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Topic: Absorption and Emission Spectra (Hydrogen)  (Read 2649 times)

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Offline hamedhaghjo

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Absorption and Emission Spectra (Hydrogen)
« on: April 22, 2018, 07:29:19 PM »
Hi,

Did not really get any help from my teacher, he just told me "The energy emitted when this electron moves between the shells corresponds to red light"

Sidenote: I translated almost everything, but I do not feel like I need to explain the images of the spectrums, they have been written on in a different language but it is easy to understand what they mean, the only thing I will explain is that Väte = Hydrogen


So to my questions:
I have already filled in the right statements. But I need further explanation regarding why it would emit red light when an electron moves from the L-shell to the K-shell.
I understand the the light emitted corresponds to the level of energy required to move the electron to a shell closer to the nucleus. But I also know that the different shells are at different levels of energy with higher energy the further away from the nucleus. (Which I find confusing also).

Please explain this well to me.

Offline Enthalpy

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Re: Absorption and Emission Spectra (Hydrogen)
« Reply #1 on: April 23, 2018, 05:28:05 AM »
Welcome, hamedhaghjo!

Maybe your confusion results from signs. The positively charged nucleus attracts the negatively charged electron. Getting closer is spontaneous down to the K level and releases energy, as  light for instance.

With the quasi universal convention sign for energy, it increases when pulling the electron away from the nucleus. That is, the electron receives energy, which is released when the electron falls down. It's like pulling a mass above the ground.

Then, you also need some convention to set the zero energy level, which is arbitrary. The most common is zero energy when the electron is far from the nucleus. Then, the electron has a negative energy. This doesn't change the variations: the energy is less negative or more positive when increasing the electron's distance to the nucleus.

The electron has also a kinetic energy, but it doesn't change the above thoughts.

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The pictured transitions, with visible light, don't involve the K level. You could double-check that.

Offline hamedhaghjo

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Re: Absorption and Emission Spectra (Hydrogen)
« Reply #2 on: April 23, 2018, 05:36:26 PM »
Thanks for the Welcome!! :)

Oh my god, I feel so stupid. I already know that!
Good lesson for others if they do not want to study super ineffectively and look dumber than they are:
Do not study when you have not slept for the last 48 hours.

I know and understand everything you just replied. One thing I was confused about last night was why it showed us 4 emission/absorption lines for hydrogen. Because I earlier thought that the amount of lines equaled to the amount of electron levels for the atom. (I was like "Hydrogen has one electron if not an ion = should have only one line??") But now I know that it still means that, but it also means the amount of possible levels the electron can reach without leaving the atom. So 4 is the amount of lines and therefor the electron can reach the fourth shell (n-shell) if it recieves enough energy for that orbital transition. More energy recieved will lead to the recieving the level of energy required for ionization.

Okay, so the other thing I was confused about was why it would emit red light, I percieved the K shell to have connections with the line furthest to the left in the screenshot I provided. And that the line in the red light was the n- shell.
Well, I now know thanks to Wikipedia, searching electromagnetic spectrum, that I was looking on it the wrong way.

Keep this in mind:

If Energy is high, Wavelength must be low and vice versa. Because h and c are constants.


So looking at the electromagnetic spectrum in the visible class, we can see that red light has a long wavelength, therefor energy must be low. And the ultraviolet part has shorther wavelengths = more energy.
And as you mentioned, just like potential energy for physical objects being lifted above the ground it is at a higher energy level the further away it is from the attracting force (center of the earth, gravity) which in our case is the nucleus. So that means Red light - K shell, "Green" light - L shell, and so on.
Therefor an electron moving from L to K emits red light (it was one energy level transition = first energy level is red light). (I have read more about this aswell, discrete values for energy levels, or maybe I should say "light is quantized into packets of energy" and these values are countable etc)

Sorry for my bad english and if I wrote a sloppy essay, just thought if some else is asking the same question as me they might find my post useful.

Thank you again Enthalpy, you guided me into the right path of solving this problem.
« Last Edit: April 23, 2018, 06:37:53 PM by hamedhaghjo »

Offline Enthalpy

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Re: Absorption and Emission Spectra (Hydrogen)
« Reply #3 on: April 24, 2018, 04:20:26 AM »
No worry with your English. But you still have to think at the shells.

All transitions to the K shell are in ultraviolet wavelengths. Most visible ones are toward the L shell. Whatever the "correct" answer tells, sorry.

There is an infinite number of energy levels available to the electron in the hydrogen atom. Transitions among these energy levels (not all pairs: there are "selection rules") make a big (really big) number of lines. But the highest levels, next to electron freedom or ionization, are close to an other and hard to discern.

As for the absorption spectrum in your example, it supposes that the electron is already on the L shell, which is quite uncommon. At rest, the electron is on the K shell. Such a spectrum would require something (like pumping from K to L with concentrated light) putting the electron often on the L shell.

My suggestion is that you don't put too much time in this particular question, which I find isn't very well formulated. But in atoms, electrons, shells, it's useful. For instance Wikipedia.

Offline Corribus

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Re: Absorption and Emission Spectra (Hydrogen)
« Reply #4 on: April 24, 2018, 09:46:59 AM »
Important to add - although there are infinite number of spectral lines (because there's an infinite number of electronic states), only a small percentage of transitions between states are allowed, due to selection rules. From a classical physics perspective, these rules basically boil down to conservation of momentum - a transition can't occur if the momentum isn't conserved during the transition. Intensity of the transition is therefore no uniform for all the lines, because some are more allowed than others.

For emission spectra, another factor that influences the intensity of any line is the relative population of excited (higher energy) state that serves as the origin of the transition. The distribution of excited states populated at time zero is temperature dependent and can be determined using the Boltzmann distribution.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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