Absorption of EM radiation

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Part 7 of a series on electromagnetic radiation: Absorption of EM radiation - explains how radiation of different frequencies is absorbed including reflection, molecular oscillations, Electron excitation, Photoelectric effect, Compton effect (including deriving the Compton formula) and pair production.

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24 июл 2024

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Комментарии : 54

@michaelfoster8929 2 года назад

I have watched these videos for years - no need for all the " bells & whistles " just straight to the point whilst maintaining a high level of interest

@xx7667 8 лет назад

national treasure ... thats the only thing to say about this person ... many thanks grate lecture...

@p1ai162 4 года назад

So many formulas and no conclusion (

@Libouh 7 лет назад

Brilliant lecture i have seen on youtube...superb...and thank you for such valuable knowledge

@simplysailingonline 7 лет назад

OUTSTANDING! Thank you very much for the clear, concise explanation. Excellent.

@Edifier1221 11 лет назад

Please never stop making these amazing videos!

@hp127 2 года назад

Amazing series: well presented and clear explanations. (2021) Years of (repeated) visits: thank you

@bhavarlalsharma9246 3 года назад

Wonderful explanation sir! I took many deep breaths to understand the derivation. Thank you very much sir, long live !!!

@FranzCalvo 8 лет назад

I found particularly useful "...radiation absorption, as opposed to radiation reflection or transmission of radiation through the material". Thank you!

@DrPhysicsA 11 лет назад

And thanks to you for your feedback.

@soi..all. 4 года назад

Hello From Morocco , I wanted to thank you for your video ! it helped me a lot !

@DrPhysicsA 11 лет назад

Thanks. What a good idea. I didn't think my camera had manual focus but just checked and it does. That should resolve the problem of it refocusing and creating times when the writing goes hazy. I've already got several videos pre-recorded (awaiting upload) but I'll definitely try it on the next one that I record. Many thanks.

@kendallbarboza6477 7 лет назад

Wonderful explanation!!! I loved this video

@peterluxus7382 9 лет назад

You are a great teacher!

@DrPhysicsA 11 лет назад

As I try to show in the video it depends on a number of factors, largely to do with the extent to which the light is reflected and absorbed.

@a3rv803 4 года назад

Thank u very much! Amazing lecture! Greetings from Mexico. :)

@hubercats 3 года назад

Very informative. Thank you!

@ranjithedirisooriya2262 Год назад

Brilliant explanation

@justinhill6206 11 лет назад

you are giving knowledge to the world thank you

@pacinpm2 11 лет назад

Very good material. Thank you. One advice: set your camera on manual focus.

@beatlessteve1010 2 года назад

I wish you would have touched on XRF x-ray fluorescence and if the principles are similar to the Compton effect.

@PinkCammy 11 лет назад

thank you, thank you, thank you!

@johnsmith-gk3ek 9 лет назад

DrPhysicsA 7:35 ish Why 2.5Ghz ish? Why not 5Ghz etc? That is a question I have never found an answer to and I really really want to know. Is there really any correlation between that particular frequency ballpark and a molecule of water?

@stijndhondt9611 7 лет назад

www.schoolphysics.co.uk/age16-19/Wave%20properties/Wave%20properties/text/Microwave_ovens/index.html This explains it rather well. Turns out the 2.5GHz isn't the resonant frequency of water, but a sort of average, for if it were the resonant frequency, it'd be absorbed by the first layers of water, and than the food wouldn't be coocked on the inside.

@unixzee 10 лет назад

Very nice video. Maybe you could do one on the electronic band structure at some point.

@DrPhysicsA 10 лет назад

You will find something along these lines in the playlist on atomic physics.

@abinavbaskar4842 5 лет назад

If the electron jumps from the higher energy state back to the lower state, then how is any energy being absorbed?

@DrPhysicsA 11 лет назад

Indeed you are right. But it was de Broglie in the 1920s who postulated that the equation might also apply to electrons. He argued that if you knew the momentum of an electron, then you would know its associated wavelength and that would mean that the electron was behaving like a wave. For that he got a PhD and a Nobel Prize. His ideas were subsequently experimentally verified.

@olivierlaurent1409 11 лет назад

Thank you for your lecture. I have one question. I don't understand why you say that the momentum of the electron is equal to the energy divided by the speed of light (22min). This definition should only apply for photons (E=pc when m=0).

@jorgepiloto7277 2 года назад

brilliant

@rgudduu 5 лет назад

excellent

@DrPhysicsA 11 лет назад

It just means that the energy of an electron has increased. We often graphically represent its as an electron with a larger radius of its orbit, but that is not necessarily the case. The location of an electron is governed by its wave function and it could be that a higher energy electron is orbiting closer to the nucleus than a lower energy electron. But high energy electrons will always want to give off energy in order to return to a low energy state.

@apuntes8883 6 лет назад

10:30 it means that your initial particle is burning material in increasing orbital cycles. The inner orbital burn has less energy and the outter orbital is the one that has the most exponential momentum of energy because from n=1 to n=5 the burning particle grows as an inverted pyramid. So in physics they would calculate the amount of photon energy to be reciprocal to the initial particle in order to not exceed nor to lack of burning mixture otherwise youll have reflecting radar waves or absorbing radar waves. But is not as much as it is calculated but I guess is about the way you simply put things toghether. And of course the video talks about the producing of energy and not the producing of elements so energy is not being understood that way.

@Gullshunter 11 лет назад

Why does some frequencies of light pass through matter and some doesn't as much?

@ArjhunSwaminathan 11 лет назад

Sir, Can you please suggest me a chemistry channel just like this?? btw, you are awesome

@DrPhysicsA 11 лет назад

Its highly frequency dependent. eg high f gamma radiation will pass thro lead but low f wont.

@dannygreat5156 4 года назад

Nice

@ajingolk7716 6 месяцев назад

I didnt get it how can this been absorption (first part)

@lookupverazhou8599 Год назад

A spherical wavefront spreads out symmetrically into space. It encounters a particle. The entire spherical wavefront, across the entire space it takes up, is absorbed. How is this possible?

@apuntes8883 6 лет назад

But theory says that any radar wave lenght beam that is directed for example to the area of a nuclear blast it gets "absorbed" by it, not turned into different charges or radiation, not bounced away, not distorted. In order to be absorbed radar waves must to become part of the chain reaction itself. So if an atomic blast is bombarded with radar wave beams maybe the reading instruments would register either a lost of burning material or of neutrons because of the radar waves getting into the fueling process. On the other hand if an airplane which has been painted with lets say boron pain may absorb instead of reflect ome of the wave beams.

@DrPhysicsA 11 лет назад

Alas I am afraid I don't know of any. And I was pretty hopeless at chemistry myself, apart from those bits which are essentially physics. Organic chemistry was very puzzling!

@olivierlaurent1409 11 лет назад

Thank you for your response. For me, you don't use (and in fact you don't need) the De Broglie equation p=h/lambda for electrons for your demonstration which is perfectly correct. You only use p=h/Lambda for photons but this formula for photons was known since 1905. Indeed, the publication of this De Broglie equation was established in his thesis in 1924 and Arthur Compton published is formula before in 1923. So, Compton couldn't know p=h/lambda.

@olivierlaurent1409 11 лет назад

I am sorry, but for me, E=pc for electron is not correct even considering the De Broglie formula. You can see in your video (22min53s) that as a consequence of p=E/c, you write p=sqrt(p^2.c^2+m^2.c^4)/c for the electron (necessary wrong when m # 0). We should always use E^2=p^2.c^2+m^2.c^4 and p=h/lambda for massive particules and E=pc (p=h/lambda) for massless particules like photons. I just gave you my idea, I don't critic your videos I discovered recently. I like the way you explain physics.

@TheAmberkathleen 8 лет назад

I'm a GED educated mom not smart but I loved this I understood it. An I want to start from beginning anyone I'm jo gillis on facebook plz send info

@DrPhysicsA 8 лет назад

+Amber Gillis Delighted to hear it. All good wishes for your future studies. You might want to start with my GCSE revision playlist, then the A level playlist which will give you a good grounding in Physics.

@Gullshunter 11 лет назад

Hmm just have another question. Does high and low frequency of light have the same chance of interacting with matter (absorption, reflection, wathever it is) ? I mean, does one travel matter more because it interact less (the EM wave doesnt ''collide'' an electron or nucleus?) or because of the way they interact (through heating, compton effect, etc.) ? Plus, when a photon makes a atom vibrate, can this atom re-emit the photon or not at all? thanks