm or not.Theres something really boggling my mind now. On a continuous spectrum like that what is a unit? Whats a single wavelength? The range between 400 and 700 nm is also the range 400 x 10googleplex and 700 x 10googleplex nm-googleplex. If an absorption corresponds to a single wavelength what is the exact wavelength that gets absorbed?

Also, I don't think that absorbtion leading to electrons jumping orbitals is the ONLY way a photon can interact with matter, but I'm not sure.

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Yeah I think thats an unanswerable question like "whats the largest number". If laser beams truly are monochromatic then there must be something other than electron jumps going on if they can burn matter.I can't answer that as I don't know. I don't know if there are actually infinite wavelengths between 400-700 nm or not.

Also, I don't think that absorbtion leading to electrons jumping orbitals is the ONLY way a photon can interact with matter, but I'm not sure.

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Drakkith

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Lasers aren't all exactly 1 wavelength, but a small band.Yeah I think thats an unanswerable question like "whats the largest number". If laser beams truly are monochromatic then there must be something other than electron jumps going on if they can burn matter.

For example, the following is from wikipedia on lasers.

Most so-called "single wavelength" lasers actually produce radiation in several modes having slightly different frequencies (wavelengths), often not in a single polarization. And although temporal coherence implies monochromaticity, there are even lasers that emit a broad spectrum of light, or emit different wavelengths of light simultaneously.

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DrDu

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You are right that absorption in only a single infinitely narrow line would not produce a change of perceived color, as the intensity of the light in that small spectral range would tend to 0. But in the case of dissolved molecules and ions the lines become broad bands due to the coupling of the electronic transitions to all kinds of molecular vibrations (so called vibronic coupling). For each vibrational state, the absorption wavelength occurs at a slightly different frequency and, as the whole system is quantum mechanically, a single molecule can nevertheless potentially absorb at different frequencies at the same time.So lets say for example the hydrogen absorption corresponding to a jump from n=1 to n=2. Is it a band of frequencies that gets absorbed there? I thought it was only a single frequency corresponding to the exact energy difference between n=1 and n=2 that gets absorbed. A band being a range of frequencies would be a range of energies.

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Drakkith

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It makes so much sense now!

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Borek

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Atoms are moving, Doppler effect makes this single frequency a band. A narrow one, nonetheless a band.So lets say for example the hydrogen absorption corresponding to a jump from n=1 to n=2. Is it a band of frequencies that gets absorbed there? I thought it was only a single frequency corresponding to the exact energy difference between n=1 and n=2 that gets absorbed. A band being a range of frequencies would be a range of energies.

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Absorption "leading to electrons jumping orbitals" certainly isn't the ONLY way a photon can interact with matter.Also, I don't think that absorbtion leading to electrons jumping orbitals is the ONLY way a photon can interact with matter, but I'm not sure.

A photon is an EM wave "packet" with some known frequency, and the electric field in the EM wave can accelerate electrons in a material (bound to nuclei, or in a "free electron gas" as in a metal) in some oscillatory manner.

Now recall: electrons accelerating (and decelerating) emit EM radiation / photons. If the oscillation is relatively uninhibited, we can get out the same photon energy (explaining reflection or refraction for instance -- although also note there are restrictions about EM waves at boundaries here using Maxwell's equations -- some math we won't get into here). But more interesting: if the electrons are at a surface or boundary, that oscillation can become nonlinear (making solving Maxwell's equations simultaneously at a boundary really fun!)... thus: nonlinear optical materials -- which can be engineered to create some of those two-or-more tone "lasers" -- which we preferred to called optical parametric oscillators in our research).

There's also generally some loss of the initial energy to heating (kinetic motion of the electrons, which can be transferred to other masses within the material). This is why cooler bodies, where the energy has been dispersed to a lower energy/particle level, emit lower energy (more red) photons, and fewer of them overall.

Probably too much info, but fun to think about.

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Drakkith

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Thx PG!

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DrDu

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We can be more quantitative: The absorption at a given frequency f is given by Lamberts Beers law:

[tex]

\log{I/I_0}(f)=\epsilon_b(f)lc[/tex]

Here l is the length of the path in the medium, c the concentration of absorbing molecules (or atoms) and [tex]\epsilon_b(f)[/tex] is the molar extinction coefficient which I assume to depend parametrically on some line broadening mechanism b.

The integral [tex] \int df \epsilon_b(f) [/tex] is approximately constant over the line and independent of b.

Then the perceived color will be somehow proportional to

[tex] \int I df/\int I_0 df=\int_{f_0}^{f_0+\Delta f} I_0(f) \exp{\epsilon_b(f)lc}/ \int I_0(f) df [/tex]

If the line is not too broad I_0 is approximately constant for white light and we can use a saddle point approximation for the integral

[tex] \int I_0(f) \exp{(\epsilon_b(f)lc)}/ \int I_0(f) df \approx \exp{(\epsilon_b(f_{max}}lc)} 1/[{\Delta f \cdot \sqrt{2\pi \epsilon_b''(f_{max})}}}][/tex],

The second derivative of the extinction coefficient being inversely proportional to the square of the linewidth. So the perceived color change is approximately proportional to the linewidth.

[tex]

\log{I/I_0}(f)=\epsilon_b(f)lc[/tex]

Here l is the length of the path in the medium, c the concentration of absorbing molecules (or atoms) and [tex]\epsilon_b(f)[/tex] is the molar extinction coefficient which I assume to depend parametrically on some line broadening mechanism b.

The integral [tex] \int df \epsilon_b(f) [/tex] is approximately constant over the line and independent of b.

Then the perceived color will be somehow proportional to

[tex] \int I df/\int I_0 df=\int_{f_0}^{f_0+\Delta f} I_0(f) \exp{\epsilon_b(f)lc}/ \int I_0(f) df [/tex]

If the line is not too broad I_0 is approximately constant for white light and we can use a saddle point approximation for the integral

[tex] \int I_0(f) \exp{(\epsilon_b(f)lc)}/ \int I_0(f) df \approx \exp{(\epsilon_b(f_{max}}lc)} 1/[{\Delta f \cdot \sqrt{2\pi \epsilon_b''(f_{max})}}}][/tex],

The second derivative of the extinction coefficient being inversely proportional to the square of the linewidth. So the perceived color change is approximately proportional to the linewidth.

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Drakkith

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