Physics Help Forum Need help with: Absorption of electrons/photons

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 Jan 14th 2016, 02:40 AM #1 Junior Member   Join Date: Jan 2016 Posts: 2 Need help with: Absorption of electrons/photons Hello! I'm not an english speaker, but I really need help with a question. I'll try to translate it as best as I can. Explain what happens when an electron with kinetic energy=1.12aJ collides with a simple mercury atom. What happens if a photon with the energy 1.12aJ collides with the atom? In the back of my book the answer says that an electron may be absorbed, while the photon will not be absorbed. I have calculated that 1.12 aJ is not the exact amount of energy, so in reality it cannot be absorbed no matter what, I think? Or does that only apply to the photon? So the electron can give some of it's energy, though it's too much/ too little amount? Hope you somewhat understand what I am wondering about. Thank you on beforehand! Last edited by MarieO; Jan 14th 2016 at 03:25 AM.
 Jan 14th 2016, 06:14 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Hello Marie, Let us start by understanding that electrons in atoms exist in set or particular energy levels. There are no photons in atoms. The easy part: if a photon hits an atom it can only be absorbed if its energy is exactly equal to the difference in energy levels for the electrons. When the photon hits, it hits the electrons first since they are on the outside. If there is a vacant energy level with a difference between the vacant level and the electron exactly equal to the energy of the phton the photon is absorbed (and disappears) by the electron and the electron jumps (is promoted) to the higher level. So the photon energy adds to the energy of some existing electron if absorbtion takes place. Otherwise the photon continues on its way and is not absorbed. I haven't calculated 1.2aJ for mercury, I will leave that for you to check. The situation with the electron is different. There are no two energies to add together. If there is a vacancy in the mercury atom at an energy level of 1.2aJ or less then the electron can be captured and form a negative mercury ion by entering this empty level. Excess energy is lost by the creation of a photon emitted as light by the mercury ion. This can be seen by light emission in mercury vapour, bombarded by electrons. https://www.google.co.uk/search?hl=e...72.tXDGqEiDXBM
 Jan 17th 2016, 02:54 AM #3 Junior Member   Join Date: Jan 2016 Posts: 2 Thank you! That was very helpful, I understand it now!!
Jan 21st 2016, 07:10 PM   #4
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 Originally Posted by studio The easy part: if a photon hits an atom it can only be absorbed if its energy is exactly equal to the difference in energy levels for the electrons. ... If there is a vacant energy level with a difference between the vacant level and the electron exactly equal to the energy of the phton the photon is absorbed (and disappears) by the electron and the electron jumps (is promoted) to the higher level.
Nit pick: I just wanted to point out that the photon doesn't need to have the exact energy as stated above, etc. Think of it like this: is it possible to generate a photon whose energy is exactly equal to a particular energy? This would mean accuracy to an infinite number of decimal places. The energy levels in atoms aren't bands with exactly zero energy width either. This is due to Doppler broadening. Let's just say that the width of the energy band has to be very small and the photon's energy very close to the energy band.

You don't find this in textbooks just like you won't find Doppler broadening in a text either. But it's good to know. I just sent an e-mail to the author of a great QM textbook to seek his help clarifying the explanation. He's pretty good a responding so I'll let you know what he said.

Jan 21st 2016, 08:40 PM   #5

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 Originally Posted by Pmb Nit pick: I just wanted to point out that the photon doesn't need to have the exact energy as stated above, etc. Think of it like this: is it possible to generate a photon whose energy is exactly equal to a particular energy? This would mean accuracy to an infinite number of decimal places. The energy levels in atoms aren't bands with exactly zero energy width either. This is due to Doppler broadening. Let's just say that the width of the energy band has to be very small and the photon's energy very close to the energy band. You don't find this in textbooks just like you won't find Doppler broadening in a text either. But it's good to know. I just sent an e-mail to the author of a great QM textbook to seek his help clarifying the explanation. He's pretty good a responding so I'll let you know what he said.
Nice explanation but probably a bit TMI for this level. (Though I certainly enjoyed reading it. I had never thought about it in this form.)

-Dan
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 Jan 22nd 2016, 12:56 AM #6 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Yes indeed, Pete. Like most things in the real world things get much more complicated when you go more deeply into them and allow for all possible effects. We still need the guiding principle for understanding though, especially when first meeting these ideas. MarieO, if you are still reading this, the next stage is complexity is fine and hyperfine structure, bonding and antibonding orbitals, the Zeeman effect and the Kronig-Penny equation. The K_P equation is important because it is one of the few cases where we can obtain analytic solutions to the Schrodinger equation. You should study these before the effects PMB is mentioning. Last edited by studiot; Jan 24th 2016 at 01:21 AM.
 Jan 23rd 2016, 11:59 PM #7 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,576 The reason that I think that it's important to mention these things at this stage is because if you don't then when left in its simple form students will think that its the absolute truth and when someone comes along and tells them the truth they have a hard, impossible, time accepting it. I found that to be the case with the concept of an "incompressible fluid." Real fluids can be compressed and that fact is important to know. Not telling this to students leads them to believe that whomever says different is ignorant and should be ignored. That's just bad juju! Lol! Another thing about this that students will need to learn is what happens when a photon whose momentum has some uncertainty to it. If there's uncertainty to momentum then there's uncertainty in energy. What happens to a photon whose energy is uncertain, collides with an atom? This has a very easy answer to it. Last edited by Pmb; Jan 24th 2016 at 12:03 AM.
Jan 24th 2016, 01:28 AM   #8
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 The reason that I think that it's important to mention these things at this stage is because if you don't then when left in its simple form students will think that its the absolute truth and when someone comes along and tells them the truth they have a hard, impossible, time accepting it.
Yes I agree you should always take a statement of 'fact' as meaning

"This is goood enough for what we are talking about", not as "gospel".

Of course, the teacher should always make that clear as well, and not all do.

I have suffered many times in my career from learning a simplified version too well.

As regards the Uncertainty Principle, there is an issue here that I have started another thread to discuss.

http://physicshelpforum.com/showthre...1509#post31509

How is the snow with you Pete?

Last edited by studiot; Jan 24th 2016 at 01:42 AM.